in this video we're going to talk about equilibrium but within the context of chemistry so what is equilibrium well let's say we have a reversible reaction where a can convert to b and b can convert to a equilibrium occurs when the rate of the forward reaction that is a going into b is equal to the rate of the reverse reaction when these two rates are equal the concentration of a and b will be constant there's not going to be any net change and so that's when we have a case of equilibrium where the concentration of a and b are not changing because the rate of the forward reaction is equal to the rate of the reverse reaction chemical equilibrium is a form of dynamic equilibrium it's not static because things are happening even though a system may be at equilibrium the reaction is still occurring a is converting into b b is converted back into a so we still have a reaction but the concentrations of a and b are no longer changing at equilibrium so we have a case of dynamic equilibrium now to illustrate this concept let's say we have two cities on the left we'll call this city a on the right city b now let's say that city a has a thousand cars and we're gonna say that city b has two thousand cars and let's say connecting these two cities is a highway and on average each hour 10 cars travel from city a to city b and each hour let's say that 10 cars travel from b to a so we can say that the rate of the four reaction is 10 cars per hour and the rate of the reverse reaction is also 10 cars per hour so if each hour 10 cars is going from a to b and 10 cars is going from b to a will the number of cars in city a and b change the answer is no because if a loses 10 cars each hour but also gains 10 cars each hour the total number of cars in a is gonna remain constant the same is true for b these two don't have to be equal but at equilibrium the concentration of a and b are no longer changing it's constant and so that's what happens when the reaction is at equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction and at that point the concentrations of a and b are no longer changing another way in which we could visually show this concept is by the use of a concentration profile so this is going to be a graph with the concentration of the reactants on the y-axis and time on the x-axis so we're going to put a in red let's say initially we put one mole of a inside a container of a fixed volume and there's no quantities of b in its container so initially if we only have a and no b the reaction has to go to the right it can't go to the left because there's no b to convert into a so that's what's happening initially if we put one mole of a let's say in a one liter container as the reaction goes to the right the concentration of a is going to decrease over time the concentration of b is going to increase so a is going to decrease it's not going to go to 0 but eventually it will become constant b starts from 0 and then it increases b becomes constant when a becomes constant when a stops decreasing b will stop increasing now the point where the concentration no longer changes that's where we have a case of dynamic equilibrium so whenever the concentration of a and b when it becomes horizontal in this graph that means it's not changing anymore at that point the rate of the forward reaction equals the rate of the reverse reaction and so they're equal at that point so now let's plot a graph that shows the rate of the forward reaction and the rate of the reverse reaction so we're going to put the reaction rate on the y-axis time on the x-axis so the rate of the forward reaction is equal to k1 times the concentration of a as the reaction goes to the right a is going to decrease and as a decreases the rate of the forward reaction is going to decrease so initially the rate of the forward reaction is going to start at a high value and then it's going to decrease over time let me draw that better and eventually it's going to be constant now as a decreases we know b is going to increase and the rate of the reverse reaction is dependent on the concentration of b it's k negative 1 that's the reverse rate constant or the rate constant for the reverse reaction times b so as b goes up the rate of the reverse reaction is also going to go up so i'm going to put that in blue now eventually the rate of the forward reaction is going to equal the rate of the reverse reaction which it does here so at that point we've reached a state of equilibrium once the rate of the forward reaction equals the rate of the reverse reaction a is no longer decreasing b is no longer increasing the concentration of a and b will be constant at that point so if we set the rate of the forward reaction equal to the rate of the reverse reaction which occurs at equilibrium we can say that k1 times a is equal to k negative one times b i'm going to divide both sides by a and k negative 1. so these will cancel and those will cancel and so k 1 over k negative 1 is going to be b over a the equilibrium constant k is equal to the concentration of the products divided by the reactants so therefore the equilibrium constant