Transcript for:
General Chemistry - Part 2: Key Concepts and Problem Solving Techniques

in this video we're going to cover the second part of general chemistry so this is a multiple choice uh video it's full of multiple choice problems and first we're going to talk about chemical kinetics how to write the rate law expression how to find the rate constant k uh the reaction mechanism problems how to uh solve problems associated with zero order first order and second order reactions and then move on to nuclear chemistry we're going to talk about how to find the missing element how to uh calculate the nuclear binding energy how to also solve half-life problems as well and then we'll move on to chemical equilibrium the shetalia's principle we're going to talk about what's going to happen if we increase the concentration of the reactants or the products you know will the reaction shift to the right or will it shift to the left what's going to happen if we change the pressure the volume the temperature of an endothermic reaction how that affects the equilibrium constant k we're going to cover all of those things and then we'll move on to acids and bases we're going to talk about how to find the ph of a strong acid strong base weak acid weak base buffer solution and also how to find the ph of the solution before after and during a titration for equilibrium we're also going to go over ice tables and how to solve those type of problems with quadratic formulas and then we'll move on to ksp we're going to talk about how to calculate the ksb and um also how to find the molar solubility both in moles per liter and in grams per liter we're also going to deal with a few le chatelier's principal problems associated with ksp so if you increase the concentration of the product will the solubility of the compound increase or decrease and how would that affect the ph of the solution so we're going to go into that as well and then move on into thermodynamics we're going to talk about delta g delta h delta s gives free energy enthalpy entropy how to calculate them both under standard and non-standard conditions and then move on to electrochemistry we're going to talk about how to find the cell potential both under standard and non-standard conditions we're going to talk about how to find the oxidizing and reduce an agent how to balance redox reactions under acidic and basic conditions and also electrolysis problems like how to find the mass that was deposited on a cathode if you're given the electric current and the time that it passed through the solution and also how to calculate the electric current if you're given the amount of mass that was plated onto the cathode so we're going to go over all of those things and so let's begin let's jump right into it and let's start with uh number one number one the average rate of appearance of nh3 is 0.215 molar per second determine the average rate of disappearance of h2 so for this problem feel free to pause the video and try it yourself and then unpause it check the answers and see if you got it right so let's begin let's start with 0.215 m for nh3 per second now molarity and moles are proportional if the volume is the same and h2 and nh3 they're going to be in the same reaction vessel so therefore the the volume or the liters is going to be the same so we can use the molar ratio but with molarity instead so for every three moles of h2 that reacts or disappears two moles of nh3 will appear so we can put 2m for nh3 and 3m for h2 so it's a 3-2 ratio and the molarity for nh3 will cancel so to find the answer it's going to be 0.215 times 3 over 2 and that's about 0.323 now is it going to be positive or is it going to be negative well we know the answer has to be negative it has to be b the reason why it's negative is because the concentration of h2 is decreasing hydrogen gas is on the reactant side and as the reaction goes from the left to the right side the reactants will decrease the concentration so they will disappear and that's why it we have a negative sign if we were calculating the rate of appearance for nh3 it would be positive because the concentration of nh3 is increasing as the reaction goes from the left to the right number two which of the statements shown below is correct given the following rate law expression is it a b c or d let's find out so all we got to do for this problem is plug in the numbers for anything that is constant we're going to plug in a 1 for anything that changes we'll plug in the respective value so for a the rate of the reaction doubles as the concentration of a doubles so what we're going to do is we're going to plug in a 1 for k it's a constant we're going to put a 2 for a because it doubles b is a constant so we're going to substitute 1 for that and the same is true for c so 2 squared is 4. so the rate should increase by a factor of four so a is a false statement the rate doesn't double b the rate of the reaction decreases by half as the concentration of c doubles true or false well let's find out so rate is going to equal k well we're going to put a 1 for k um a is going to be one b is one as well and for c we're going to plug in two but it's raised to the zero power anything raised to the zero power is one so if the rate is equal to one that means that it doesn't change it remains the same so b is eliminated c the rate increases by a factor of 3 as the concentration triples in value so rate is going to equal one for k we don't have to put one for a or c because it's going to be the same but we can plug in three for b and b is raised to the first power so three to the one is equal to three so c is a true statement c is the answer but now let's check the last statement the rate of the reaction increases by a factor of four as the concentration of b and c increases by a factor of two simultaneously so rate is equal to we're gonna put one for k a stays the same and b increases by a factor of two and c increases by a factor of two but c is raised to zero power anything raised to the zero power is one so this is going to be two times one which is equal to two so the rate will not increase by a factor of four if we double the concentration of a and b in fact the rate increases by a factor of two so d is eliminated and therefore c is the right answer for this problem number three use the following experimental data to determine the rate law expression and the rate constant k for the following chemical equation so we're going to find the rate law expression two ways we're gonna do it the fast way conceptually and then we're gonna actually use an equation we're gonna write it out so pick two trials in which the concentration of o2 is constant and the concentration of no changes we're gonna find the rate order for no first so this is trial one two and three if we pick trials one and two notice that the concentration of no it doubles from point one to point two it's not constant the concentration for o2 is constant or remains point one uh for the first two trials so um by choosing trials one and two we can find the rate order for no and as the concentration doubles from point one to point two notice the effect that it has on the reaction rate the rate increases by factor four from one point two five to five five divided by one point two five is four so therefore two to the x is equal to four if two to the x is equal to four what is the value of x we know that two squared is four so x has to equal two two squared means 2 times 2 which is 4. so therefore in the rate law expression the exponent of reactant n o is going to be 2. so it's second order with respect to no so now let's find the order for o2 pick two trials in which the concentration of no is constant but the concentration of o2 is not so we're gonna have to pick trials one and three in those two trials no has a constant value of point one but the concentration of o2 it doubles from point one to point two and as that occurs notice the effect on the reaction rate it doubles from 1.25 to 2.5 2.5 divided by 1.25 is 2. so the equation that we need to write is 2 to the y we're using y for a different variable is equal to 2 because the rate doubles so what is the value of y whenever there isn't an exponent it's always assumed to be a one so since the twos are the same that means that y is equal to one two to the first power is two so what this tells us is that it's first order for o2 so let's put in one here so this is the rate law expression rate is equal to k times no square times o2 to the first power and the only answer that has that option is answer choice d so we know d has to be the correct answer so let's find out if there's another way that we can get the answer because maybe you don't see conceptually maybe you prefer to use equations so we're still going to use the first two trials to find the order for no so the way we can write the equation it's going to be like this rate for trial two divided by the rate for trot one is equal to the concentration of no raised to the x power and then the concentration of o2 raised to the y power divided by no raised to the x and o2 raised to the y you can write it this way but another way you can write it which i actually think might be easier now that i'm doing this problem let's write it like this no2 and no one it's really no but the 2 means like for trial 2 and then raised to the x power and then here we're going to have o2 over o2 this is for trial two that's for trial one and that's going to be raised to the to the y so the rate for trial two it's uh five times ten to the minus four and for trial one to rate is one point two five times ten to the minus four the value for no2 for trial two it's uh point two and for trial one it's 0.1 and that should be raised to the x power now the rate of o2 for trial 2 it's 0.1 and for trial 1 is 0.1 so it cancels so this cancels out to one one raised to the y is just one so we don't have to worry about it anymore and on the left sides the 10 to the negative four cancels when we divide them so 5 divided by 1.25 is 4 and .2 divided by 0.1 is 2 so we get 2 raised to the x now if you're having trouble figuring out what x is x is going to be equal to log 4 divided by log 2. let me separate those two equations so that's how you can find x and then x is going to equal 2. so if you want to see why it works that way let me first make some space so let's say if we have 4 is equal to 2 to the x the first thing you need to do is take the log of both sides so log 4 is equal to log 2 raised to the x a property of logs allows you to move the x to the front now for this problem we really don't need to use logs but in the case that you do need to use it let's say if you can't figure out what the exponent is um this is what you can do to figure it out so now we could divide both sides by log 2. so log 4 divided by log 2 is equal to x which we know it to be 2. so you can type it in your calculator whenever you don't see a base for log the base is always assumed to be 10 by the way so we know it's second order for no so now let's use the same equation um but for o2 so we have to use trials three and one so the rate for trial three divided by the rate for trial 1 is equal to the concentration of no raised to the x which we know it to be 2 times the concentration of o2 this should be a three for trial three and then raised to the y so let's plug in the numbers that we have so r3 the rate for trial 3 is 2.5 times 10 to the minus 4 and for r1 it's 1.25 times 10 to the minus 4. for now we know it's going to cancel but i'm going to write it anyway so it's going to be 0.1 over 0.1 raised to the x which is going to be 1 to the x which is just 1. and