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## Chemistry library

### Course: Chemistry library > Unit 15

Lesson 4: Gibbs free energy- Gibbs free energy and spontaneity
- Gibbs free energy and spontaneity
- Gibbs free energy example
- More rigorous Gibbs free energy / spontaneity relationship
- A look at a seductive but wrong Gibbs spontaneity proof
- Changes in free energy and the reaction quotient
- Standard change in free energy and the equilibrium constant
- 2015 AP Chemistry free response 2c

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# Changes in free energy and the reaction quotient

The relationship between Gibbs free energy, standard Gibbs free energy, and the reaction quotient Q. Predicting the direction of the reaction when we don't have standard conditions.

## Want to join the conversation?

- I have been keen to watch all the videos in the suggested order (Chemistry playlist). However, to understand this particular video, it seems that there is a missing one that explains why deltaG = deltaG0 + RTln(Q) ! Any suggestion where can I learn about this relationship between Gibbs energy and the reaction quotient?(78 votes)
- At 9.35 how did he know that K = 6.1 * 10^5?

How was it calculated?(36 votes)- The equilibrium constant has been determined experimentally. For this video, the author looked up the number.

This reaction is called the Haber Process.(18 votes)

- The math on the last calculation seems off. When I calculated RTlnK I got 33004.2674. Why did he round this number to (33.0 x 10^3)? When he does that and adds (-33 x 10^3) the equation balances to zero, but why would he round mid-equation? If you don't round that number mid-equation you end up with deltaG = 4.27.(1 vote)
- He didn't have to round off, but he would have gotten the same result.

Your precision is limited to 3 significant figures by the 298 K term (it's really 298.15 K).

So the best value you can get for RTlnQ is 33 000 J = 33.0 × 10³ J.

Δ G = -33.0×10³ J + 33.0×10³ J = 0.0 × 10³ J.

A value of 4.27 J is the same as 0.0 × 10³ J within the limits of uncertainty.(9 votes)

- For this equation for Gibbs Free Energy, dG= dG0 + RTlnQ, or dG=-RTlnK, will it matter what equilibrium constant, Kp or Kc, is used for gases? My reasoning is that Kp=Kc(RT)^n, so whichever K value is used, different Free Energy values would be calculated, right?(5 votes)
- Lets say we have an ideal gas A , then (Pa)V= nRT <-> n/V = (Pa)/RT = [A] , and the same for another gas B and C .

Supose we have this reaction x A + y B ---> z C

Kc = [A]^x . [B]^y / [C]^z = (Pa/RT)^x . (Pb/RT)^y / (Pc/RT)^z

Kc = (Pa^x) . (Pb^y) / (Pc)^z . (1/RT)^x . (1/^RT)^y . (1/RT)^-z

Kc = Kp . (1/RT)^(x+y-z)

I think in this way.(0 votes)

- 9:30How do you figure that kp out?(3 votes)
- If Q<K, does this ALWAYS mean spontaneous reaction?

Also what's the difference between the reaction quotient (Q) and the ion activity product (IAP)?(2 votes)- Yes, and if Q>K then the reaction is not spontaneous (Delta G = + ). In fact, the reaction may head in the reverse direction to equilibrate(2 votes)

- Why is 8.314 used for R? Isn't the pressure unit atm? Shouldn't it be 0.082?(1 vote)
- The atm numbers are all that really matter here rather than the units. R = 8.314 J K^-1 mol^-1 will show up much more often in physical chem.(3 votes)

- Is the atm standard measuring unit for this type of calculation?

Shouldn't he be dealing with Pa when calculating Q?(1 vote)- The SI unit is Pa, but Atm is also a common unit.(3 votes)

- How did you know the value of kp in the second question (written in red marker)(2 votes)
- If delta G is greater than 0, the reaction is non-spontaneous in the forward direction, but is it spontaneous in the reverse reaction?(1 vote)

