Question

The equilibrium constant of the reaction $2AB\rightleftharpoons A_2+B_2$ is $49$. The equilibrium constant of
$AB\rightleftharpoons \frac{1}{2}{A}_2+\frac{1}{2}{B}_2$ is:

• A. $0.01$
• B. $2401$
• C. $9$
• D. $7$

Answer 1

Answer: (D)

$7$

Given $2AB\leftrightharpoons A_2+B_2{K}_1-=49$

$K_1=\frac{\left[A_2\right]\left[B_2\right]}{{\left[AB\right]}^2}⟶i$

$AB\rightleftharpoons \frac{1}{2}{A}_2+\frac{1}{2}{B}_2$

$K_1=\frac{\left[A_2{]}^{\frac{1}{2}}\right[B_2{]}^{\frac{1}{2}}}{\left[AB\right]}⟶ii$

From $i$ and $ii$,

$K_2^2=K_1=49$

$K_2=\sqrt{49}=7$