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# Changes in free energy and the reaction quotient

## Video transcript

the relationship between the change in free energy and Q the reaction quotient is very important to understand the change in free energy or Delta G is the instantaneous difference in free energy between the reactants and the products q is our reaction quotient it tells us where we are in the reaction and remember it has the same form as the equilibrium constant K Delta g0 is the standard change in free energy so the change in free energy under standard conditions R is the gas constant and T is the temperature in Kelvin remember when Delta G is less than zero so when Delta G is negative the reaction is spontaneous in the forward direction when Delta G is greater than zero so when Delta G is positive the reaction is non spontaneous in the forward direction and when Delta G is equal to zero the reaction is at equilibrium let's apply this equation to an example so here we're trying to synthesize ammonia so our goal is to calculate the change in free energy so we're trying to find Delta G for the following reaction at 25 degrees C given the following set of partial pressures so nitrogen gas plus hydrogen gas gives us ammonia gas and at 25 degrees C the standard change in free energy Delta G zero is equal to negative 33.0 kilojoules and remember for gases it's easier to measure pressures than it is to measure concentrations so at this instant in time the partial pressures for all these gases are one atmosphere so one atmosphere for nitrogen one atmosphere for hydrogen one atmosphere for ammonia so at this instant in time calculate the change in free energy so we're trying to find Delta G so let's write down our equation Delta G is equal to Delta G zero plus R T natural log of Q so we need to find Q we need to find our reaction quotients remember it has the same form as the equilibrium constant for our equilibrium constant we had the concentration of products over the concentration of reactants and we raised the concentration to the power of the coefficient but here we're dealing with partial pressures right instead of concentration so I'm gonna write Q sub P here to remind ourselves that we're working with partial pressures so that would be the partial pressure of our products so the partial pressure of ammonia raised to the power of the coefficient so raised to the second power over the partial pressure of nitrogen so the partial pressure of nitrogen raised to the power of the coefficient in our balanced equation here the coefficient is a 1 so this is raised to the first power times the partial pressure of hydrogen raised to the power of the coefficient which is 3 so raised to the third power so everything is at one atmosphere at this moment in time so this would be one point zero squared over one point zero to the first power times one point zero to the third power and all that of course is equal to one so at this moment in time our reaction quotient Q is equal to one so to solve for the change in free energy so define Delta G we take Q and we plug it into our equation so what is the natural log of one that of course is equal to zero so this this goes through this is equal to zero and we find that the change in free energy Delta G is equal to the standard change in free energy Delta G zero which at this temperature is equal to negative 33.0 kilojoules so the change in free energy Delta G is equal to negative 33.0 kilojoules all right let's talk about a few things here first we got so we got a negative value for Delta G we got negative 33.0 so our reaction is spontaneous in the forward direction so we're going to make more of our products we're going to make more ammonia here since Delta G is negative it's our driving force to make more of our products notice that Delta G in this example Delta G is equal to Delta g0 and that makes perfect sense because delta g0 is the standard change in free energy it's the change in free energy under standard conditions which is what we have here we're under standard conditions everything is at one atmosphere so that makes sense and finally you could write if you wanted to you could write kilojoules per mole of reaction indicating for for this reaction how it's written right here that is the change in free energy so when one mole of nitrogen combines with three moles of hydrogen to give us two moles of ammonia Delta G the change in free energy for that reaction is negative 33.0 kilojoules now let's get a new set of partial pressure so now let's say all the gases have a partial pressure of 4.0 atmospheres and when all the gases have that partial pressure what is the change in free energy at that moment for our reaction at 25 degrees C so we're trying to find the change in free energy we're trying to find Delta G and we know Delta G is equal to Delta G 0 plus RT natural log of Q so we need to find Q the reaction quotient so this is QP and from the previous example we know how to find Q so I'll go through this a little bit faster this would be 4.0 to the second power so 4.0 squared over 4 point 0 to the first power so 4.0 to the first times 4.0 to the third so 4.0 to the third 4.0 squared is 16 over 4 times at C 4 cubed is 64 so what is Q equal to we get out the calculator 16 divided by 4 times 64 is equal to point zero 6 to 5 so at that moment the reaction quotient is equal to point zero 6 to 5 at the moment when all of our partial pressures are 4 atmospheres so this is our reaction quotient we're going to plug this into our equation we're going to solve for Delta G so Delta G is equal to Delta G 0 the standard change in free energy at 25 degrees C Delta G 0 is negative 33 kilojoules so we have negative 33.0 kilojoules we need to make that joules so I'm going to say times 10 to the third joules plus R is the gas constant so that's 8.314 so 8.314 and that's joules over mole times K since the gas constant is in joules per mole we need to make Delta G 0 in joules per mole - next we have the temperature so the temperature of our reaction is 25 degrees C that needs to be in Kelvin so 25 degrees C is 298 Kelvin so we write in here 298 Kelvin so Kelvin