## Video transcript

- [Voiceover] 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 G zero 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 nonspontaneous
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. 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 hydrogren
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 gasses it's
easier to measure pressures than it is to measure concentrations so at this instant in
time the partial pressures for all these gasses are one atmosphere, so one atmosphere for nitrogen, one atmosphere for hydrogen, one atmosphere for ammonia. So at this instant in time calculatethe change in free energy. We're trying to find delta G so let's write down our equation. Delta G is equal to delta G zero plus RT natural log of Q. So we need to find Q, we need
to find our reaction quotient. 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 instead of concentrations. So I'm gonna right Q sub
P here to remind ourselves that we're working with partial pressures. So that would be the partial
pressure of our product, 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, and our balanced equation
here, the coeffecient is a one so this is raised to the first power times the partial pressure hydrogen raised to the power of the
coefficient, which is three, so raised to the third power. So everything is at one
atmosphere at this moment in time. So this would be 1.0 squared over 1.0 to the first power times 1.0 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, to find delta G, we take Q and we plug
it into our equation. What is the natural log of one? That of course is equal to zero. So 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. Let's talk about a few things here. First, 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 gonna make more of our products. We're gonna 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 G zero, and that makes perfect
sense because delta G zero 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 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 pressures. Now let's say all the gasses have a partial pressure of 4.0 atmospheres, and when all the gasses
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 zero plus RT natural log of Q. So we need to find Q,
the reaction quotient. 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
4.0 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
four times-- Let's see. Four cubed is 64, so what is Q equal to? We'll get out the calculator. 16 divided by four times 64 is equal to .0625. So at that moment the reaction
quotient is equal to .0625, at the moment when all of our partial pressures are four atmospheres. So this is our reaction quotient. We're gonna plug this into our equation and we're gonna solve for delta G. So delta G is equal to delta G zero, the standard change in free energy. At 25 degrees C delta G zero
is negative 33 kilojoules. So we have negative 33.0 kilojoules. We need to make that joules so I'm gonna say times 10
to the third joules. Plus R is the gas
constant, so that's 8.314. 8.314, and that's joules over mole times K. Since the gas constant
is in joules per mole we need to make the delta G
zero in joules per mole too. Next we have the temperature. 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. Then we're going to multiply
that by the natural log of Q. The natural log where Q is equal to .0625. .0625. Let's do the math. Let's start with the natural log of .0625. The natural log of .0625 is equal to that value, which we
need to multiply by 298 and multiply by 8.314. We're gonna add that to delta G zero, so plus negative 33.0 times 10 to the third. And that should give us
our change in free energy, which is equal to negative
39.9 kilojoules per mole, so I'll round that. So delta G is equal to negative 39.9 kilojoules per mole. It's just easier to think in kilojoules. And 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. Delta G is negative so
we know our reaction is spontaneous in the forward direction. 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. Q, our reaction quotient, is .0625. For this reaction at 25 degrees C the equilibrium constant,
which would be KP, is equal to 6.1 times 10 to the fifth. Our value for the reaction
quotient was .0625. 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. So we're gonna make more of our products. The reaction proceeds to the right, we make of-- overall, I should say. We make more of our products and we lose some of our reactants. What happens to Q? As the reaction progresses
we make more of our products so the numerator should increase, and we lose some of our reactants so the reactants should decrease. And so you can think about that, an increase in the numerator,
a decrease in the denominator, that means that as the
reaction progresses, as the reaction moves to the
right to make more products, Q should increase. So as Q increases what happens to delta G? Let's think about that next. If we get an increase in Q what happens to the change in free energy? Let's just make up a number here. Let's say that Q is now equal to 100. So we're at a different moment in time. Here Q was equal to .0625. Let's increase that to Q is equal to 100. So if we increase Q
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 give us a little bit more room. That's delta G zero, plus
8.314, our gas constant, times our temperature,
we're still at 298K. So the only difference now is we're substituting 100 in for Q. So we've increased the value for the Q. What happens to delta
G? Let's do the math. We need to find the natural log of 100 and we multiply that by 298 and we multiply that by 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 negative
21.6 kilojoules per mole. So delta G is equal to negative 21.6 kilojoules per mole of reaction. So we increased Q, we increased
Q, what happened to delta G? We went from a delta G of negative 39.9 to a delta G of negative 21.6. 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. Q is 100 and K is 6.1
times 10 to the fifth. So we're still gonna go to the
right and make more products, our reaction is still spontaneous. What happens at equilibrium? What happens at
equilibrium? We know already that delta G should be equal
to zero at equilibrium. So equilibrium, delta G
should be equal to zero. 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. We have delta G is equal to delta G zero, negative 33.0 times 10 to the third plus the gas constant is 8.314. Our temperature is still 298. But instead of writing
Q we're gonna plug in K, we're gonna plug in the
equilibrium constant, which I already gave to you up here. It's 6.1 times 10 to the fifth. So now we're doing the natural log of K. We plug in the natural log of K which is 6.1 times 10 to the fifth. And let's see what happens to delta G. Now we're going to find the natural log of 6.1 times 10 to the fifth. And we need to multiply that by 298 and then multiply that by 8.314. So we get positive 33 kilojoules. So positive 33 kilojoules, or
33.0 times 10 to the third. So notice what happens. 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 33.0 times 10 to the third. So that of course is equal to zero. So the change in free energy is now zero. And we are at equilibrium. 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. 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.