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.

  • $\begingroup$ Make an I.C.E, table first. $\endgroup$ Jul 5, 2019 at 17:00
  • $\begingroup$ 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. $\endgroup$ Jul 5, 2019 at 17:35

1 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 $K_p$:

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

  • $\begingroup$ 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 ? $\endgroup$ Jul 6, 2019 at 2:30
  • $\begingroup$ @user5721565 I added corresponding passage to the answer. $\endgroup$
    – andselisk
    Jul 6, 2019 at 8:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.