k is also equal to the ratio of the forward rate constant divided by the reverse rate constant equilibrium constant k is represented by capital k the rate constant is indicated by lower case k so make sure you remember this the equilibrium constant k is equal to the forward rate constant divided by the reverse rate constant and it's also equal to the ratio of the products divided by the ratio of the reactants coefficients in this case is one the coefficients become exponents in the equilibrium expression now let's work on some practice problems number one write the equilibrium expression kc and kp for the following reaction now there's two types of equilibrium constants you need to be familiar with kc and kp kc is for concentration kp is associated with partial pressure so this is the equilibrium concentration constant that's the equilibrium partial pressure constant now to do this you need to be familiar with the law of mass action so let's say we have a plus b reacting to produce c and d and let's say the coefficients are j k l m to write the equilibrium expression when you see k or k eq it's typically associated with k c so keq is going to be the concentration of c times the concentration of d divided by the concentration of a times the concentration of b is basically the concentration of the products divided by the concentration of the reactants now the coefficients become exponents in the equilibrium expression so that's the equilibrium expression for this reaction using the law of mass action now let's do the same thing for kc and kp so kc is going to be equal to the concentration of the products divided by the concentration of the reactants and then the coefficients will become exponents so there's a 2 in front of nh3 so we're going to put that on the exponent of nh3 and this is going to be to the first power and for h2 we have a 3. so that's how we can write the equilibrium expression for kc the square brackets indicates concentration now to write the equilibrium partial pressure constant kp it's going to be the partial pressure of the products in this case nh3 and that's going to be raised to the second power due to uh that coefficient divided by the partial pressure of n2 that's gonna be raised to the first power and then times the partial pressure of h2 which will be raised to the third power so that's how we can write the equilibrium expression for kp based on this reaction number two nitrogen reacts with chlorine to produce a nitrogen trichloride at equilibrium the concentrations of each gas were found to be 0.15.25 and 0.50 molar calculate the value of the equilibrium constant kc by the way if you see problems that ask you to calculate k or k eq this typically refers to k c unless specified otherwise always go for kc if it specifically asks for a kp or the equilibrium partial pressure constant then you use kp if you see k or k eq it typically corresponds to kc so just keep that in mind so before we can calculate kc we need to write a balanced reaction nitrogen is a diatomic gas chlorine is also a diatomic gas and this is going to become nitrogen trichloride so ncl3 now we need to balance the chemical equation we have two nitrogen atoms on the left we need to put a two in front of ncl3 so we have two nitrogen atoms on both sides now we have a total of six chlorine atoms on the right side if we divide that by two that tells us that we need to put a three in front of cl2 so now we have a balanced chemical equation so let's write the expression for kc it's going to be the concentration of the products divided by the concentration of the reactants and the coefficients will become exponents now remember at equilibrium the concentrations of the reactants and the products are no longer changing and since we have the concentrations at equilibrium we can just go ahead and plug it in to get the value for kc so the concentration of ncl3 that's 0.5 the concentration of n2 is 0.15 and the concentration of cl2 is 0.25 moles per liter and that needs to be cubed so when you're plugging this in you may want to put an extra set of parentheses in the denominator otherwise your calculator may divide by this number but may multiply by .25 to the third power so i recommend enclosing the entire thing in parenthesis so let's go ahead and get the answer so kc is going to be 106.6 repeated so that's the value of the equilibrium constant kc in this example problem number three sulfur dioxide reacts with oxygen gas to produce sulfur trioxide at equilibrium the partial pressures of each gas were found to be 0.10 atm 0.30 and 0.45 for so2o2 and so3 respectively calculate the equilibrium constant kp well let's begin by writing a balanced chemical equation so we have sulfur dioxide so2 it's going to react with oxygen gas o2 and it's going to produce sulfur trioxide so3 so all of these are in the gas phase so to balance it right now we have equal number of sulfur atoms but we're going to have to multiply it by two here we have four oxygen atoms plus two that's six and two times three is six so we have six oxygen atoms on both sides let's put a one in front of