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Suppose that $n$ moles of four generic perfect gases are made to react in a container, whose volume can vary, at a constant temperature and pressure, according to the formula $$\ce{A(g) + B(g)<=>C(g) + D(g)}$$ and suppose also the the equilibrium constant $K_c$ is known.

Now, if temperature and pressure are constant, then so will the density of the gases be. But if the density is constant, then the concentration of the gases will be constant as well, regardless of the number of particles present in the container (this is more or less the same argument that my chemistry textbook makes to justify the exclusion of concentration of solids and liquids from the expression of the equilibrium constant).

But if the concentrations are constant, then at the outset of the reaction, when there are $n$ moles of each gas, the reaction is already at equilibrium, because the reaction quotient $Q=K_c$.

My question is the following: is there something wrong about this argument? If not, how will the number of moles of each gas vary with time?

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You have yourself all wrapped around the axle with odd conditions.

Suppose that $n$ moles of four generic perfect gases...

This is a really odd way to specify the initial concentrations of four gases. Say I have 5 moles total. There are four gases, so I have four unknowns and only two equations (total moles and equilibrium equation) which is unsolvable.

This is also ambiguous. It is $n$ moles total, or $n$ moles for each gas?

Now, if temperature and pressure are constant, then so will the density of the gases be. But if the density is constant, then the concentration of the gases will be constant as well, ...

Since you're holding temperature and pressure constant, and given the reaction: $$\ce{A(g) + B(g)<=>C(g) + D(g)}$$ then the density will always be the same. Even if you have all A and B which are totally converted to C and D, the density of the gas mixture will be the same (assuming ideal gas behavior which you stipulated).

If at a constant volume and the reaction had been: $$\ce{2A(g) + B(g)<=>C(g) + D(g)}$$ then the density would change as the equilibrium shifted.

But since you stipulated constant temperature and pressure then the volume would also have to change as $\ce{2A(g) + B(g)}$ was converted to $\ce{C(g) + D(g)}$ to maintain constant pressure, so the overall density of the gas remains constant.

But if the concentrations are constant, then at the outset of the reaction, when there are n moles of each gas, the reaction is already at equilibrium, because the reaction quotient $Q=K_c$.

Let's break this down. If the concentrations are constant, then the reaction is already at equilibrium.

It does not necessarily follow that when there are $n$ moles of each gas then the reaction is at equilibrium. That would only be true if $K_c = 1$, but $K_c$ has not been specified.

If not, how will the number of moles of each gas vary with time?

Ok, let's assume $n$ moles of each of the four gases. You'd also need to provide the rate constants for the forward and reverse reactions to calculate concentrations over time. Without the rate constants it isn't clear if the equilibrium will be established in seconds or years.

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  • $\begingroup$ Density is mass over volume. We do not know that the molar mass of A+B=C+D so hence the mass would change. Now since we do not know what kind of container it is being held in, we cannot assume that the volume would change in a way as to keep the density of the entire container the same. $\endgroup$ – Clangorous Chimera Jan 1 '17 at 22:10
  • $\begingroup$ @ClangorousChimera - you have the formula for density right. Since we don't have real numbers to work with then the actual value of the density is unknown. But given the reaction, the density of the gas mixture won't change. Let's say I have 1 mole of A and 1 mole of B which is wholly converted to 1 mole of C and 1 mole of D. For both the starting case and the products I have two molar volumes. Chemical reactions don't change masses, so the total volume and the total mass are both some unknown constants. $\endgroup$ – MaxW Jan 1 '17 at 22:17
  • $\begingroup$ Ah! You're right I forgot about that. Thanks! I deleted my answer $\endgroup$ – Clangorous Chimera Jan 1 '17 at 22:36
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    $\begingroup$ Sorry, but either you are misunderstanding me or I am misunderstanding you. What I meant to say was that at the outset of the reaction you have n moles of A, n moles of B, n moles of C, and n moles of D. Now, we agree that in the conditions specified density is constant, but since density is directly proportional to concentration, then the concentrations will not change, hence Q will be a constant. But then either 1)Q=K, which entails that the reaction is at equilibrium right in the beginning or 2)Q=/=K, in which case it never reaches equilibrium $\endgroup$ – Nicol Jan 1 '17 at 23:02
  • $\begingroup$ 2) does not seems plausible. Therefore we're left with 1), which is still kind of odd $\endgroup$ – Nicol Jan 1 '17 at 23:10

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