Question

The equilibrium constant for the synthesis of $HI$ at ${490}^{o}C$is $50.0$. The value of $K$ for the dissociation of $HI$ will be:

• A. $0.02$
• B. $50.0$
• C. $0.50$
• D. $0.20$

Answer 1

Answer: (A)

$0.02$

The equilibrium constant for the synthesis of $HI$ at ${490}^{o}C$is $50.0$. The value of $K$ for the dissociation of $HI$ will be $\frac{1}{50.0}=0.02$

Note: If $K_1$ is the equilibrium constant for $H_2+I_2\rightleftharpoons 2HI$, then the equilibrium constant for $2HI\rightleftharpoons H_2+I_2$
will be $K_2=\frac{1}{K_1}$