o2 so now the chemical equation is balanced now let's write an expression for kp so kp is going to be the partial pressures of the products over the reactants so we're going to have the partial pressure of so3 and this is going to be raised to the second power divided by the partial pressure of so2 raised to the second power times the partial pressure of o2 raised to the first power so the partial pressure for so3 that's going to be point 45 atm now the partial pressure for so2 that's 0.10 atm and for o3 it's 0.30 so let's go ahead and plug this in and let's put this in one set of brackets or parentheses so it's .45 squared divided by 0.10 squared and then you could take that result divided by 0.30 so you should get 67.5 so that's the value of kp in this example problem number four if kc for the reaction shown below is 9.6 at 300 celsius what is the value of kp at this temperature now this is a formula that we could use to go from kp to kc or vice versa but let's talk about how to get that formula so let's say we have a generic reaction a plus b turns into c plus d and we're going to have our generic coefficients j k l and m let's write the equilibrium partial pressure expression for this reaction so kp is going to be the partial pressure of the products so the partial pressure of c and that's going to be raised to the l power times the partial pressure of d raised to the m power and that's going to be divided by the partial pressure of a raised to the j power and then times the partial pressure of b raised to the k power now based on the ideal gas law equation we know that pv is equal to nrt so if we divide both sides by the volume v we get p is equal to nrt over v but instead of dividing it that way we can write it like this n over v because concentration is moles divided by volume so we can replace n over v with c so we get the pressure is equal to the concentration times rt so i'm just going to rewrite that here so now what we're going to do is we're going to replace p with crt in this expression so the partial pressure of c we can replace that with the concentration of c times rt and it's going to be raised to the same exponent l so let's do the same for the rest so the partial pressure of d is going to be the concentration of d times rt the partial pressure of a is going to be the concentration of a times rt and the same is true for the partial pressure b that's going to be the concentration of b times r t and let's put the exponents m j and k so now what we're going to do at this point is we're going to separate the concentration from rt the concentration of c we can simply put it in brackets it will be just c but let me rewrite this over here so keep in mind this whole thing is still equal to kp we can delete this as well actually let's keep that for now so this is going to be the concentration of c and then times rt this is going to be the concentration of d times rt now keep in mind this is raised to the l so this is the concentration of c raised to the l and the concentration of r t raised to l and then this is the concentration of d raised to the m times the concentration of r t raised to m you have to distribute the m to both d and r t now the concentration of a we can simply just put that in brackets as well and then times rt and those two will be raised to the j power and then we have the concentration of b which is raised to the k power times rt which is also the raised to the k power and this is still equal to k p now what i'm going to do is i'm going to break this fraction into two smaller fractions in one fraction i'm going to collect all of the concentration terms and then in the other fraction all of the rt terms so here we have the concentration of c times the concentration of d divided by the concentration of a and b and then let's put the exponent so this is l m j and k now let's multiply these two if we were to multiply two terms with the same base we can add the exponents x squared times x cubed this is going to be x raised to the 2 plus 3 which is x to the fifth power so if we multiply these two we can add the exponents l and m together so this is going to be times so we're going to have rt raised to the l plus m and then we could do the same thing with those two we can multiply them and then we can add these two exponents together j and k so we're gonna have r t raised to the j plus k so all of this is still equal to kp so we could say that kp is equal now this this is the equilibrium expression for kc products over reactants but in terms of concentration so we can replace this whole thing with kc so we have kp equal kc now that part what happens when we divide by if we divide two common bases let's say if we have x to the seven over x to the fourth we can write that as a single x variable by subtracting the exponents seven minus four is 3. so we can write this as a single rt value by subtracting l plus m and j plus k so we're going to get times rt raised to the l plus m minus j plus k and this is in parentheses now we can define delta n as being the sum of the coefficient of the products minus the sum of the