then for o2 it's going to be 0.2 divided by 0.1 raised to the y power so on the left the 10 to the negative fours will cancel because they're the same 2.5 divided by 1.25 if you type that in your calculator you should get 2. now this cancels to 1 and then 0.2 divided by 0.1 is 2 but raised to the y so we know that y is one but using logs is going to be like log two over log two which is one so therefore we have the rate expression which is rate is equal to k times uh no squared times o2 raised to the first power which we don't have to put the one there so it's second order for nl first order for o2 but third order overall you added two plus one is three so now let's calculate uh the value of k so we can use any trial it's easier to use trial one because one is easier to deal with than two so let's plug in the information for trial one so we're going to use this so the rate is equal to 1.25 times 10 to the minus 4 and then for trial 1 the value of no is 0.1 squared and the value of o2 is 0.1 so 0.1 squared times 0.1 that's equal to .001 times k so we need to divide both sides by .001 so 1.25 times 10 to the minus 4 divided by 0.001 is equal to 0.125 which we can see d is the correct answer it has the correct rate law expression and the correct value for k so that's it that's all you got to do for this problem number four which of the following will give a straight line plot in the graph of ln a versus time is it a a zero order reaction b a first order reaction or a second order or third order reaction now to answer this question you need to know uh the equations that are associated with chemical kinetics so this is a good time to go over it so first we have a zero order reaction and first order and a second order reaction so for a zero order reaction the rate is equal to k for a first order reaction rate is equal to k times a or a to the first power for second order reaction rate is equal to k times a squared now that's the rate law expression you need to know that if you plot a versus t or the concentration of a versus time you're going to get a straight line graph if it's a zero-order reaction for first order reaction if you plot the natural log of the concentration of a versus time it's going to be a straight line and for a second or reaction if you plot 1 divided by the concentration of a against time that is going to give you a straight line plot so the answer for this problem is b you simply have to know that if you plot the natural log of a versus time you'll get a straight line plot now there's some other things you need to know for a zero-order reaction the slope is equal to negative k for first order reaction the slope is also equal to negative k but for a second order reaction the slope is equal to positive k and not negative k now the last thing or the next thing you need to know is the half-life equation the half-life equation for zero-order reaction is is the initial concentration of a divided by 2k where k is the rate constant for a first order reaction the half-life which is the time it takes for the substance to change by half is the natural log of two divided by the rate constant k and for second order reaction the half-life is equal to one over the rate constant times the initial concentration of a now there are some other things that you need to know in addition to that and that is the the straight line equation which is in the form y equals mx plus b so the straight line equation for a zero-order reaction it's a final is equal to kt plus a initial and notice how it's in the form y is equal to mx plus b so as you can see um here we're plotting a versus t or like y versus x x is the time so time is on the x axis and the concentration of a which is is going to be on the y-axis the slope is equal to positive k and the y-intercept is a initial b is the y-intercept for any straight-line equation but for a zero-order reaction the y-intercept is the initial concentration of a now for first order equation it looks very similar but it involves the use of natural logs so the natural log of a is equal to oh by the way this should be negative kt i forgot the negative sign the slope is equal to negative k so ln a is equal to negative kt the negative means that the concentration of a is decreasing over time so it's negative kt plus the natural log of a initial or the initial concentration of a now for a second-order reaction the equation that you need looks like this it's 1 over a final i'll put f of final is equal to positive kt plus 1 over a initial so the reason why we have a positive slope is because as a decreases one over a increases whenever you increase the denominator of a fraction the value well whenever you decrease the denominator of a fraction the value of the fraction increases so that's why we have a positive slope in this case because it's the reciprocal of a so those are the equations that you need to know for chemical kinetics but the answer to this problem is answer choice b a first order reaction will give you a straight line plot if you were to graph the natural log of the concentration of a versus time number five which of the following units of the rate constant k correspond to a first order reaction is it a b c or d so in the first order reaction we have the rate law expression rate is equal to k times the concentration of a raised to the first order so overall it has to be first order so let's solve for the units of k so the rate is usually molarity divided by seconds it's concentration over time or the change in concentration over the change in time so that's it has units m over s and a it's going to be just m so a is going to be since a is a form of concentration is measured in molarity so to solve for k we're going to multiply both sides by 1 over m so here m cancels and the same is true on this side so for first order reaction we have one over s which is the same as s to the minus one now for any order you can solve it the same way but here's some things that you need to know for a zero-order reaction the units will be m to the first power s to the minus one now the time could be minutes it can be hours but it's always going to be to the negative one it shouldn't be negative two or negative three so for a first order reaction it's gonna be m to the zero s to the minus one or which is the same as s to the minus one for a second order reaction it's m to the minus one s to the minus one for a third order reaction it's m to the minus two s to the minus one do you see a pattern here notice that as the order increases the exponents for m decreases by one however for time it doesn't change it's always going to be to negative one power but keep in mind it doesn't have to be s it could be minutes it can be hours it can be days it can be years but it's always time raised to the negative one so if you want to write a general equation that you can just plug in and get the units the units is going to equal m raised to the 1 minus n and then i'm going to put time raised to the negative one or t to negative one so n is going to be the order of the reaction the overall order so let's say if we have a rate law expression that looks like this it's k a squared b so for b it's first order but overall if you add the exponents the overall order is its third order overall so we need to plug in three for n so the units is going to be uh m one minus three and let's say the time is in minutes minutes to the negative one so it's going to be m to the minus two minutes to the minus one let's try another example so let's say if we have the rate law expression rate is equal to k a squared b c to the third so overall the overall order is six so we're going to plug in n for six or replace n with six so it's gonna be m one minus six s to the minus one let's say the time is in seconds this time so the units is going to be m to the minus five s to the minus one so in the case for our problem use an equation it was first order so if we plug in one for n it will be one minus one s to the minus one so one minus one is zero so it's m to the zero s minus one and anything raised to the zero power is one so this is equal to s to the minus one for a first order reaction so now you can find the units of k whenever you have the rate law expression as long as you have the overall order of the reaction you can now find the units of k and just keep in mind that the time can change it could be seconds minutes hours days years so you have to look at the problem to figure out what the unit of time will be so the right answer for this problem is answer choice b number six the initial concentration of a reactant is .453 for a zero-order reaction calculate the final concentration of the reactant after 64.4 seconds if the rate constant k is .00137 m over s so we went over the equation for this particular problem and for a zero-order reaction the straight-line equation is the concentration of a or the final concentration of a is equal to negative kt plus the initial concentration of a so we can see that this is similar to y equals mx plus b the slope is negative k the y intercept b is the initial concentration of a so now all we need to do is plug in the information into this formula and let's solve for the missing variable so what are we looking for in this problem so this question says calculate the final concentration so we're looking for a final k is given to us it's negative point zero zero well it's positive point zero zero one three seven but there's a negative sign in front of it and the time is 64.4 seconds and the initial concentration is 0.453 by the way when you look at this problem make sure the units match notice that you have seconds in for time and for k so the seconds will cancel giving you molarity and this is in molarity so that works out so make sure your units match if you see seconds and minutes you may have to do a conversion so now at this point let's just plug in these numbers into the calculator it's a negative point zero zero one to be seven times sixty four point four that's negative point zero eight eight two three and we're gonna add point four five three to that value the final answer should be 0.3648 which rounds to 0.365 so c is the right answer for this problem number seven the initial concentration of a reactant is 0.738 m for a zero-order reaction the rate constant k is 0.0352 m per minute calculate the time it takes for the final concentration of the reactant to decrease to 0.255 so it's a zero-order reaction again so the equation is going to be the same it's a final is equal to negative kt plus a initial this time we can see that the final concentration is 0.255 k is positive point zero three five two we're looking for t and a initial is uh point seven three eight so first let's subtract both sides by 0.738 we need to solve for t so 0.255 minus 0.738 that's going to equal negative point four eight three and that's equal to negative point zero three five two t so let's divide both sides by negative point zero three five two so these cancel so if you type in negative point four eight three divided by negative 0.0352 you should get a time value of 13.72 and notice that k has unit minutes so time is going to be in minutes as well so b is the right answer for this problem number eight calculate the rate constant k for second order reaction if the half-life is 243 seconds the initial concentration of the reactant is 0.325 the equation that we need is this one half-life is equal to 1 over the rate constant times the initial concentration now we need to solve for k so what we're going to do is we're going to multiply both sides by k and at the same time we're going to divide it by 1 over the half-life