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

Question

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.

Answer 1

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 $K_p$:

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