coefficients of the reactants so j and k represents the coefficients of the reactants l and m represents the coefficients of the products so delta n is basically the sum of the coefficients of the products minus reactants so now we have our equation that relates kp and kc so this is the equation that you want to write down in your notes kp is equal to kc times rt delta n and remember delta n is the sum of the coefficient of the products let's say cp for coefficient of the products minus the sum of the coefficient of the reactants or you can write it this way if you want the important thing is that you know how to apply it in a problem so now in part a we're given kc and we want to find the value of kp so kp is going to equal kc which is 9.60 times r now there's two r values there's the r value that is associated with energy 8.3145 the units of which are joules per mole per kelvin and then there's this r value 0.08206 liters times atm divided by moles times kelvin now kp which is described in terms of partial pressure it's usually in units of atm therefore you want to use this r value which contains the unit atm so it's going to be .08 206 and the temperature has to be in kelvin so 300 celsius to convert that to kelvin we need to add 273 to it so it's going to be times 573 kelvin and then raised to the delta n so let's calculate delta n for this reaction so delta n is going to be the coefficient of the products which is two minus the sum of the coefficients for the reactants so that's one plus three so delta n is two minus four which is negative 2 so we're going to raise this to the negative 2 power and that's going to give us kp so let's go ahead and plug that in so this is going to be 4.34 times 10 to the negative 3. so that's the value of kp for part a now let's calculate the value of kc so how can we do that to get kc kc is going to be kp divided by rt raised to the delta n so that's the formula we need to get kc in part b we're told that the value of kp is 1.45 times 10 to the negative three and we're given the temperature as well the temperature this time is 500 celsius r is still going to be the same it's 0.08 206 and the temperature is going to be 500 celsius plus 273 which is 773 kelvin delta n is not going to change because the coefficients for this reaction remain the same so delta n is still negative two thus k c for this reaction is 5.83 so now you know how to calculate kp given kc and you also know how to calculate kc given kp number five given the value of k for the reaction shown below what is the value of k for the adjusted reaction or more specifically k prime so notice that we doubled the coefficients of the first reaction how is that going to affect the value of k when you double the reaction the new value of k is going to be the square of the original value of k so k prime is going to be k squared thus that's going to be a hundred squared which is ten thousand so that's going to be the new value of k for adjusts the reaction now if you're wondering why k prime is equal to k squared here's the reason so let's get the equilibrium expression for the adjusted reaction so that's associated with k prime and it's going to be products over reactants so we have no2 on top no and o2 on the bottom and then the coefficients become exponents so what we're going to do is we're going to factor out a 2 from every exponent if we do that if we factor out a 2 from four this becomes two this also becomes two and if we factor out two from two that becomes one now this whole thing is going to be raised to the second power since we factored out a 2. now notice what we have in here inside the parentheses is the equilibrium expression for the original reaction we can see the exponents 2 2 1 correlates to the exponents that we see here in that reaction so this is equal to the original k value so we have k prime we're going to replace this with k so this equals k and then we have a 2 outside of that so that's how we can show why k prime is equal to k squared now let's try another example so what's going to be the new k value if we divide the original reaction by one-half so let's say the coefficients are one one-half and one what's going to be the new k value so when we doubled the coefficients it was k squared if we multiply the coefficients by half this is going to be k raised to the one half and you can prove it with the same process so this is going to become the square root of 100 which is 10. so that's going to be the k value for this particular reaction now what's going to be the new k value let's say if we reverse the reaction what's going to happen then if we reverse the reaction the new k value is going to be 1 over k so this is going to be 1 over 100 which is point zero one now if you wanna prove that simply write the expression for k prime so it's no to the second power times o two to the first power divided by no2 squared now this whole thing since it equals k prime we want to adjust that expression k prime is equivalent to one over one over k prime and if you're not sure about that here's how you can show it let's put this over one if we multiply