# Chemistry - Effect on mass when decreasing pressure in equilibrium reaction

## Solution 1:

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.

## Solution 2:

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 $$\ce{A}$$).