in this video we're going to talk about chemical kinetics we're also going to talk about reaction rates and also how to write the rate law expression using the initial rates method so you might be wondering what is chemical kinetics well chemical kinetics is the study of reaction rates it tells us how fast a reaction can occur so we're concerned with the speed of a reaction which is quite different than thermodynamics in thermodynamics we're concerned if a process is spontaneous or not will it occur or will it not occur on its own a good example is the conversion of diamond to graphite this is a spontaneous process the free energy change of the reaction is negative which tells us that it's spontaneous however in real life if you were to take a piece of diamond you wouldn't notice any change from diamond to graphite you can have a piece of diamond for 10 years it will still look like diamond and so even though this reaction is energetically favorable according to thermodynamics the process is so slow that it's not very useful in terms of uh everyday application so chemical kinetics would describe how fast this reaction is going it says basically it's a slow reaction thermodynamics basically tells us that this reaction is spontaneous it's energetically favorable another example is the combustion of glucose when glucose reacts with oxygen it produces carbon dioxide and water delta g for this reaction is also negative so this reaction is spontaneous is energetically favorable energy is released during the combustion of glucose with oxygen however if you were to put sugar on the table it's not going to spontaneously combust glucose doesn't react with oxygen very fast at room temperature so this too is a slow reaction the activation energy is relatively high so even though it's spontaneous it's not a fast reaction unless you heat it up or add a spark to it same thing is true with gasoline gasoline doesn't always spontaneously combust within air unless you give it a spark or something to get the reaction started so these chemical reactions at room temperature is relatively slow so make sure you understand that chemical kinetics deals with how fast a reaction can occur whereas thermodynamics tells you if a reaction is spontaneous if it's energetically favorable now consider the reaction where a converts to b so a is the reactant b is the product now as the reaction proceeds from left to right the concentration of the reactant will decrease the concentration of the product will increase now we can describe how fast this reaction is going by measuring the rate of the reaction the rate of the reaction is going to be equal to the change in the concentration of b divided by the change in time so the square brackets indicates concentration so we have the unit's molarity over some unit of time like seconds it could be minutes or hours so typical unit of rate is molarity per second it could be molarity per minute and so forth now we have a positive sign here because the concentration of b is increasing as the reaction progresses but we can describe the rate of the chemical reaction with respect to a it's also equal to the change in the concentration of a divided by the change in time but we're going to put a negative sign in front so by putting the negative sign in front the reaction rate is assumed to have a positive value by definition because the change in a is negative times another negative which will make this positive so for reactants you're going to have a negative sign if you relate it to the rate of the overall chemical reaction with products you're going to have a positive sign so let's draw some graphs so if we were to plot the concentration of a with respect to time the concentration will decrease over time now how it decreases depends on the reaction it can be a straight line or it can be a curved line but generally speaking the reactant will decrease in concentration over time whereas the product let's use b for example the product b will increase in concentration over time now let's erase a few things the change in concentration of b divided by the change in time this is known as the average rate of appearance of b b is apparent the concentration of b is increasing if we wish to calculate the average rate of disappearance of a it's simply the change in the concentration of a divided by the change in time so what we would do is we would take the final concentration of b subtracted by the initial concentration of b and then divide that by the change in time that is the time difference between these two concentrations and the same is true for calculating the average rate of appearance for a reactant a it would be just the final concentration of a minus the initial concentration over the change in time now let's say if we have a reaction where one molecule of a reacts with two molecules of b to produce three molecules of c and four molecules of d now we can express the rate of the overall reaction in terms of the change in concentration of a b c or d and here's how you can do it so the rate is going to equal negative the change in the concentration of a divided by the change in the concentration of i mean the change in time so that's the rate with respect to a now what about with respect to b this is going to be negative change and the concentration of b divided by the change in time but there's more to it we need to incorporate the two so you need to divide by two so it's going to be negative one half times the change in concentration of b over the change in concentration of time so this is how we can relate the rate of the chemical reaction with respect to the rate in the change of the concentration of b now with respect to c the rate of the chemical reaction is going to equal this time it's going to be positive since we're dealing with a product the product increases in concentration the coefficient is a 3. so instead of 1 over 2 it's going to be 1 over 3 times the change in the concentration of c divided by the change in time now for d it's also product so