let's see if i can fit it here ignore the equal sign so if we multiply both sides by k and by 1 over t notice that t cancels on the left side but on the right side k cancels so the equation that we now have is k is equal to 1 over half-life times the initial concentration of a so basically k and t they trade places but for those of you who want to see how it happened now you know so now let's plug in the values that we have so the half-life is 243 seconds and the initial concentration is 0.325 so 243 times 0.325 that is equal to 78.975 so 1 divided by 78.975 you should get a final answer of point zero one two six six which you can round that to one point two seven times ten to the minus two as you move the decimal from left to right if you move it two spaces to the right that's where we get the negative two from as you move it to the right uh the exponent decreases by two the exponent of base ten so the correct answer is answer choice d we don't have to worry about the units because all of the units are the same so that's it for this problem number nine a reaction has a rate constant of one point four six times ten to the negative three s to the minus one at 298 kelvin and four point three three times ten to the minus two as to minus one at 421 kelvin calculate the activation energy of this reaction so the equation that you need for this problem is the activation energy is equal to r times natural log of k2 divided by k1 that's the two rate constants over one over t1 minus 1 over t2 by the way what is the order of this reaction is it first order second order zero order can you figure out based on this problem based on the information that's given to you look at the units s to the negative one based on the units you can tell what the order is now the equation that i gave you which was units which is equal to m 1 minus n s minus 1. so here we have s to negative 1 m 1 minus n and then so forth notice that these two are the same which means that if we divide both sides by s we should get one is equal to m to the one minus n and anything to the zero power is one so one is the same as m to the zero since the bases are the same the exponents must equal so if you solve for n you'll get n is one so anytime it's just s to the minus one without any molarity units it's going to be a first order reaction so those are some things to know you can find the order of the reaction based on the units alone so now let's actually finish this problem let's actually work through it so this the first value 1.46 that is the value for k1 and the temperature t1 is 298 kelvin k2 is 4.33 and t2 is 421 kelvin which corresponds to k2 now r you don't want to plug in 0.08206 for r you need to use the energy constant the 8.3145 joules per mole per kelvin and then it's going to be natural log k2 over k1 so k2 is 4.33 times 10 to the minus 2 and k1 is point four six times ten to the minus three and that's a terrible looking three t1 is 298 kelvin and t2 is 421 kelvin now if you wish to type this in directly all at once in your calculator make sure you put parentheses in the bottom now for those of you who like to do it step by step i'm going to just do it that way just for you in this video so let's start with the natural log let's type in ln 4.33 times 10 to the minus 2 divided by 1.46 times 10 to the minus three so you should get three point three eight nine seven and then that number is multiplied by eight point three one four five but we'll do that later now if you type in one over 298 that's about like point zero zero three three five five seven subtract that by one over four twenty one you should get point zero zero zero nine eight zero four so three point three eight nine seven times eight point three one four five that's about 28.1837 and then we're going to divide it by the number in the bottom so you should get 28 747 but because r has the units joules per mole per kelvin ea activation energy is going to have the units joules per mole the kelvin unit cancels with the kelvin temperatures on the bottom but we need kilojoules per mole so to convert from joules to kilojoules divide by a thousand so then you should get uh 28.7 kilojoules per mole so d is the right answer number 10 which of the following particles is equivalent to an electron is it a b c or d this is something you have to know so the answer is a a beta particle is the same as an electron it has a symbol zero negative one e you can also write it as zero minus one with a beta symbol so the top number for nuclear chemistry is the mass number and the bottom number is the charge for an alpha particle an alpha particle can be written as 4 2 with a helium nucleus you can also write a 4 2 alpha with an alpha symbol so an alpha particle has a mass of 4 and a charge of 2. a positron is the anti-particle of an electron so you can write it as 0 1 e so it has a positive 1 charge but it still has a mass of virtually zero just like an electron so it's different from a proton you can also write it as 0 1 with a beta symbol a gamma particle is basically a very energetic light particle so it has a massive zero and a charge of zero and it has the gamma symbol by the way the symbol for proton it's one one p a proton has a massive one and a charge of plus one for a neutron it's one zero and a neutron has a massive one a charge of zero because it's neutral neutrons and protons their masses are approximately equivalent but for this problem answer choice a is the right answer number 11 identify the missing element so for now let's call it x we need to find the mass and the atomic number to find the mass number we need to make sure that the equation is balanced on the left side the total mass is 226. on the right side we have a mass of 4 for the alpha particle and we need to find the missing mass so 4 plus what number is 226 or 226 minus 4 is we know to be 222. to find a missed atomic number we can subtract 88 by 2 and we will get 86. now the atomic number is really the important number that we need here that's going to identify the element so if you go to the periodic table and look for which element has an atomic number of 86 it's radon so the right answer is b number 12 the half-life of cesium-137 is 30 years calculate the rate constant k for the first order decomposition of isotope cs137 so the equation that we need that connects half-life with the rate constant k is this equation k is equal to the natural log of 2 divided by the half-life if it's a first order reaction so we just have to plug this in the calculator so ln 2 divided by 30 is equal to point zero two three one years to the minus one so the right answer for this problem is b number thirteen the half-life of iodine 131 is about 8.03 days how long will it take for a 200 gram sample to decay to 25 most decay problems are usually first order reaction problems and we can figure out the answer conceptually so starting with 200 grams it's going to take 8.03 days to decay to 100 grams because that's the half-life the time it takes to decay by half so in another 8.03 days we're going to have 50 grams of iodine 131 left over and it's going to take another eight days for it to go down to 25. so 8.03 times 3 is about 24.09 or 24.1 days that's how you could find the total time but now let's say if we want to use an equation to get the same answer the equation that you could use is ln a final over a initial is equal to negative kt you've seen this equation earlier in this video in this form ln a final is equal to negative kt plus ln a initial this is for a first order reaction so let's plug in everything that we have and let's solve for t so the final amount is 25 and the initial amount is 200 the units grams will cancel now we don't know the value of k but we got to find it so we can use this equation k is going to be ln 2 divided by half-life which is uh 8.03 days so ln2 divided by 8.03 you should get .08632 for the value of k so now we could plug it in into the equation so now we need to solve for t ln 25 divided by 200 is about negative 2.0794 and if you take that number and divide it by negative 0.08 6 you should get the value for t and you should get about the same answer which is 24.1 days so c is the right answer so now 14 which of the following shows the correct equilibrium expression for the reaction shown below now what you need to keep in mind is that any reactant or product in the liquid or solid phase should not be included in the expression for k so we can eliminate copper oxide and copper because they're both in a solid phase so a cannot be the right answer the equilibrium constant k is equal to the ratio of the products divided by the reactants and that is the concentration of the products and reactants so h2o is the product it's on the right side and h2 is a reactant it's on the left side now the coefficients of the balanced reaction will become the exponents of the equilibrium expression so the reaction is already balanced that means that there's a 1 in front of h2o and h2 so the exponents are 1 as well respectively so the correct answer for this problem is answer choice b 15 calculate k p for the following reaction at 298 kelvin kc is 2.41 times 10 to the minus 2. for this problem you need to know the right equation to use kp is equal to kc times rt raised to the n or the change in n delta n is the sum of the products or the sum of the coefficient of the products minus the total sum of the reactant coefficients so for the products we have one methane molecule and one h2o molecule so we have two product molecules and for the reactant side we have three hydrogen gas molecules and one carbon monoxide molecule so there's four molecules on the left side so delta n is two minus four which is negative two we have the value of kc we know it's 2.41 times 10 to the minus 2 and r is the gas constant 0.08206 and the temperature is 298 kelvin and so we're going to raise that to the minus 2. so if you type in 0.08206 times 298 that's 24.45 and if you raise that to the negative 2 you should get 1.672 times 10 to the negative 3 and then we're going to multiply that number by 2.41 times 10 to the minus two so the final answer is equal to four point zero three times ten to the minus five so answer choice c is the right answer sixteen use the following information below to calculate the missing equilibrium constant of the net reaction so we have to solve this problem as if we're doing the hassle of problem but with equilibrium constants instead so you need to know that if you double a reaction like if you increase the coefficients by a factor of two the equilibrium constant will be changed not by a factor of 2 so it won't become 2k in fact it's going to become k squared and if you multiply the coefficients by half the equilibrium constant k will become k raised to the half if you reverse it it turns into 1 over k so our goal is to get the equilibrium constant for the third reaction so we need to focus on the right species in the first reaction let's call this number one which species should we not focus on which reactants and products should we not focus on is it co c or o2 you do not want to focus on any reactant or product that is found in both equation 1 and 2. you want to focus on the ones that is only found in equation 1 and not in equation two so oxygen is found in both equations one and two so we don't want to focus on oxygen carbon monoxide is found in equations one and two so we don't wanna focus on that if you do it's going to make it harder to adjust the two equations such that they would add to the third equation we want to focus on carbon it's only found in equation one and we want to focus on co2 it's only found in equation two and not