Question

If the equilibrium constant for $A_2+B_2\rightleftharpoons 2AB$ is $K$, then equilibrium constant for
$\frac{1}{2}{A}_2+\frac{1}{2}{B}_2\rightleftharpoons AB$ is $\sqrt{K}$.
If true enter 1, if false enter 0.

Answer 1

If the equilibrium constant for $A_2+B_2\rightleftharpoons 2AB$ is $K$, then equilibrium constant for
$\frac{1}{2}{A}_2+\frac{1}{2}{B}_2\rightleftharpoons AB$ is $\sqrt{K}$.

For the first reaction, the equilibrium constant is $\frac{[AB{]}^2}{\left[A_2\right]\left[B_2\right]}=K$.

For the second reaction, the equilibrium constant is $\frac{\left[AB\right]}{\left[A_2{]}^{1/2}\right[B_2{]}^{1/2}}=\sqrt{\frac{[AB{]}^2}{\left[A_2\right]\left[B_2\right]}}=\sqrt{K}$.