it's going to be plus and then it's going to be 1 over 4 the change in the concentration of d divided by the change in time so let's consider an example problem let's say that b has an average rate of disappearance of negative 0.5 molarity per second what is the average rate of appearance of c and also what is the rate of the reaction for this example so let's calculate the average rate of appearance of c which should be positive so we're going to start with 0.5 molarity per second or you can express that as moles per liter per second if you want to but this is going to be substance b and we can convert it to substance a the molar ratio is 2 to 3. so the molarity ratio so to speak will be two to three so we could say one molarity per second for b and that's actually two the coefficient in front of b is two so when the concentration or rather when the molarity changes by two for b every second the molarity for substance c will change by 3 every second so we can cancel the unit molarity per second for substance b and we can get the average rate of appearance for substance c so it's going to be 0.5 times 3 which is 1.5 divided by 2. so that's going to be positive 0.75 molarity per second for c now technically this should be a negative value because b is decreasing but this two should also be negative because it's still decreasing and this should be positive so overall these two negative signs will cancel giving us a positive result so make sure you understand what this means the average rate of disappearance for substance b is negative 0.5 molarity per second the average rate of appearance for substance c 0.75 molarity per second now let's put this number here so we have positive 0.75 molarity per second for c and we're going to use both numbers to calculate the rate of the chemical reaction so the rate of the chemical reaction with respect to b is this it's negative one-half times the change in concentration of b divided by the change in time so this part right here delta b over delta t that is the rate of disappearance or the average rate of disappearance for b which is this value so we're going to replace that with negative 0.5 molarity per second so negative one half times negative point five this is going to be positive 0.25 molarity per second so notice that the rate of the chemical reaction is going to be equivalent to the rate of the change in concentration of a substance with a coefficient of 1. in the case of a it would simply be negative 0.25 so the rate of disappearance and the rate of appearance for each species is going to be proportional to the coefficients in front of them as you can see c has a change in concentration per change in time that's three times the amount relative to a because the molar ratio is one to three now let's calculate the rate of the chemical reaction but with respect to c it's going to be positive we're using this equation here 1 over 3 delta c over delta t so that's 1 over 3 times 0.75 so here we have two positive values the overall rate of the reaction will be positive and we're going to get the same answer it's going to be positive 0.25 molarity per second so i want you to understand how this equation relates to the rate of the chemical reaction so now you can calculate the rate of the chemical reaction with respect to any of the substances in the reaction now let's work on some practice problems so in this example we're given a table that shows the concentration of no with respect to time calculate the average rate of disappearance of no in the first 20 seconds so what should we do in this problem to calculate the average rate of a reaction we need to calculate the change in concentration and divided by the change in time so to calculate the average rate of disappearance of no it's going to be the change in the concentration of nl which is the final concentration minus the initial concentration divided by the time difference so we need to do it for the first 20 seconds so that's from 0 to 20. the final concentration at 20 seconds is 0.586 moles per liter which is molarity minus the initial concentration which is a 0.75 and the time difference is 20 seconds so 0.586 minus 0.75 that's the change of negative 0.164 divided by 20 and so the average rate of disappearance of no is negative 8.2 times 10 to the minus 3 molarity per second so that's the average rate of disappearance now let's move on to the next problem dinitrogen pentoxide decomposes into nitrogen dioxide and oxygen gas as shown in the equation below if the average rate of appearance of no2 is 4.6 times 10 to the minus 3 moles per liter per second what is the average rate of disappearance of n205 so how can we find the answer for this problem it turns out that the rate of appearance for no2 is directly related to the rate of disappearance of n205 based on the balanced chemical equation in fact you can use the molar ratios to get the answer you simply need to convert it so let's start with what we're given 4.6 times 10 to the minus 3 and that is moles of no2 per liter per second now let's convert moles of no2 to moles of n2 o5 now for every 4 moles of no2 that appears or that is produced 2 moles of n2o5 disappears now you need to understand that the rate of appearance is always positive and the average rate of disappearance is always negative which means that the concentration of the reactants always decreases and the concentration of the products increases as the reaction moves forward so the rate of the reaction in terms of reactant should always be negative and for a product it should be positive so i'm going to put a negative sign since we're dealing with a reactant so these units will cancel and so it's going to be 4.6 multiplied by negative 2 divided by 4. and so the answer is going to be negative 2.3 times 10 to the minus 3 moles of n2o5 per liter per second so that's how you can calculate the average rate of disappearance of n205 number three the graph below shows the concentration of fe plus 3 with respect to time estimate the average rate of disappearance