an equation one so let's start with carbon we have two carbons on the right side but we need to match it so that it can look like equation three we only have one carbon on the left so therefore we need to reverse equation one and multiply by a factor of one half so first let's reverse it well let's do it simultaneously two times a half is one so we're going to have one carbon atom plus half of an oxygen molecule and then we have two carbon monoxide molecules multiplied by a half this is going to turn into one so we're going to get one carbon monoxide molecule so now what is the effect on k1 since we reversed it it's going to be 1 over k1 and we multiply the coefficients by one half so we're going to raise the exponent of k1 to a half now let's focus on equation two so we're going to focus on co2 we have half of co2 on the left but we need one on the right so we need to reverse it and multiply the equation by a factor of two so we're going to have one whole co2 molecule on the left and on the right half i mean 1 4 times 2 is a half oh wait we need to reverse it i almost forgot about that so we have to multiply by two and reverse it so one-fourth of o2 times two is a half of o2 and half of co times two will give us one co molecule and half of co2 times two is one co2 molecule so now since we reverse it it's going to be 1 over k2 but now since we doubled the coefficient we need to square k2 as well so 1 over k2 squared now let's add these two reactions let's see if it's going to give us the net reaction so notice that we have a carbon monoxide molecule on the right and one on the left so they cancel on the left we have one carbon atom now a half o2 plus the half a half plus a half is a whole so we get one oxygen molecule and plus one co2 molecule the co2 molecule just dropped down so now what is the equilibrium constant for the third reaction do we add one over k1 raised to the one-half plus one over k2 squared or do we do something else now if it was hess's law using enthalpy we would add the enthalpy values but because we're dealing with the equilibrium constant we have to multiply them and not add so it's going to be 1 over k1 raised to the half times k2 squared so first let's go ahead and make some space so now let's plug in everything into that equation so kc is now equal to 1 divided by k1 is 0.143 raised to the one-half i converted the scientific notation into its standard value and k2 is point zero two three five raised to the second power so point zero two three five squared that's like five point five two times 10 to the minus 4 and if we multiply that by 0.143 raised to the half you should get point zero zero zero two zero eight eight four so one divided by that number is approximately forty seven eighty eight so this rounds to about forty 47.90 so d is the right answer so that's how you can find the k value of a third equation using the first two equations so just to review let's say if you have this reaction a plus b and it converts into c and let's say uh the equilibrium constant is just k if you double the coefficients of the reaction it will turn into k squared if you multiply by a half it's going to be k raised to the one half and finally if you reverse it it's going to be 1 divided by k and whenever you add 2 equations it's going to be like k1 times k2 so if we have a let's say d plus e and becomes f let's call that uh k2 so if we add equations one and this equation that will give us a plus b plus d plus e turns into c plus f so the k value that's associated for that reaction is it's k one times k two it's the product of these k values so make sure you understand that in case um if you're if you see it on your next exam or something so that's it for this problem number 17 at equilibrium the partial pressures of xenon chlorine and xenon tetrachloride were found to be 215 315 and 723 millimeters of mercury respectively calculate kp if you want to find the equilibrium constant and if you have the equilibrium partial pressures or concentration values you can simply plug it into the equation you do not need an ice table only when you have initial values or initial concentration values and that's when you uh need an ice table to solve it so let's write the expression for kp it's going to be products over reactants so xenon tetrachloride is on the product side and it has a coefficient of one so it's raised to the first power xenon is on the reactant side and chlorine is also on the reactant side but chlorine has a coefficient of two so it's going to be raised to the second power so now all we need to do because we have the partial pressures at equilibrium we can plug in these numbers directly into this expression to get the value for kp so for xenon tetrachloride it's going to be 723 and for xenon it's 215 and for chlorine it's 315 but squared 315 squared is 99225 and if you multiply that by 215 you should get this number on the bottom two one three three three three seven five so 723 divided by 21 million three hundred thirty three thousand three seventy five you should get three point three eight nine times ten to the minus five so d is the right answer for this problem eighteen a reaction mixture initially contains 0.75 of hi at equilibrium the concentration of i2 was found to be 0.3 calculate the equilibrium concentration of hi in the mixture so we have initial concentration values and equilibrium concentration values so we need an ice table now initially we have 0.75 of hi so that means that there's no hydrogen gas or iodine gas in the reaction chamber now at equilibrium the concentration of i2 was found to be 0.3 so if we add the missing values that means that iodine if it went from 0 to 0.3 it increased by 0.3 and that means a hydrogen gas must also increase by 0.3 because those two are in a one-to-one ratio now if we look at iodine and h-i notice that it's a one-to-two ratio so for every mole of iodine that's produced two moles of hi reacts so if the products are increasing the reactors must be decreasing but by twice the amount so if one mole of iodine is produced two moles of hr reacts so if point three moles of iodine is produced point six moles of hi reacts and that's why we have negative 26 because it's decreasing so the final concentration of h i is simply 0.75 minus 0.6 which is going to be 0.15 and for this problem we don't need kc we could find kc if we want to using these three values but the question only wanted the equilibrium concentration of h on a mixture and that's 0.15 so a is the right answer for this problem 19 the partial pressures of dinitrogen oxide or monoxide oxygen gas and dinitrogen tetroxide are currently 0.134 0.265 and 0.483 atmospheres respectively determine if the reaction is at equilibrium or if it will shift to the right or to the left so what we need to do is find q which is the reaction quotient and compare it to k the equilibrium constant so if q is greater than k it's going to shift to the left one way you can visualize it is you can put k on a number line and put q to the right because q is greater than k it's going to shift towards k so it shifts uh to the left so now if q is less than k it's going to shift to the right so if it's less than k we need to put it on the left side and since it wants to go wherever k is it's going to move towards the right now the last thing we need to know is if q is equal to k the system is at equilibrium so that's great but how can we use this to find q the reaction quotient we already have the value of k we know that it's 56.8 but how can we find q q can be found the same way as k k is products over reactants but you have to use the equilibrium concentration or partial pressure values for q you have to use the initial concentration or partial pressure values the partial pressure values that were given to us are not at equilibrium because it didn't state that it was the equilibrium it says it's currently at 0.134 0.265 and point four eight three so we're gonna assume that those are initial values unless if it's stated it's an equilibrium which there'll be no point for this problem so q just like k is going to be products over reactants and the reaction is already balanced so the only product that we have is n2o4 and it has a coefficient of one so it's raised to the first power o2 has a coefficient of one but it's on the reactant side so we're going to put it in the bottom n2o has the coefficient of 2 so for this species it's going to be squared so let's plug in the values so n2o4 has a value of 0.483 and for o2 the value is 0.265 and for n2o it's 0.134 squared so you can plug this in directly in your calculator by the way if you do if you want to type it in all at once just make sure to put the numbers in the bottom within one uh large parenthesis otherwise your calculator may divide by 0.265 but it may multiply by 0.134 squared so you might get a different answer unless you put it in parentheses by the way you should be careful with that because and that's where a lot of students make a mistake is when you're when you have two things in your denominator um if you divide it without the parentheses your calculator is going to divide by the first number then multiply by the second and a lot of students lose a lot of points because of that problem so just make sure you enclose it in parentheses and you can you'll get the right answer so you don't want to lose points due to calculated errors so the value that i got for this is 101.5 if i typed it in correctly so now that we have q we need to compare it to k so k is equal to 56.8 and q is 101.5 so we can say that q is greater than k and because q is greater than k we know that it's going to shift to the left so the correct answer for this problem is answer choice b the reaction will shift to the left 20 which of the following statements is correct let's look at a the reaction shifts to the right if q is greater than k is that true or false well let's find out so if we make a number line let's put k in the middle and since q is greater than k let's put it on the right side if q is greater than k is going to shift to the left so a is a false statement b the reaction is product faded when k is significantly less than one is that true or false when k is significantly less than one let's say when it's like .001 it's react in favor not product favored k is the ratio of products over reactants for it to be product favored we have to have a lot of products at equilibrium compared to the reactants so let's say if we have a thousand products to one reactant this would be product favored k would be very large k would equal a thousand but let's say if we have more reactants than products let's say if we have a thousand more reactants than products so it's react in favor whenever the denominator of a fraction is very large you're going to get a small value like .001 whenever the numerator of a fraction is very large the value of the fraction is large as well but since the denominator is very large the value of the fraction is very small and therefore when k is very small like 0.001 we can see that we have a lot more reactants than products so it's reacting favored so b is a false statement now what about part c increasing the temperature for an endothermic reaction causes the equilibrium constant k to decrease is that true or is that false let's find out so let's say if we have a reaction a plus b turns into c and it's endothermic so delta h is positive what's going to happen if we increase the temperature of an endothermic reaction will the reaction shift to the right or will it shift to the left for an endothermic reaction you can treat delta h as if it's a reactant anytime you increase the