Question

Consider following equilibrium at $400K$ and $30atm$
$3A\left(g\right)\rightleftharpoons 2B\left(g\right)+2C\left(g\right)+D\left(g\right)$
If the equilibrium pressure of $A$ is $5atm$ then $K_c$ will be:

• A. $9.5L^{-2}mo{l}^2$
• B. $0.39L^{-2}mo{l}^2$
• C. $0.95L^{-2}mo{l}^2$
• D. $3.9L^{-2}mo{l}^2$

Answer 1

The correct option is B $0.39L^{-2}mo{l}^2$

$$\begin{array}{ccccccc}& 3A\rightleftharpoons & 2B& +& 2C& +& D\\ \text{Initial}& P^{\circ }& 0& & 0& & 0\\ \text{Equilibrium}& P^{\circ }-3x& 2x& & 2x& & x\end{array}$$
Equilibrium pressure $=P^{\circ }-3x+2x+2x+x$
$\Rightarrow 30=P^{\circ }+2x$...(1)
From question $5=P^{\circ }-3x$...(2)
From equation (1) and (2)
$25=5x\Rightarrow x=5K_p=\frac{P_B^2{P}_C^2{P}_D}{P_A^3}=\frac{\left(10{)}^2\right(10{)}^{2}5}{5^3}\Rightarrow K_p=400at{m}^2$
We know,
$K_p=K_c(RT{)}^{\Delta n}\Rightarrow K_c=\frac{K_p}{(RT{)}^{\Delta n}}=\frac{400}{(0.08\times 400{)}^2}=0.39L^{-2}mo{l}^2$