of fe plus 3 in the first 30 seconds to calculate the average rate of change from a graph you need to calculate the slope of the secant line so we need two points a secant line touches the curve at two points so the slope of this line will give us the average rate of change now if we want to find the instantaneous rate of change we need to find the slope of a tangent line and the slope of the tangent line touches the graph at one point as opposed to the secant line which touched it at two points so keep that in mind so whenever you want to find the slope of a line it's the change in y divided by the change in x and in this problem the change in y values represents the concentration so it's going to be c2 minus c1 the change in x values is the change in time and the change in concentration divided by the change in time can give us the rate of disappearance of fe plus 3. so this is the initial concentration which we'll call c1 and this is the final concentration c2 so it's going to be the final concentration which is about 0.4 minus the initial concentration which is 1.0 divided by the change in time which is the first thirty seconds so point four minus one that's negative point six divided by thirty and so it's going to be negative point zero two m per second or molarity per second so that's the average rate of appearance or rather disappearance of fe plus three so now let's briefly talk about the rate law expression this is also known as the differential rate law which is different from the integrated rate law the differential rate law shows how the rate of the chemical reaction depends on concentration the integrated rate law which is not really covered in this video it shows how concentration depends on time so here's a typical rate law expression in this equation the rate of the chemical reaction is equal to sum rate constant k times the concentration of a let's say to the first power times the concentration of b to the second power times the concentration of c to the third power times the concentration of d let's say to the zero power when you see a rate law expression you need to know what this means the rate of the reaction is first order with respect to a it's second order with respect to b third order with respect to c zero order with respect to d so the order for each reactant is simply the exponent associated with that respective reactant now another thing you need to be familiar with is something called the overall order for the reaction to find this simply take the sum of all of the exponents one plus two plus three plus zero is six so we can say that the reaction is six order overall now what does it mean when it's first order with respect to a this means that if we double the concentration of a the rate of the reaction is going to increase by a factor of 2. and the reason for this is 2 to the first power is 2. if we were to increase a by a factor of three the rate of the reaction will increase by a factor of three three to the first power is three now what about b what does it mean that it's second order with respect to b if we were to increase the concentration of b let's say by a factor of 2 the rate is going to increase by 2 squared so it's going to increase by 4 because 2 squared is 4. if we were to increase the concentration of b by a factor of 3 the rate is going to increase by a factor of 3 squared three squared or three times three is nine so the rate is going to increase by nine if we were to increase the concentration of b by factor four four squared is sixteen so the rate is going to increase by a factor of 16. so that's what it means when it's second order overall any changes in the concentration of b the changes that the rate will fill will be squared of what the changes that b feels now let's focus on c it's third order with respect to c so if we increase c by a factor of two the rate is going to go up by two to the third power two to the third power is eight so the rate of reaction will increase by a factor of eight if we increase the concentration of c by a factor of three three to the third power is 27 so the rate is going to increase by a factor of 27. i'm giving you a lot of examples because i really want you to understand this concept now let's focus on d it's zero order with respect to d so if we increase the concentration of d by a factor of two what's going to happen to the rate what would you say two raised to the zero power is what anything raised to the zero power is one so there's not going to be any change for the rate it doesn't matter how much we change d let's say if we were to increase d by a factor of 3 the rate will still not change there's going to be no change in the rate of the chemical reaction 3 to the 0 is 1. so the rate of the reaction does it depend on a concentration with a zero order let me say that again the rate of a reaction doesn't depend on a reactant that is uh zero order so in this case for this particular example the rate doesn't depend on the concentration of d if you increase d or if you decrease it the rate of the chemical reaction will still be the same so a zero order reactant has no effect on the rate of the reaction the one that has the greatest effect is reactant c because it has the highest order if you double c the rate will greatly increase by a factor of 8. so c has the greatest effect on the rate of the chemical reaction but d has no effect on it now let's work on some practice problems that involve finding the rate law expression by experiment so in this problem we're given a table which shows the concentration of reactants a and b and units of molarity and we also have the initial rate of reaction with the unit's molarity per minute so how can we determine the rate law expression that fits the experimental data so first we need to find the order of the reaction with reactants a and b so what you want to do is pick two trials in which one reactant is held constant but the other changes so if we pick trials one and two notice that a is held constant in concentration but the