concentration of a reactant the reaction will shift to the right according to the shotalia's principle if you increase the concentration of the reactant the system is going to try to decrease the concentration of that reactant and the way it decreases it is it's going to shift the reaction to the right as the reaction shifts to the right the concentration of the products will go up and the concentration of the reactants will go down so that's how it brings the reactant back down to its uh equilibrium level so to speak so the same will be true if we increase the temperature we can treat delta h as if it's a reactant and so the reaction is going to shift to the right and any time it shifts to the right the um the concentration of the products will increase in value because it's going towards the products and since the reaction is moving away from the reactants as it moves to the right the concentration of the reactants go down and since k is the ratio of products to reactants k is going to increase in value since the products go up the value of the fraction will increase anytime you increase the numerator of a fraction the value of the entire fraction goes up and the reactants because they decrease in value k will also increase in value whenever you decrease the value of the denominator the value of the entire fraction also goes up so increasing the products will increase k the numerator increases decrease in the reactants will also increase k because the denominator decreases which increases the whole value of the fraction so k greatly increases when you increase the temperature so c is a false statement d the presence of an inert gas has no effect on the equilibrium constant k is that true or false by elimination you know it has to be the right answer so it's a true statement if you were to add an inert gas to the system because it's inert that means it's non-reactive it's not going to affect or react with the reactants or the products and so it will not affect the equilibrium constant k also k will not change if you change the reactants or the concentration of the products and it will not change if you were to change the pressure or the volume of the reaction the only way cable change is if you change the temperature so i just want to make this clarification if you change the concentration of the reactants or products without changing the temperature the equilibrium constant k will remain the same it's going to stay constant it is a constant to begin with however if you change the temperature which will change the reactants and the products that's the only case when the value of k will change if it changes due to temperature so make sure you remember that fact k depends only on temperature nothing else 21 which of the following is a weak acid so to answer this question you need to know the six common strong acids which are hdl hbr hi hydro iotic acid sulfuric acid nitric acid and perchloric acid usually it's if it's not one of these six common strong acids it's safe to say that it's not a strong acid so we can eliminate nitric acid that's strong sulfuric acid is a strong acid and h i is a strong acid so b hno2 is the weak acid 22 which of the following salt will decrease the ph of an aqueous solution acids have low ph values bases have high ph values so we're looking for the acidic salt let's start with c potassium chloride so we have the k plus ion and the chloride ion alkali metal ions like sodium and potassium are neutral ions when placed in an aqueous solution they will not change the ph of the solution significantly now chloride is the conjugate base of a strong acid hcl the stronger the acid the weaker the conjugate base hcl is such a strong acid that the conjugate base the chloride ion is so weak that it's considered to be neutral so kcl is a neutral salt if you put it in water the ph will remain approximately seven now let's consider sodium fluoride the sodium plus ion is an alkali metal ion and so it's neutral however the fluoride ion is not neutral hf is a weak acid it's not a strong acid so because hf is the weak acid the conjugate base will be a relatively stronger base so fluoride will be considered a basic so or basic ion if you put sodium fluoride in water the ph will increase it's going to be above seven because the conjugate acid is a weak acid so we can write basic but we can still eliminate d so what about answer choice a sodium nitrite is it an acidic cell basic salt or neutral salt well the sodium ion is neutral but the nitrite ion is a basic salt and the reason for that is because a channel 2 is a weak acid notice that both hf and a channel 2 are weak acids therefore the conjugate base fluoride and nitrite are basic ions but because hcl is a strong acid the conjugate base chloride is going to be a neutral ion so a is eliminated it's a basic cell so by elimination we know the answer is nh4 br so we have two ions nh4 plus the ammonium ion and the bromide ion so let's start with bromide is bromide acidic basic or neutral if it has a negative charge it's rare that's going to be acidic this is considered to be a neutral salt because or neutral ion hbr is a strong acid so the conjugate base bromide is going to be neutral now nh4 plus this is something you have to know it's a weak acid if you were to put nh4 plus in water it's going to uh donate a proton to water and generate an h3o plus and nh3 the conjugate base and this reaction is reversible so as you can see when you put nh4 plus in water it generates h3o plus so it increases the hydrogen ion concentration in water thus making water more acidic or making the solution more acidic rather and so the ph will decrease so nh4br is considered to be an acidic cell if you put it in water the ph will decrease below seven and so b is the right answer for this problem twenty-three the ph of a solution is three point seven eight determine the h2o plus concentration in order to determine the hydronium ion concentration there is an equation that we need and it's equal to 10 raised to the negative ph so all we have to do is type in 10 raised to the negative 3.78 into our calculator and you should get a concentration of one point sixty six times ten to the minus four so the right answer for this problem is answer choice c 24 calculate the ph of a 0.25 molar acetic acid solution the ka of acetic acid is 1.8 times 10 to the minus 5. so for this problem you could use an ice table or you can just use an equation but let's start with nice table so we have a generic weak acid which i'm going to write as h a and when it reacts with water it's going to generate h3o plus and the conjugate base a minus so initially we have 0.25 for the acid concentration because water is a liquid we don't have to include it in the equilibrium expression for ka and initially we have a concentration of 0 for h2o plus and for the conjugate base a minus so this is going to increase by x and this is going to be negative x so we're going to have x x and 0.25 minus x so when we write the expression for k a it's going to be products over reactants so it's h3o plus it's times a minus divided by the concentration of h a so therefore k a is equal to x squared divided by .25 minus x when ka is significantly small which it is 10 to the minus 5 we could ignore this x value so we can avoid using the quadratic equation so let me make some space first so the simplified expression that we now have is uh we have x squared divided by 0.25 and that's equal to ka which we can replace it with 1.8 times 10 to the minus 5. so let's put the ka value over 1 and let's cross multiply so 1 times x squared is x squared and then point two five times one point eight times ten to the minus five that's about four point five times ten to the negative six now this point we can square root both sides so x is about 2.12 times 10 to the minus 3. now keep in mind x represents the concentration of h3o plus so therefore um we can now find a ph so the ph is equal to negative log of the h2o plus concentration and so we can take this value and plug it in for h3o plus so if you type in negative log of 2.12 times 10 to the minus 3 you're going to get an answer of about 2.67 so b is the right answer for this problem by the way if you're wondering if ka is small enough such that we can neglect the value of x test it out plug it back into the equation and see if the answer you get is very close to the true value of k so let me give you an example so the equation that we had was ka is equal to x squared over 0.25 minus x and the value of x that we calculated when we assumed that x was negligible was about 2.12 times 10 to the minus 3. so to see if that answer is valid what we're going to do is we're going to take this x value and plug it in to the equation so go ahead and type this in your calculator this is going to be 0.25 minus the value of x 2.12 times 10 to the minus 3. and this is squared so if you type in 2.12 times 10 to the minus 3 squared and then divide that by .25 minus 2.12 times 10 to the minus 3 all in parentheses you should get 1.813 times 10 to the minus 5. so notice that the answer is it's slightly different from the true value of ka the true value of ka for this problem was 1.8 times 10 to the minus 5. so it really didn't deviate that much because they're close to each other that means this is a good estimation of x which means we made the right decision to neglect the value of x now this answer was significantly different from the true value of ka then that's the indication that you shouldn't neglect the value of x and you may have to do a quadratic equation so now you know how you can tell whether or not if you should neglect the value of x just plug it back in and see if the ka value that you calculate is very close to the true value of ka 25 calculate the ph of the solution made by dissolving 30.5 grams of sodium fluoride in enough water to make a 650 milliliter solution the ka of hydrofluoric acid is 6.8 times 10 to the negative four so go ahead and try this problem feel free to pause the video and then unpause it when you're ready to see the solution now what you should not do in this problem is you don't want to use the equation ka is equal to x squared over the acid minus x the reason being is we don't have an acid dissolved in a solution instead we have a base or a basic cell sodium fluoride if we had hf dissolve in the solution then we should use ka but because we have sodium fluoride a base dissolve in this solution we should use kb instead so first let's write the reaction fluoride when we dissolve sodium fluoride in water uh sodium is going to be the specs of the ions so we don't have to worry about it fluoride is going to react with water to produce hf and hydroxide and it's reversible hf is a weak acid and so let's make our ice table the we need to find the initial concentration of fluoride and we're given 30.5 grams so let's convert that so 30.5 grams of sodium fluoride over one n a has a molar mass of 23 and fluorine has a molar mass of 19. so the molar mass for naf is 42 grams per 1 mole and we need to convert it to uh molarity so we need to divide by liters now that we have moles to convert milliliters into liters simply divide it by a thousand or move the decimal three units to the left so 650 milliliters is the same as 0.650 liters so 30.5 divided by 42 and then take that result divided by 0.65 you should get a molarity of 1.117 so that's the initial concentration of the fluoride ion and we're going to have zero and zero now because we have nothing on the right side the reaction has to shift to the right it can't shift to the left if it were to go to the left then we would get negative values