Question

Consider following equilibrium at $400\text{K}$ and $24\text{atm}$.
$3A\rightleftharpoons 2B+2C+D$
If the equilibrium pressure of $A$ is $4\text{atm}$ then $K_c\left({\text{in mol}}^2{\text{L}}^{-2}\right)$ will be:
(Take $R=0.08{\text{L atm K}}^{-1}{\text{mol}}^{-1}$)

• A. $0.5$
• B. $0.25$
• C. $2.5$
• D. $5.0$

Answer 1

The correct option is B $0.25$

$$\begin{array}{ccccccc}& 3A& \rightleftharpoons 2B& +& 2C& +& D\\ \text{Initial}& P^{\circ }& 0& & 0& +& 0\\ \text{Equilibrium}& P^{\circ }-2x& 2x& & 2x& & x\end{array}$$
Equilibrium pressure $=P^{\circ }-2x+2x+2x+x24=P^{\circ }+3x.........\left(1\right)$
From question $4=P^{\circ }-2x.........\left(2\right)$
From equation (1) and (2)
$20=5xx=4K_p=\frac{P_B^2{P}_C^2{P}_D}{P_A^3}=\frac{\left(8{)}^2\right(8{)}^{2}4}{4^3}=256{\text{atm}}^2{K}_p=K_c(RT{)}^{\Delta n}{K}_c=\frac{K_p}{(RT{)}^{\Delta n}}=\frac{256{\text{atm}}^2}{(0.08{\text{L atm K}}^{-1}{\text{mol}}^{-1}\times 400\text{K}{)}^2}=0.25{\text{mol}}^2{\text{L}}^{-2}$