# Calculating partial pressure equilibrium constant Kₚ given initial pressure and equilibrium pressure

A reaction

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

happens in constant volume and constant temperature. The reaction starts only with gas $$\ce{A}$$ (no $$\ce{B}$$ or $$\ce{C}$$) with given pressure $$P_1 = \pu{6 atm}$$, in equilibrium the pressure of all three gases is $$P_2 = \pu{10 atm}$$. Calculate $$K_p.$$

It seems to me like a very simple question, however it seems that I don't understand a basic concept regarding gas equilibrium. As for my understanding it is suppose to be

$$K_p = \frac{P_2 \cdot P_2}{P_2} = P_2 = \pu{10 atm}$$

But the given solution is $$\pu{8 atm}$$ and I really don't get what miss.

• Make an I.C.E, table first. Jul 5, 2019 at 17:00
• I don't see how it make sense, since the initial amount of $A$ is $6\cdot \frac{V}{RT}$ and the final amount is $10\cdot \frac{V}{RT}$ (V,T are constants), but the amount of material can't grow. Jul 5, 2019 at 17:35

The problem is that you are using wrong pressures. By definition for the reaction at equilibrium partial pressures can be expressed via initial partial pressure $$P_1$$ and conversion factor $$α$$

$$\begin{array}{ccc} \ce{&A(g) &<=> &B(g) &+ &C(g)}\\ &(1 - α)P_1& & αP_1& & αP_1 \end{array}$$

equilibrium constant $$K_p$$ is to be found as

$$K_p = \frac{P(\ce{B})\cdot P(\ce{C})}{P(\ce{A})} = \frac{α^2P_1^2}{(1 - α)P_1} = \frac{α^2P_1}{1 - α}$$

Unknown $$α$$ can be found by equating total pressure at equilibrium $$P_2$$ to the sum of partial pressures of all gaseous components at equilibrium:

\begin{align} P_2 &= P(A) + P(B) + P(C) \\ &= (1 - α)P_1 + αP_1 + αP_1 \\ &= (1 + α)P_1 \end{align} \quad\implies\quad α = \frac{P_2}{P_1} - 1 = \frac{\pu{10 atm}}{\pu{6 atm}} - 1 = \frac{2}{3}

Finally, all the values can be plugged into the expression for $$K_p$$:

$$K_p = \frac{\left(\frac{2}{3}\right)^2\cdot\pu{6 atm}}{\left(1 - \frac{2}{3}\right)} = \pu{8 atm}$$

• Thanks, may you please explain why the statement in the question that $P_2(A)=P_2(B)=P_2(C)=10atm$ in equilibrium it translated to $P_2 = P(A)+P(B)+P(C)?$ I mean, how could I deduce it from the question ? Jul 6, 2019 at 2:30
• @user5721565 I added corresponding passage to the answer. Jul 6, 2019 at 8:10