the top and bottom by one over k prime k prime times one over k prime they cancel we get one and then just multiply those together you get one over k prime so based on that this expression is equivalent to one over one of that expression now how can we show that this is 1 over k let's multiply this part of the fraction by no squared that is the bottom of this fraction highlighted in blue and we're going to multiply the top of that fraction by l squared so if you multiply a fraction by n squared over n squared you're multiplying it by one you're not changing the value of that fraction here n l squared will cancel and so what we're going to have left over is k prime is equal to 1 and then that's this line in white divided by no squared and i'm going to draw the fraction line that's highlighted in blue which is this line over no squared times o2 now this expression here where no squared is on top and on the bottom we have i mean no2 squared is on top and on the bottom we have no square times o2 the reactant side this is the equilibrium expression for the original k so we can replace this with k so that's how we could show that k prime is one over k when you reverse the reaction so now that we have the value for k prime for this one let's try a new one so let's say we have this reaction but the coefficients are six six and three what's the k value for this reaction so if we compare it to the original expression the reaction has been reversed so we know k prime is going to be 1 over k and if we compare the coefficients we've multiplied the reaction by three whatever you multiply the reaction by you need to raise k to that exponent since we multiplied it by three we're going to raise it to the third power so k prime is going to be 1 over k cubed so that's 1 over 100 to the third power which is 100 times 100 times 100 that's one with six zeros so that's a million a million is one times ten to the six since we have six zeros if we move ten to the six to the top the exponent changes from positive six to negative six so this becomes one times 10 to the negative 6. so that's the new k value that corresponds to this reaction so now you know how to calculate k when the reaction is adjusted either you reverse it multiply by 2 or multiply by half and so forth 20 moles of nocl is placed inside an empty 4 liter container at equilibrium 8 moles of cl2 was found to be in the container calculate the value of kc for this reaction in order to calculate kc we need to find the equilibrium concentrations for every species in this reactant we don't have that what we need to do is write an ice table so at the beginning the initial value of nocl can be calculated notice that we place 20 moles in the 4 liter container so to calculate the concentration it's moles divided by liters so we have 20 moles of solute divided by 4 liters so that's going to be five moles per liter that's the initial concentration of nocl now the container is empty so we don't have anything for no or cl2 so we're going to place a zero there now because the products are equal to zero the reaction has to shift towards the right it cannot shift to the left and so cl2 is going to increase by some amount 1x now notice the mole ratio between no and cl2 it's two to one so if cl2 increases by x and o is going to increase by 2x and nocl is going to decrease by 2x as the reaction shifts to the right the products will increase in value but the reactants will decrease in value so if we add the first two rows this is going to be x 2x and 5 minus 2x now the concentration of cl2 at equilibrium is x and we also know that at equilibrium 8 moles of cl2 was found to be in the container so the concentration of cl2 at equilibrium is going to be 8 moles divided by the 4 liter container so it's 2 moles per liter so now we know the value of x x is equal to 2m because they're both equal to cl2 so now we could find the equilibrium concentrations for everything else for no it's equal to 2x and since x is 2 that's going to be 2 times 2 so no is equal to 4m now we can also calculate the equilibrium concentration for nocl which is equal to five minus two x so that's gonna be five minus two times two so that's five minus four so it's one mole per liter at equilibrium so now that we have the equilibrium concentrations for every reactant and product in this example we can now write the expression for kc so products over reactants on a product side we have no and cl2 and for the reactant side we only have nocl the coefficient for no is two the coefficient for cl2 is one and for nocl that's two so now let's plug in the values so for no that's gonna be four for cl2 it's 2 and for nocl it's 1. so 4 squared is 16 and 1 squared is 1 16 times 2 is 32 so kc in this example is 32 and that's the answer carbon monoxide reacts with oxygen gas to produce carbon dioxide at equilibrium the concentrations of o2 and co2 are 0.10 at 0.75 moles per liter respectively so we're given kc what is the concentration of co at equilibrium the first thing we need to do is write a balanced chemical equation so we have carbon monoxide reacting with oxygen gas