concentration of b doubles so as b doubles from point 20 to point 40 notice that the rate increases by a factor of 4 from 0.4 to 1.6 so you can write this expression two to the what power is equal to four we know that two squared is four so x is two this means that it's second order with respect to b so when we write the rate law expression let's say for a we don't know what the order is so i'm just going to put a y variable but for b we know it's second order so any time the concentration doubles and if the rate quadruples you know it's second order but if you wish to calculate the two if you want a step-by-step process here's what you could do we can use the formula b2 over b1 raised to the x is equal to r2 of r1 now a is constant so we don't have to worry about aim but for trials 1 and 2 b changes so that's why we have b in this expression so b2 or the concentration of b for trial 2 is 0.4 the concentration of b for trial 1 is 0.2 and the rate for trial 1 is 0.4 and the rate for trial 2 is 1.6 so 0.4 divided by 0.2 is 2 and 1.6 divided by 0.4 is 4. now if you wish to calculate x it's going to be log 4 divided by log two you just gotta take the log of those two numbers and this number is always going to go on the bottom this one's gonna go on the top log four divided by log two is equal to two and so that's how we know it's second order with respect to b so now let's find the order with respect to a so notice that if we pick trials one and three a doubles from point 20 to point 40 but b is held constant and notice what happens to the rate the rate doubles from 0.4 to 0.8 so since a doubles we're going to put a 2 and the rate doubles we put a 2. so 2 is the same as 2 to the first power so y is equal to 1. so it's first order with respect to a now for those of you who want to use the formula here's what you can do it's going to be a2 divided by a1 raised to the y and that equals r2 over r1 but i'm going to change a little because we use trials 1 and 3 so i'm going to write a3 over a1 raised to the y and that's equal to r3 over r1 so a3 is 0.4 for trial 1 the concentration of a is 0.2 and that's raised to the y for trial 3 the rate is 0.8 and for trial 1 it's 0.4 so 0.4 divided by 0.2 is 2 and .8 divided by 0.4 is also 2. so therefore y is going to be log 2 divided by log 2 which is equal to and so that's how we know it's first order with respect to a so now what is the overall order of the reaction it's first order with respect to a second order with respect to b to calculate the overall order of the reaction all you have to do is simply add the exponents so 1 plus 2 is 3 that means that the reaction is third order overall so this is the rate law expression this is the answer to part a now what about b calculate the value of the rate constant so pick any trial i'm going to choose trial 1. for trial 1 the rate is 0.4 and a is 0.20 or 0.2 and b is 0.2 as well so point 20 squared times point two that's equal to eight times ten to the minus three or point zero zero eight so k is going to be point four divided by eight times ten to the minus three so in this example k is 50. now we need to determine the units of k so in this expression solve for k so k is going to be the rate divided by everything on the right except k so k is equal to the rate divided by a to the first power times b to the second power now we're given the units of the initial rate it's molarity times mean minutes to the minus one so let's write that molarity to the first power and then minutes to the negative one power now a is in molarity and b is in a molarity but it's squared so i'm going to write m squared now we could cancel m and i'm going to take this m move it to the top so the units of k you can write it as molarity to the minus 2 times minutes to negative one now keep in mind that molarity is moles per liter so you can write molarity as moles to the first power leaders to negative one so if we raise everything to the negative two that means we need to multiply every exponent by negative two and so what we now have is that molarity to the negative two power is moles to the negative two times liters to the positive two negative one times negative two is positive two so we could replace m to the negative two with moles to the negative two times liters to the positive two so therefore we could write the final answer for k like this so we could say k is equal to 50 times moles to the negative two times liters to the positive two times minutes to the minus one it might be better to put the positive exponent first so you could say it's 50 liters squared times moles to the minus two times minutes to the negative one and there's a lot of ways in which you can adjust these units now you could write it in terms of positive exponents you could move the negative exponents to the bottom so you could say it's 50 liters squared per mole squared times minutes where these two are in the denominator fraction so now you know how to calculate the value of the rate constant and you know how to determine the units of the rate constant k so now let's focus on part c what will be the initial rate if the concentration of a is 0.5 moles per liter and b is 0.8 moles per liter so now we just got to plug it into the equation so we have the value of k k is going to be 50 and the units is m to the negative 2. which is basically you could put that on the bottom m squared and then it's uh minutes on the bottom as well which is minutes to minus 1 on top so that's k and then the concentration of a is what we have here 0.5 m and the concentration of b is 0.8 m squared so 50 times 0.5 times 0.8 squared will give us a rate of 16. now m squared will cancel and so the units for rate will be what we have here molarity per minute and so it's 16 molarity per minute or moles per liter per minute you can describe it either way and so that's it for this problem so now you know how to determine the rate law expression how to calculate the value of the rate constant and how to calculate the initial rate given different concentrations you