would cancel out here and then we're going to multiply that by the natural log of Q the natural log which where Q is equal to point zero six - five point zero six two five let's do the math so let's start with the natural log of point zero six two five so the natural log point zero six two five is equal to that value which we need to multiply by 298 and multiplied by eight point three one four we're going to add that to Delta G zero so plus negative 33.0 times ten to the third and that should give us our change in free energy which is equal to negative thirty nine point nine kilojoules per mole so I'll round that so Delta G Delta G is equal to negative 39 point nine kilojoules per mole so it's just easier to think in kilojoules remember this is moles of our reaction so for this specific reaction where one mole of nitrogen combines with three moles of hydrogen to give us two moles of ammonia so how it's written Delta G for the reaction how it's written is equal to negative 39.9 kilojoules so Delta G is negative and so we know our reaction is spontaneous in the forward direction so the reaction is spontaneous in the forward direction which means we're going to make more of our products let's think about how far away we are from equilibrium all right so Q our reaction quotient is point zero six to five for this reaction at 25 degrees C the equilibrium constant which would be KP is equal to six point one times 10 to the 5th our value for the reaction quotient was point zero six to five that's much much smaller that's much smaller than the equilibrium constant so Q is much smaller than K so we know we have too many reactants and too few products we have a driving force Delta G is negative we have a driving force to make more of our products so the reaction moves forward to make more products here so we know that by just looking at the sign for Delta G which we know is spontaneous we also know that from comparing the reaction quotient Q to K so because Q is not equal to K we're not at equilibrium here we are away from equilibrium and we have a spontaneous reaction all right so we're going to make more of our products so the reaction the reaction proceeds to the right we make more of overall I should say we make more of our products and we lose some of our reactants what happens to Q so as the reaction progresses we make more of our products so the numerator should increase we lose some of our reactants so the reactants should decrease and so if you think about that an increase in the numerator a decrease in the denominator that means that as the reaction progresses as reaction moves to the right to make more products q should increase so as q increases what happens to Delta G so let's think about that next so if we get an increase in Q what happens to the change in free energy well let's just make up a number here let's say let's say that Q is now equal to 100 so at a different moment in time here Q is equal to point zero six to five let's increase that to Q is equal to 100 so if we increase Q what happens what happens to Delta G let's just plug the numbers into the equation and let's see for ourselves Delta G is equal to negative 33.0 times 10 to the third that's joules per mole actually let's just leave out units here to make it a little bit to give us a little bit more room all right plus that's Delta G zero plus eight point three one for our gas constant times our temperature or still at 298 K so the only difference now is we're substituting 100 in for Q so we've increased the value for Q what happens to Delta G let's do the math so we need to find the natural log of 100 and we multiply that by 298 and it'll try that by 8.314 8.314 and then we add that to Delta G zero Delta G zero is negative 33.0 times 10 to the third so we get in kilojoules per mole that would be that would be negative negative twenty one point six kilojoules per mole so Delta G is equal to negative twenty one point six kilojoules per mole of reaction so we increased Q and we increase Q what happened to Delta G we went from a Delta G of negative 39.92 a Delta G of negative twenty-one point six so we're getting closer to zero as the reaction proceeds to the right we're still spontaneous we still have a negative value for Delta G so we still have a driving force to make more products q is still less than K right Q is 100 and K is 6.1 times 10 to the fifth so we're still going to go to the right and make more products our reaction is still spontaneous what happens at equilibrium alright so what happens at equilibrium we know already that Delta G should be equal to zero at equilibrium so at equilibrium Delta G should be equal to 0 and Q should be equal to K the reaction quotient is equal to the equilibrium constant let's plug that into our equation and see if that's true so we have Delta G is equal to Delta G 0 negative 33.0 times 10 to the third plus the gas constant is 8.314 our temperature is still 298 but instead of writing instead of writing Q we're going to plug in K we're going to plug in the equilibrium constant which already gave to you up here is six point one times 10 to the fifth so now we're doing the natural log of K so we plug in the natural log of K which is six point one times 10 to the 5th and let's see what happens to Delta G so now we're going to find the natural log of six point one times 10 to the 5th and we need to multiply that by 298 and then multiply that by 8.314 so we get we get positive thirty-three kilojoules alright so positive 33 kilojoules or thirty three point zero times ten to the third so notice what happens right over here we have Delta G zero is negative 33.0 times 10 to the third and all of this on the right gives us positive thirty three point zero times ten to the third so that of course is equal to zero so the change in free energy right is now zero and we are at equilibrium right Q is equal to K we plugged into our equation and we found that Delta G is indeed equal to zero so there's no more driving force to make more of our products right there's no more driving force to make more of our products because we are at equilibrium the free energies of the reactants and the products are equal