# Pressure dependence of the equilibrium constant for an ideal gas

• ussername
In summary, the conversation discusses the equilibrium constant for an ideal gas and how it is not a function of pressure. However, this claim is not generally true and it is shown that the partial derivative of lnK with respect to pressure, at constant temperature and moles, is equal to the reaction volume. This is derived by considering the standard state of an ideal gas at the same temperature and pressure as the reaction, and the reaction volume is then calculated. This shows that the partial derivative is not exactly zero, but it is very small. The conversation also addresses how this concept may have been misunderstood in the original question and clarifies that the equilibrium constant is not a function of the number of moles, but only of pressure and temperature.
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.

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.

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

ussername said:
How can anybody claim that (∂lnK/∂p)T,⃗n=0(∂ln⁡K/∂p)T,n→=0(\partial \ln K / \partial p)_{T,\vec{n}}=0 for any reaction in ideal gas?
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.

DrDu
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.

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.

DrDu and Chestermiller
ussername said:
I read in some scripts that equilibrium constant for an ideal gas is not a function of pressure:
View attachment 207509
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.
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.

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.)

mjc123 said:
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.)
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.

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.

jim mcnamara and Lord Jestocost

## 1. What is the pressure dependence of the equilibrium constant for an ideal gas?

The pressure dependence of the equilibrium constant for an ideal gas is a measure of how the equilibrium constant changes when the pressure of the gas is altered. This relationship is described by the ideal gas law, which states that the pressure of a gas is directly proportional to its concentration.

## 2. How does the pressure affect the equilibrium constant for an ideal gas?

The pressure of a gas has a direct effect on the equilibrium constant for an ideal gas. As the pressure increases, the concentration of the gas also increases, leading to a higher equilibrium constant. Conversely, when the pressure decreases, the concentration and equilibrium constant also decrease.

## 3. What is the relationship between pressure and equilibrium constant for an ideal gas?

The relationship between pressure and equilibrium constant for an ideal gas is described by the ideal gas law. This relationship shows that as the pressure of a gas increases, the equilibrium constant also increases, and vice versa.

## 4. How do changes in temperature affect the pressure dependence of the equilibrium constant for an ideal gas?

Changes in temperature can also affect the pressure dependence of the equilibrium constant for an ideal gas. As temperature increases, the pressure of the gas also increases, leading to a higher equilibrium constant. However, the effect of temperature on pressure dependence may vary depending on the specific gas and reaction being studied.

## 5. Is the pressure dependence of the equilibrium constant for an ideal gas always the same for all gases?

No, the pressure dependence of the equilibrium constant for an ideal gas can vary for different gases. This is because the ideal gas law is based on the assumption that the gas molecules behave ideally, which may not be the case for all gases. Additionally, the equilibrium constant for a specific gas may also be affected by other factors such as temperature and the specific reaction being studied.

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