# Pressure dependence of the equilibrium constant for an ideal gas

1. Jul 20, 2017

### ussername

I read in some scripts that equilibrium constant for an ideal gas is not a function of pressure:

But that is not generally true!

Since:
$$\left (\frac{\partial \Delta_{R} G}{\partial p} \right )_{T,\vec{n}}=\Delta_{R} V$$
and
$$\Delta_{R} G^{0}=-RT\cdot \ln K$$
it should be:
$$\left (\frac{\partial \ln K}{\partial p} \right )_{T,\vec{n}}=-\frac{\Delta_{R} V^{0}}{RT}$$
If the standard state is ideal gas with the same temperature $T$ and pressure $p$, it is:
$$V^{0}=\frac{RT}{p}(n_{1}+n_{2}+...+n_{N})$$
and the reaction volume is then:
$$\Delta _{R}V^{0}=\sum \upsilon _{i}\cdot \left (\frac{\partial V^{0}}{\partial n_{i}} \right )_{T,p,n_{j\neq i}}=\frac{RT}{p}\cdot \sum \upsilon _{i}$$
$$\left (\frac{\partial \ln K}{\partial p} \right )_{T,\vec{n}}=- \frac{\sum \upsilon _{i}}{p}$$

How can anybody claim that $(\partial \ln K / \partial p)_{T,\vec{n}}=0$ for any reaction in ideal gas?

Edit: Usually $p=100000 \, Pa$ thus the derivation is small enough but not principally zero. Maybe that's what the script is saying.

2. Jul 20, 2017

### Staff: Mentor

Could you please tell us what exact text (like who wrote it, name of text, etc) you copied the image from, and where (looks like chapter 3 maybe 25th formula).
Context helps us old fogies figure out what is going on. Thanks.

If we knew what the script was, someone could probably agree with that statement. I do not know enough to comment reasonably.

3. Jul 20, 2017

### Lord Jestocost

4. Jul 20, 2017

### mjc123

Maybe because it's experimentally true?
Your fallacy seems to be in assuming
dlnK/dp = -1/RT*dΔG0/dp
But ΔG0 does not vary with the pressure of the system. It is a constant for a standard state of defined pressure and temperature, not necessarily equal to the actual pressure of the system. Of course if you change the definition of the standard state pressure, ΔG0 and K will change, but that is a different thing. For example, for the reaction
A ↔ 2B
K = (PB/P0)2/(PA/P0)
If pressures are measured in atm and the standard state is P0 = 1 atm, then numerically K = PB2/PA
If instead you use a standard state of P0 = 2 atm, then K' = PB2/2PA = K/2
But if you consistently use the same standard state , say P0 = 1 atm, then PB2/PA will always equal the same value of K (at the same temperature), irrespective of the actual pressure of the system.

5. Jul 23, 2017

### ussername

You are right, for standard state with $p^0$ it is $G^0 = f(n_1,n_2,...,n_N) \neq f(p)$ and
$$\left( \frac{\partial \Delta_{R}G^0}{\partial p} \right)_{T,\vec{n}}=0$$ $$\left( \frac{\partial \ln K}{\partial p} \right)_{T,\vec{n}}=0$$
Just for standard state with $p$ this derivation is not generally zero but it is almost zero as I showed above.

6. Jul 24, 2017

### mjc123

What do you mean by "standard state with p" (as distinct from p0)? ΔG0 is defined for a defined standard pressure, which I have called p0. Are you suggesting a "standard state" with standard pressure = the pressure of the system, whatever that happens to be? That's not a proper standard state; you can't define an equilibrium constant that way.

7. Jul 25, 2017

### DrDu

I was a bit puzzled at the beginning, but I am convinced now that you are right. mjc123 is right in that $K_p$ does not change with temperature when you hold both the standard P constant and the partial pressure of the components. But this is not the point, as we are deriving with respect to P not keeping the partial pressures constant, but the number of moles n, or, what amounts to the same, the concentrations. Now $\ln K_p=\ln K_c+\ln (P/P_0) \sum_i \nu_i$, from which your claim follows.

8. Jul 25, 2017

### mjc123

I don't understand you. We're talking about varying pressure, not temperature. Kp does not vary with P because ΔG0 does not vary with P, being defined for a particular standard pressure. Kp is constant, whatever the partial pressures of the components - that is the point of an equilibrium constant.
Can you clarify what is the P you are varying here? At constant temperature, you can't vary P and keep the concentrations constant. (Nor the number of moles, if the reaction involves a change in the number of moles; changing the pressure will shift the position of equilibrium - but not the equilibrium constant.)

9. Jul 25, 2017

### DrDu

You are obviously right.
Already the original question is wrong:
"How can anybody claim that [...]"
as Kp is not a function of $\vec{n}$ at all, but only of P and T.
Maybe what I and the OP had in mind is the quotient $Q= \prod_i (P_i/P_0)^{\nu_i}$, which is a function of P and $\vec{n}$. In equilibrium $Q=K_p$, but when P is changed with n fixed, Q will be out of equilibrium.

10. Jul 25, 2017

### Staff: Mentor

I agree with mjc123. $\Delta G^0(T)$ is defined as the change in free energy between the following two thermodynamic equilibrium states:

1. Pure reactants in stoichiometric proportions in separate containers at temperature T and pressure 1 bar

2. Pure products in corresponding stoichiometric proportions in separate containers at temperature T and 1 bar

So $\Delta G^0(T)$ is a function only of T (and not pressure). Since $RTln{K_p}=-\Delta G^0(T)$, $K_p(T)$ is a function only of temperature.