Question

For the reaction, $2 A(g)+B_{2}(g) \rightleftharpoons 2 A B_{2}(g)$ the equilibrium constant, $K_{p}$ at $300\, K$ is $16.0$. The value of $K_{p}$ for $A B_{2}(g) \rightleftharpoons A(g)+1 / 2 B_{2}(g)$ is

• A. 8
• B. 0.25
• C. 0.125
• D. 32

Answer 1

Answer: (B)

0.25

Answer 2

For the reaction,
$2 A+B_{2} \rightleftharpoons 2 A B_{2},$
Equilibrium constant,
$K_{p}=\frac{p_{A B_{2}}^{2}}{p_{A}^{2} \cdot p_{B_{2}}}=16$ ... (i)
For the reaction,
$A B_{2} \rightleftharpoons A+1 / 2 B_{2},$
$K_{p}'=\frac{P_{A} \cdot p_{B_{2}}^{1 / 2}}{p_{A B_{2}}}$ ...(ii)
On squaring E (ii), we get
$\left(K_{p}'\right)^{2}=\frac{p_{A}^{2} p_{B_{2}}}{p_{A B_{2}}^{2}}$ ...(iii)
From E (i) and (iii), we get
$K_{p} \cdot\left(K_{p}^{\prime}\right)^{2}=1$
$16 \cdot\left(K_{p}^{\prime}\right)^{2}=1$
$\left(K_{p}'\right)^{2}=\frac{1}{16}$
$K_{p}'=\frac{1}{4}=0.25$