Question

The equilibrium constant $K$, for the given reaction $2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$ at room temperature is $2.85K$ and that at $698K$ is $1.4\times {10}^{-2}$. Hence the reason that $HI$ exists as a stable compound at room temperature is____________.

• A. It decomposes so slowly that equilibrium is not readily achieved
• B. The bond $HI$ has a large covalent contribution
• C. The standard free energy of the dissociation reaction at room temperature is $-0.62kcal$
• D. It is uncatalytic reaction

Answer 1

The correct option is A It decomposes so slowly that equilibrium is not readily achieved

At room temperature even though $K>1$ forward reaction is favoured, it kinetic rate is low. So it decomposes slowly to achieve equilibrium. Hence stable at room temperature.

The correct answer is (A)