and it's going to produce carbon dioxide so to balance in we just need a 2 in front of co and co2 and so we have kc for this reaction 10 to the third is a thousand so this is going to be four thousand now we have the concentrations of o2 and co2 at equilibrium so we don't need an ice table to find the equilibrium concentration of co let's write the expression for kc so it's products over reactants on a product side we have co2 and on the reaction side we have co and o2 the coefficient in front of co2 is a 2 so we need to put that here and for co it's also a two and it's a one for o2 now kc is 4000 the equilibrium concentration of cl2 is 0.75 and the value for co is what we're looking for so let's call that x and the concentration for o2 is point 10. so let's cross multiply so first on the left side we're going gonna have one times point seven five squared which is point five six two five and on the other side we're gonna have four thousand times point one which is 400 multiplied by x squared now we need to divide both sides by 400 so 0.5625 divided by 400 that's point four zero six two five times ten to the minus three so that's equal to x squared so now we gotta take the square root of both sides and so x is .0375 so that's the concentration of carbon monoxide at equilibrium number eight at a certain temperature ammonia partially decomposes into nitrogen gas and hydrogen gas ammonia was added to an empty container until the partial pressure of nh3 reach 0.85 atm at equilibrium the partial pressure of h2 is found to be 0.45 atm what is the partial pressure of nh3 at equilibrium well first we need to write a reaction so ammonia is decomposing into nitrogen gas and hydrogen gas and we've seen this reaction plenty of times so we know how to balance it now we have partial pressures at equilibrium and initially before the reaction begins the 0.85 atm that's the initial partial pressure of nh3 and then we have the partial pressure of h2 at equilibrium whenever you have information at equilibrium and before the reaction begins you need to write out an ice table initial change equilibrium so initially the partial pressure of nh3 is 0.85 we don't have anything else in that container so the partial pressure of the products is zero now our next step is to determine what direction the reaction is going to go is it going to go to the right or to the left now anytime any one of the products is zero the reaction has no choice but to go to the right because if you don't have any products it cannot go to the left otherwise these will become negative but since we don't have any nitrogen or hydrogen gas they can't react together to form ammonia so the reaction has to go to the right as the reaction goes to the right we know that the products will increase in value the reactants will decrease in value so this is going to be positive 1x because it's a product it's going to increase and this is going to be positive 3x let me write that better the one two three ratio is based on the molar ratio that we see here as the reaction shifts to the right the reactants are going to decrease in value so we're going to have a negative sign for the change this is going to be negative 2x so we have negative 2x 1x 3x based on a mole ratio of 213 now let's add the first two rows so at equilibrium the partial pressure of nh3 is going to be 0.85 minus 2x for n2 it's going to be x for h2 it's going to be 3x now that we've completed the ice table what should we do next now looking at the problem we're told that at equilibrium the partial pressure of h2 was found to be 0.45 atm so we can write that at equilibrium we also know that the partial pressure of h2 is equal to 3x therefore we could set point 45 atm equal to 3x and that will help us to solve for x so let's divide both sides by 3. so x is going to be .45 divided by 3 which is point 15. so that's the value of x now that we have that we can calculate the equilibrium partial pressure of any one of the reactants or products so our goal is to calculate the equilibrium partial pressure of nh3 so that's going to be 0.85 minus 2x so let's replace x with point 15. 0.85 times 2 times 0.15 that's point 55 my handwriting is terrible today so this is the final answer for this problem that's the partial pressure of nh3 at equilibrium if we want to find the equilibrium partial pressure of n2 it's simply x which is point 15. and if we want to we can calculate kp for this uh entire process so kp is going to be h2 raised to the third power times n2 raised to the first power divided by nh3 to the second power it's products over reactants so we know the equilibrium partial pressure actually i wrote this wrong this should be the partial pressure of h2 raised to the third power times the partial pressure of n2 to the first power divided by the partial pressure of nh3 to the second power so the partial pressure of h2 that's 0.45 for n2 it's simply x which is 0.15 and for nh3 we calculated to be 55 based on that expression and we need to square it so kp is going to be .0452 so that is the equilibrium partial pressure constant if we want to find it you