# Effect on mass when decreasing pressure in equilibrium reaction

For example, if we have the reaction:

$$\ce{A + B <=> C}$$

And the volume is decreased, the pressure would increase. Hence, according to Le chatelier's principle, the system will partially oppose the change by favoring the forward reaction to produce more $$\ce{C}$$. I have a few questions:

1. Would the mass of $$\ce{C}$$ increase?

I think yes, since the forward reaction is favored

1. Would the concentration of $$\ce{[A]}$$ be greater when compared to the original concentration of $$\ce{[A]}$$?

I think yes, since the volume is decreased.

1. Would the amount of $$\ce{A}$$ be reduced?

I think yes, since it's being consumed for the forward reaction.

• Is this in gaseous or aqueous medium? If aqueous, then using pressure wouldn't make much difference since water is practically incompressible. If gaseous, there is no concept of concentration only partial pressures. – Safdar Faisal Jul 26 '20 at 7:45
• You don't mean 'mass' but the amount of C otherwise you are correct. Point 2 will be true initially but may not be as the reaction reaches equilibrium. – porphyrin Jul 26 '20 at 8:39

Assuming $$\ce{A, B,}$$ and $$\ce{C}$$ are gases and behave like ideal gases we can conclude following by applying $$P_xV=n_xRT$$ where $$x$$ is either $$\ce{A, B,}$$ or $$\ce{C}$$, and $$V$$ and $$T$$ are constants:

$$[\ce{A}] = \frac{n_A}{V}=\frac{P_A}{RT}; \quad [\ce{B}] = \frac{n_B}{V}=\frac{P_B}{RT}; \quad \text{and } \quad [\ce{C}] = \frac{n_C}{V}=\frac{P_C}{RT}$$

Since $$\frac{1}{RT}$$ is a constant, above relationships mean concentration of each species id directly proportional to its partial pressure.

1. Question 1: Would the mass of $$\ce{C}$$ increase?
• Assuming you means amount of $$\ce{C}$$, your reasoning and conclusion are correct. When the volume is decreased in gaseous system, the pressure would increase. Hence, according to Le Chatelier's principle, the system will partially oppose the change by acting to reduced the pressure. To do so, it favors the forward reaction producing more C (the major reason is $$\pu{2 mol} \rightarrow \pu{1 mol}$$).
1. Question 2: Would the concentration of $$[\ce{A}]$$ be greater when compared to the original concentration of $$[\ce{A}]$$?
• It would be greater just after you increase the pressure by reducing the volume. However, when the Le Chatelier's principle applies, amounts of $$\ce{A}$$ and $$\ce{AB}$$ would decrease (see answer to Q1). However, as I prove above, concentration is directly proportional to partial pressure. When $$n_A$$ decreases, $$P_A$$ decreases so does $$[\ce{A}]$$. However, whether $$[\ce{A}]$$ reduces past its original value or not can only be determine by knowing the reaction conditions, specially its $$K_c$$ or $$K_p$$ values.
1. Question 3: Would the amount of $$\ce{A}$$ be reduced?
• Answer is yes as you figured out. See answer to Q2 for more details ($$n_A$$ means amounts of $$\ce{A}$$).
• For question 2, with a small decrease in volume, I can prove that, at the new equilibrium, the increase in concentration of A due to the volume decrease wins out over the decrease in concentration of A from the reduction in the number of moles of A. This conclusion is independent of the equilibrium constant. So the concentration of A will increase with a decrease in volume. – Chet Miller Jul 26 '20 at 20:15
• @Chet Miller: I clearly said it depends on the condition of the reaction. So volume change isa condition. The equilibrium constant cannot be disregarded, It plays a significant role. Why don't you put your thoughts and proofs as an answer? That'd be helpful. – Mathew Mahindaratne Jul 26 '20 at 22:06
• @MathewMahindaratne Thank you for the proof and the direct answers to my questions. Really appreciate the clarification!!! Thank you again! – Munchies Jul 27 '20 at 6:44
• Done. See my Answer. – Chet Miller Jul 27 '20 at 12:13

This is an approach to Question 2.

The reaction equilibrium condition for the system is: $$\frac{V}{RT}\frac{n_C}{n_A n_B}=K_P\tag{1}$$Let $$n_{A0}$$, $$n_{B0}$$, and $$n_{C0}$$ be the number of moles at equilibrium for each species when the volume is equal to $$V_0$$. Then we have $$\frac{V_0}{RT}\frac{n_{C0}}{n_{A0} n_{B0}}=K_P\tag{2}$$. For a small change in volume dV, we can write the new number of moles of A, B, and C at equilibrium as $$n_A=n_{A0}-dn\tag{3a}$$ $$n_B=n_{B0}-dn\tag{3b}$$ $$n_C=n_{C0}+dn\tag{3c}$$Combining Eqns. 1-3 yields:$$\left(1+\frac{dV}{V_0}\right)=\frac{(1-dn/n_{A0})(1-dn/n_{B0})}{(1+dn/n_{C0})}$$And, linearizing this equation with respect to dn yields:$$\frac{dV}{V_0}=-\left(\frac{1}{n_{A0}}+\frac{1}{n_{B0}}+\frac{1}{n_{C0}}\right)dn$$This equation indicates that if dV is negative, then dn is positive, and the reactants decrease while the product increases.

If we apply this same methodology to the concentration of A, we obtain $$dC_A=-\frac{n_{A)}}{V_0}\left(\frac{dn}{n_{A0}}+\frac{dV}{V_0}\right)$$Combining this with the previous equation then yields: $$\frac{dC_A}{dV}=-\frac{n_{A0}}{V_0^2}\left[\frac{1/n_{B0}+1/n_{C0}}{1/n_{A0}+1/n_{B0}+1/n_{C0}}\right]$$The term in brackets is always positive, so, when the volume decreases, the concentration of A increases at all points along the equilibrium contour. This conclusion is independent of the specific value of the equilibrium constant.

• Thank you for the kind response. I really appreciate the proof. This has clarified a lot! Thank you again! @Chet Miller – Munchies Jul 27 '20 at 6:44
• You got my vote. – Mathew Mahindaratne Jul 27 '20 at 15:17