for hf and hydroxide so that's not possible so because it shifts to the right initially we're going to have positive x on the product side and negative x on the reactant side so if we add the first two rows we should get 1.117 minus x and on the other side x and x so kb is the ratio of the products over the reactants so the products are hf and the concentration of hydroxide divided by the concentration of we don't include water because it's a liquid liquids and solids you should not include it in the equilibrium expression everything else is in the aqueous phase which you can include that in the equilibrium expression so in terms of x kb is equal to x squared divided by 1.117 minus x now notice that we don't have the value of kb instead we have the value of ka but we can find kb from ka so let's go ahead and do that but first let's make some space so ka times kb is equal to 1 times 10 to the minus 14. so that means that kb is equal to one times ten to the minus fourteen divided by ka which is six point eight times ten to the minus five or i meant to say 4. so if we divide those two numbers you should get a value for kb which is 1.471 times ten to the minus eleven ten to the negative eleven is very very very small it's much smaller than ten to negative five so if we can neglect x when ka was ten to negative five we can certainly ignore this x value if it's at very small such like negative eleven that's extremely small so let's plug in kb and let's solve for x so let's put this over one and let's cross multiply so we're going to have x squared which is equal to the kb value 1.471 times 10 to the minus 11. we're going to multiply that by 1.117 and you should get uh 1.643 times 10 to the minus 11. so now we're going to take the square root of both sides so x is equal to 4.05 times 10 to negative 7 i mean six now because we calculated x from kb x represents the concentration of hydroxide and not h3o plus so therefore we need to find the ph by finding the poh first poh is negative log of hydroxide so let's plug in the value of x for the hydroxide concentration in the poh equation so if you type in negative log of 4.05 times 10 to the minus 6 the poh value is therefore equal to 5.39 so using poh we can now calculate the ph value so we said poh was 5.39 now the ph plus the poh in an aqueous environment these two numbers add up to 14. so therefore ph is equal to 14 minus poh so if you subtract 14 by 5.39 you'll get the ph value and it turns out that it's 8.61 so therefore c is the right answer now makes sense early in this video we said sodium fluoride is a basic cell the ph will be above seven and it is it's 8.61 if we were to add a acidic salt to water to make a solution the ph would be less than seven it could be like five to six somewhere or maybe four ish but in this case it's 8.61 so that's the answer for this problem and that's how you find it so just keep in mind uh look at the substance that is dissolved in water if it's an acid make sure you use k8 to find x and x will represent h3o plus if it's a base make sure you use kb sometimes you may have to find kb from ka but use kb and keep in mind that x represents hydroxide so let's review the equations uh that we have so let's say if an if you have an acid like hf is dissolved in water so step one use this equation ka is equal to x squared divided by the acid concentration minus x if ka is very small which is usually the case you could ignore the x value so if it's like 10 to negative 4 or less it's safe to say that you could ignore x if it's like 10 to the minus 2 or 10 to the minus 3 um you have to make a decision but if it's like 10 to the minus 2 i probably won't take a chance of ignoring x i would try to use the quadratic equation but once you use this equation and once you find x you need to know that x represents the concentration of h3o plus so your next step is to find a ph which is negative log of the h plus or h2o plus concentration so that's if you have an acid dissolve in water now let's say if you have a base dissolving water like sodium fluoride first you need to use the kb equation which is uh x squared divided by the concentration of the base minus x and if kb is small you could ignore the x on the bottom now once you have uh x keep in mind that for a base x is equal to the hydroxide concentration and not the h2o plus concentration so you got to find poh by taking the negative log of the hydroxyl concentration and then you can find ph by subtracting 14 by poh and keep in mind ka times kb is equal to 1 times 10 to the minus 14. so those are some equations that you need to know for the chapter on acids and bases 26 a solution contains a mixture of 0.755 molar hf and 0.125 molar sodium fluoride calculate the ph of the solution the ka of hydrofluoric acid is 6.8 times 10 to the minus 4. so notice that we have the weak acid and its conjugate weak base together they create a buffer solution and to find a ph of a buffer solution you can use the henderson-hasselbach equation which is ph is equal to pka plus log base divided by acid so let's find the pka first pka is equal to the negative log of the ka value and the ka value is 6.8 times 10 to the minus 4. so if you type in negative log 6.8 times 10 to the minus 4 you should get a pka value of 3.167 so now we can take that and plug it into the buffer equation so it's going to be ph is equal to 3.167 plus log base over acid so which one is the base well the acid is the one that has more hydrogen so hf is the acid and naf is the base so it's going to be 0.125 divided by the acid concentration which is 0.755 so you can type this in directly into your calculator just the way you see it and if you do that you should get before i give you the answer is the ph going to be greater than or less than the pka value what do you say so if we were to have equal amounts of acid and base the ph will equal the pka so it would be 3.167 if we have more of the base less of the acid the ph will be above 3.167 and if we have more of the acid less of the base the ph will be below 3.167 notice that we have more of the acid concentration so therefore the ph should be less than 3.167 and it is if you type this in the calculator you're going to get 2.39 because we have more of the conjugate acid and less of the conjugate base and so d is the right answer for this problem 27 which of the following is a buffer solution is it a b c or d now keep in mind that a buffer solution is the combination of a weak acid and its conjugate weak base so let's look at answer choice a hf is a weak acid and hcn is a weak acid we don't have any weak bases so a is eliminated if we look at c um h i is a strong acid and k i is neutral so we don't have a weak acid weak base pair so c is eliminated looking at b hcl is a strong acid and sodium chloride is neutral so b is eliminated acetic acid is a weak acid and sodium acetate is a weak base and also notice that acetic acid and the acetate differs by one hydrogen so that means that it's a conjugate acid-base pair which is a requirement for buffer solution so d is the right answer 28 which of the following is the lewis acid is it boring bh3 is it aluminum chloride alcl3 or iron iii chloride fecl3 or is it all of the above the correct answer is it's all of the above if you were to draw the lewis structure of bh3 if you add up the valence electrons you'll see that it has six valence electrons and boron likes to form three bonds it's going to have an incomplete oxide so because boron doesn't have eight electrons it's electron deficient and so it can accept a pair of electrons and lewis acids are electron pair acceptors lewis bases are electron paired donors a bronze lary acid is a proton donor and a bronstellari base is a proton acceptor and our heinous acid is a substance that generates h plus or hydronium ions in solution and our heinous base is something that generates hydroxide ions in solution but bh3 is a lewis acid it can accept a pair of um excuse me a pair of electrons so let's say if we have a lewis base like nh3 which has a lone pair nh3 can donate a pair of electrons to bh3 and we're going to get this product which looks like this so as you can see the nitrogen atom donated a pair of electrons which makes it a lewis base or an electron paired donor boron was able to accept a pair of electrons which makes it a lewis acid or electron pair acceptor so as you can see um this bond that was created between a boron and the nitrogen atom uh nitrogen provided the two electrons to generate that bond so that's why nitrogen or ammonia is the lewis base electron paired donor and boron didn't provide any electrons to create that bond so boron was the electron pair acceptor or the lewis acid now aluminum chloride has a very similar geometry it's trigonal planar and most trigonal planar structures are usually lewis acids because they can accept a pair of electrons fecl3 has a very similar structure so all of these three compounds they're very common lewis acids so d is the right answer it's all of the above 29 write the equilibrium expression for ksp based on the following reaction ksp stands for solubility product constant now there's different types of case you've seen k c which is the equilibrium concentration constant and then there was kp which is equilibrium pressure constant and ka the acid dissociation constant kb the base dissociation constant there's kw auto constant of water and if you see keq is just a generic equilibrium constant so ksp works the same way as any type of k value it's going to be products over reactants so the products are calcium two plus and fluoride now the coefficient of fluoride uh is two so we need to square it now typically would divide this by the reactant but notice that the reactant is in a solid phase so therefore we cannot or should not include it in the equilibrium expression so we're going to put a 1 instead so the right answer is answer choice b it's the only one that matches what we have and so that's how you can write the equilibrium expression for ksp the solubility product constant number 30 calculate the molar solubility of magnesium hydroxide the ksp or solubility product constant of magnesium hydroxide is 1.8 times 10 to the minus 11. so if we were to write an ice table this would be 0 0 plus x plus 2x because it's a one to two ratio and for the left side we can just put nothing there or just a one because it's a solid it's not going to be included in the equilibrium expression and so for the last row 0 plus x is x 0 plus 2x is 2x so if you were to write the expression for ksp it's going to be the concentration of mg2 plus multiplied by the concentration of hydroxide but squared since we have a coefficient of two in front of hydroxide so the molar solubility of magnesium hydroxide is basically the value of x it really represents the amount of magnesium hydroxide that is soluble and so basically we want to find the concentration of mgoh2 that's in the aqueous phase notice that mgoh2 and mge plus two is in a one-to-one ratio and x represents the concentration of mg plus two so if we could find x we're gonna have the concentration of mg plus two that's dissolve in a solution which is equal to the concentration of mgoh2 that's in the aqueous phase so basically anytime you need to calculate the molar solubility you're looking for the value of x so we can plug in 1.8 times 10 to the minus 11 for ksp the mg plus 2 concentration is the same as the value of x hydroxide is equal to 2x but we have to square it so 2x squared 2x times 2x is 4x squared and 4x squared times x is 4x cubed so let's make some space first so 1.8 times 10 to the minus 11 is equal to 4x cubed the first thing we should do is divide both sides by 4. so 1.8 times 10 to the minus 11 divided by 4 is about 4.5 times 10 to the minus 12 and that's equal to x cubed so to solve for x you need to take the cube root of both sides you can also write it this way you can raise both sides to the one-third power that's the same as taking the cube root so whenever you raise one exponent to another you multiply so 3 times 1 3 the 3's cancel so you get x to the first power which is x and the cube root of 4.5 times 10 to the minus 12 is about 1.651 times ten to the negative four so this is the molar solubility of magnesium hydroxide so that's the amount of magnesium hydroxide that is dissolved in the solution so the answer for this problem is answer choice a 31 a saturated solution of lead ii chloride contains 0.0159 m of pb plus 2 and .0317 m of cl minus calculate the solubility product constant of pbcl2 so in this problem notice that we have a saturated solution that means that the system is at equilibrium so we can simply plug in these values for pb plus 2 and chloride and that's going to give us the ksp value for lead 2 so let's write the expression of ksp first so ksp is the ratio of products to reactants so the products are pb2 plus and chloride now we have a coefficient of two for chloride so let's put a two um in the exponent position and we don't have to worry about the solid we can put a one in the denominator if we want to so let's plug in 0.0159 for the lead 2 plus concentration and 0.0317 for chloride but don't forget to square it so if you plug these numbers into your calculator you should get 1.598 times 10 to the minus 5 for the ksp value of let2 chloride so we can round that to 1.6 so c is the right answer for this problem 32 for which of the following reactions is delta s or the change in entropy positive is it a b c or d now what you need to keep in mind is that gases have a very high entropy value entropy is associated with disorder and gases are very disorderly solids in a solid the atoms are arranged in a nice neat orderly pattern so solids have a very low entropy because they have a very low disorder among the crystal lattice so they're very orderly now let's look at antichoice a on the left side we have three gas molecules and on the right side we have two gas molecules so three gas molecules have more entropy than two gas molecules so we're going from a high entropy system to a low entropy system whenever you go from height alone the value is decreasing so delta s is negative not positive so a is eliminated now let's look at answer choice b we have a solid and a gas on the left and a solid on the right well because the left side has a gas the left side has a higher entropy than the right side so we're going from high to low again so anytime you go from high to low the change in entropy is negative so b is eliminated now looking at answer choice c we're going from a gas to a liquid a gas has more entropy than a liquid so we're also going from high to low so delta s is also negative now for d we have two carbon atoms plus oxygen gas and it turns into two carbon monoxide gas molecules so we don't want to say it's three to two and assume that entropy is negative because a solid the entropy of a solid is very very low compared to the entropy of a gas in terms of numbers the entropy of a solid could be like maybe 5 10 or 20 joules per mole some of them could be 30 or 40. but the entropy of a gas could be like 200 joules per mole so a gas is significantly higher than entropy relative to a solid so when comparing a gas in a solid you could ignore the solids generally speaking so on the left side we have one gas molecule and on the right side we have two guest molecules so we're going from a low entropy system to a high entropy system so delta s is positive because the entropy is increasing as we go from left to right so d is the right answer 33 calculate delta g or the change in gibbs free energy using the following information at 340 kelvin so we're given the enthalpy value and the entropy value so the change in enthalpy is negative which means that it's an exothermic reaction and a change in s or entropy is positive so there's an equation that we can use to find delta g and it looks like this delta g is equal to delta h minus t times delta s now notice that delta h is in kilojoules but s is in joules so for this equation to work you have to convert joules into kilojoules if you don't you will not get the right answer and that's important you don't want to make this mistake on the exam so let's plug in the numbers that we have so delta h is negative 64.2 kilojoules per mole minus the temperature which is 340 kelvin now we're going to convert 105 joules into kilojoules so we can divide 105 by a thousand or move the decimal point three units to the left so it's going to be 0.105 kilojoules per mole per kelvin so the units kelvin cancel and we're left with kilojoules per mole and kilojoules per mole so we can now add the two numbers so negative 64.2 minus 340 times 0.105 you should get negative 99.9 kilojoules per mole and so that's delta g and because delta g is negative the reaction or this system is spontaneous at 340 kelvin so b is the right answer for this problem by the way the video that you're currently watching is a two hour trailer version of a longer seven hour video this video currently has 40 multiple choice problems but i'm gonna post a link so you can access the remaining portion of this video which contains a total of 150 multiple choice problems so when you get a chance feel free to check that out and let's continue with the next problem 34 which of the conditions shown below will cause a reaction to be spontaneous for all temperatures is it a b c or d so let's make a table we can have enthalpy entropy gibbs free energy and temperature so this is going to be negative positive positive negative well let's make this positive positive and then negative negative and then negative positive it turns out that when both enthalpy and entropy is positive delta g will be negative under high temperature conditions when enthalpy is positive and entropy is negative delta g will be positive under all temperature conditions so at any temperature delta g will be positive keep in mind when delta g is positive it's non-spontaneous when it's negative it's spontaneous meaning the reaction will work it will occur in the reaction uh or in the direction that it's given when both delta h and delta s are negative delta g will be negative only under low temperature conditions and when enthalpy is negative and entropy is positive delta g will be negative at any temperature so using this information the reaction will be spontaneous at all temperatures if delta h is negative and delta s is positive so that's answer choice b as delta s is positive and delta h is negative so b is the right answer now you might be wondering how you can get this information without committing this to memory if you can commit this to memory that's great but we need to come up with a way to figure this out so let's say delta h is about 10 000 joules per mole or let's say 10 kilojoules per mole and let's say delta s is about 0.1 kilojoules per mole per kelvin so h and s is going to be the same but the only thing that's going to vary is the sine of h and s and the temperature so we're going to plug in 10 for a low temperature value and a thousand for a high temperature value so let's prove the first case so let's say that delta h is positive 10 and then we have a minus sign we're going to plug in the high temperature value so let's plug in a thousand for a t and delta s is 0.1 so this is 10 minus 100 which is going to be negative 90. so delta g is negative which is true so now let's plug in a low temperature so h is still going to be positive 10 and then minus the low temperature which we're going to plug in as 10 times the entropy value of 0.1 so 10 minus let's see 10 times 0.1 is 1 so we're going to get 9 and notice that it's positive so delta g is positive at low temperatures when delta h and delta s are both positive but when the temperature was high like a thousand delta g was negative and this proved true so when delta h and delta s are both positive when the signs are the same delta g is spontaneous only at high temperatures at low temperature conditions it's non-spontaneous so now let's prove the second case so it has to be non-spontaneous or positive at all temperatures meaning if we plug in a low temperature or high temperature value it has to remain positive so let's consider the second case let's prove that delta g will always be positive at any temperature if delta h is positive and if delta s is negative so using the same equation delta h is still positive 10 minus let's try a high temperature so let's plug in a thousand and delta s is negative so we're gonna plug in negative point one so a thousand times point one is a hundred but because we have two negative signs it's going to be plus 100 so we get positive 110 so we can see delta g is positive at high temperatures so now let's try it at low temperatures so using the same equation um h is going to be positive 10 and now we're going to plug in a low temperature value of 10 and s is still negative 0.1 so this is going to be 10 plus one ten times point one is one so we get positive eleven so as you can see it's positive under low and high temperature conditions so we could say it's positive or non-spontaneous at any or at all temperature conditions so now let's confirm the third scenario when they're both negative so this time we're going to plug in negative 10 for delta h and it's supposed to be true at low temperatures but let's start with a high temperature value so delta h minus t so let's plug in a thousand the high temperature value for t times negative point one so that's negative ten and then negative a thousand times negative point one that's positive a hundred so that's ninety which is positive so we can see that delta g is positive at high temperatures we need to confirm that it's negative under low temperature conditions so notice that it didn't say all so on the table because it says negative at low temperature conditions that means that it's positive at high temperature conditions is the opposite so now let's try a low temperature so delta g is equal to delta h which is uh negative 10 minus a low temperature of 10 times negative 0.1 so this is negative 10 negative 10 times negative 0.1 that's going to be positive one so overall that's negative nine so as you can see it's negative only at low temperature conditions and not at high temperature conditions so now let's look at the last case so it has to be true at any temperature so let's start with a low temperature so h is going to be negative 10 and then minus t so we're going to try a low temperature of 10 times entropy is positive so we're going to plug in a 0.1 so negative 10 and negative 10 times positive 0.1 that's negative 1 so we get negative 11. so as we can see it's negative under low temperature conditions under high temperature conditions let's plug in a thousand this is going to be negative 10 minus 100 which is negative 110 and it's still negative under high temperature conditions so when enthalpy is negative and the change in entropy is positive delta g is negative at any temperature because two negatives when you add them will always be negative but for this problem b is the right answer so the reaction will be spontaneous at all temperatures when enthalpy is negative and entropy is positive 35 the equilibrium constant k is 1.83 times 10 to the negative 12 for a particular reaction at 372 kelvin calculate the change in gibbs free energy for this reaction is this reaction spontaneous at this temperature so the equation that you need to find delta g using k is this delta g is equal to negative rt ln k so r you don't want to use the value 0.08206 for r that's for gases rather you want to use the energy constant negative 8.3145 and for that value r has units joules per mole per kelvin so i'm going to write that in red so joules per mole times kelvin and the temperature is going to be in kelvin so 372 kelvin so the units kelvin cancel so initially this equation is going to give you the units joules per mole so we're going to have to convert that to kilojoules and then times the natural log of the value of k which is 1.83 times 10 to the negative 12. so if you type that in you should get 83 593 joules per mole now we wish to convert that to kilojoules so we need to divide it by a thousand so if we can also move the decimal three units to the left so it's going to be 83.6 kilojoules per mole and it's positive so d is the right answer whenever delta g is positive you have a non-spontaneous reaction so let's review something when k is significantly greater than one let's say like ten thousand delta g is going to be negative and therefore it's going to be a spontaneous event it's product favored so the reaction proceeds as written it goes from left to right when k is significantly smaller than one like point zero zero zero one or like ten to the negative eight if it's very close to zero delta g will be positive and therefore you have it's going to be non-spontaneous in the direction that is written so as it goes from left to right it's non-spontaneous and if k is approximately one typically delta g is approximately zero and the system is at equilibrium so those are some things that you want to keep in mind so when k is significantly larger than one it's a product favored and whenever it's product favored that means you have the reaction is it's spontaneous so to speak and when it's reacting favored it's non-spontaneous for the most part so those are some concepts that you want to keep in mind 36 which of the following statements is slash r correct is it a b c or d so these are some things you need to know oxidation occurs at the anode and not the cathode so one is a true statement and three is a false statement reduction occurs at the cathode and not the anode so number four is a true statement and two is false so therefore the answer is a one and four oxidation always occurs at the anode reduction occurs at the cathode and keep in mind whenever the oxidation number of an element increases its oxidation if the oxidation number decreases its reduction and whenever an atom loses electrons the oxidation will go up and it's oxidized reduction involves the gain of electrons oxidation is associated with the loss of electrons so those are just a few things that you want to remember for electrochem 37 what is the coefficient of zinc when the following redox reaction which occurs in acidic conditions is balanced so what we need to do in this problem is that we need to separate it into half reactions so let's start with zinc elemental zinc in its natural form or natural state has an oxidation state of zero and it's going to become zinc plus two which has an oxidation state of two so once we balance the particles we need to balance the charge we have a total charge of zero on the left and plus two on the right so how many electrons do we need to add the difference of zero and two is two so we gotta add two electrons and you should always add it to the side with the higher charge or the higher oxidation state so we're gonna add two electrons to the right side so if we add up all the charges on the right plus 2 and 2 times negative 1 which is negative 2 positive 2 and negative 2 adds up to 0 which is equal to the net charge on the left so the charge is a balance this half reaction is known as an oxidation half reaction because the oxidation state was increased from zero to plus two also there was a loss of two electrons whenever the electrons are on the right side it's an oxidation half reaction if the electrons are on the left side it's a reduction half reaction so now let's balance the other half reaction so we have bromite converting into bromide so the bromine atoms are balanced so we need to add two oxygen atoms under acidic conditions you can add h plus and h2o to balance the reaction so since we have two oxygen atoms on the left side we need to add two h2o molecules on the right side so that the number of oxygen atoms are balanced notice that we have four hydrogen atoms on the right side so we need to add four h plus ions on the left now the last thing we need to do is balance the charges the net charge on the right side is -1 and the net charge on the left side is 4 times 1 which is plus 4 plus the negative charge so plus 3. so what is the difference between three and negative one if you subtract three minus negative one that's like three plus one so that's four so we need to add four electrons to the left side if you add up all the charges negative four plus four plus negative one has the net charge of negative one so the charges are balanced the charges on the left side is now equal to the right side so to get the net reaction the electrons must be equal since they're not we'll need to multiply the first equation by 2. so this is going to be 2 2 and this is going to change to a 4. so now we can add up the reactions so the electrons will cancel because we have four electrons on both sides and we can bring everything down so we have two zinc atoms let me change the color plus four h plus ions plus one uh bromite ion which will convert to two zinc plus two ions one bromide ion and two water molecules so as you can see this reaction is balanced we have four hydrogen atoms on both sides one bromine atom two oxygen atoms and two zinc atoms on both sides and if we look at the charges we have plus four minus one plus 4 minus 1. so the charges are the same so the net reaction is balanced so now let's look at the question they wanted to know what is the coefficient of zinc when the redox reaction is balanced so we have a 2 in front of zinc now we could have seen that here we didn't really need to add the two reactions to get the net reaction but i wanted to do it for the sake of practice so the coefficient is 2 so b is the right answer for this problem so that's it 38 which of the following statements are true one is a true statement whenever the cell potential is greater than zero or if it's equal to a positive number the reaction is spontaneous two is false if it's less than zero it should be non-spontaneous and three is true if the cell potential is equal to zero then the system is at equilibrium so the correct answer for this problem is one and three so c is the right answer so those are just some facts that you have to commit to memory 39 calculate the cell potential of a reaction if the change in gibbs free energy is negative 41.3 kilojoules per mole and n has a value of 2. the equation that you need for this problem is this delta g is equal to negative nfe now delta g has to be in joules per mole not kilojoules per mole so we need to multiply this value by a thousand so we're going to plug in negative 41 300 joules per mole n has a value of two f is faraday's constant which is 96 485 columns per mole of electrons so let's solve for e so if we multiply 2 by 96 485 we should get one hundred ninety two thousand and nine seventy so let's divide both sides by negative one hundred ninety two thousand nine seventy so these cancel and therefore the cell potential of this reaction is positive 0.214 the two negative signs cancel so c is the right answer now keep in mind when delta g is negative we said it's spontaneous and when e is positive it's also spontaneous the reason why the signs are opposite is due to the negative sign in the equation number 40 calculate the cell potential of the reaction using the information shown below at 298 kelvin under non-standard concentration values so we have the standard cell potential which is the cell potential if the concentration of the reactant and the product is one molar however we want to find the cell potential when the concentration of aluminum is not one molar it's one point five times ten to minus five and when the concentration of the copper plus 2 ion is about 35.4 now notice that the concentration of the product ion was decreased the standard value is 1 but it was decreased to 1.5 times 10 to minus 5. now if whenever the reaction shifts to the right the cell potential will increase and whenever it shifts to the left the cell potential will decrease so since we decrease the concentration of the product will the reaction shift to the left or to the right according to le chatelier's principle if you decrease the concentration of the product it's going to shift to the right to increase the concentration of the product and because it shifts to the right the cell potential should go up now for the reactant the copper plus two ion we've increased the concentration of the reactant so it should shift to the right and the cell potential should increase as well so the cell potential at these uh concentrations should be greater than two volts so what this means is that we could eliminate answer choice a and answer choice b now the equation that we need the non-standard cell potential is equal to the standard cell potential value minus rt divided by nf natural log of the reaction quotient q the reaction quotient is the ratio of products to reactants it's very similar to k but it's associated with initial concentration values so q is the ratio of the aluminum plus three ion squared over cu plus two raised to the third power the coefficients turn into exponents now the next thing we need to know is the value of n if we write out the half reactions we have two aluminum atoms turning into two ar plus three ions and so to balance it we need six electrons and then for the other half reaction where we have three copper plus two ions turning into three copper atoms we also need six electrons to balance it so therefore n is equal to six so now let's plug everything into the equation so e is 2 volts r is the energy constant 8.3145 for gases it's 0.0826 the temperature is 298 kelvin ns6 faraday's constant is 96 485 and then times the natural log so for aluminum which is the product ion it's 1.5 times 10 to the minus 5 raised to the second power by the way we can't include the solid copper or the solid aluminum because liquids and solids should not be included in the equilibrium expression or in the reaction quotient just in case you were wondering now the copper plus two ion concentration is 35.4 but we have to raise that to the third power and all of this is inside the natural log so if you type this information into your calculator you should get answer choice e 2.14 votes now if you prefer to type in one at a time here's what you get let's multiply 8.3145 by two ninety eight so that's going to be twenty four seventy seven point seven and then six times ninety six thousand forty five that's five hundred seventy eight thousand nine ten and then the natural log of one point five times 10 to the minus 5 squared divided by 35.4 raised to the third power that is going to equal negative nine one 32.915 so negative thirty two point nine one five times negative twenty four seventy seven point seven that's eighty one five five three and then divide that by five seven eight nine ten um you're going to get positive 0.14 so what we have now is this is going to be 2 plus 0.14 which will give you 2.14 so d is the correct answer