The equilibrium constant, $K$ for the reaction $2{\rm{HI}}\left( {\rm{g}} \right)\;\leftrightharpoons\;{{\rm{H}}_2}\left( {\rm{g}} \right) + {{\rm{I}}_2}\left( {\rm{g}} \right)$ At room temperature is 2.85 and that at $698\;K$, it is $1.4 \times {10^{ - 2}}$. This implies that

${\rm{HI}}$ is less stable than ${{\rm{H}}_2}$ and ${{\rm{I}}_2}$

${\rm{HI}}$ is resonance stabilised

${\rm{HI}}$ is exothermic compound

${\rm{HI}}$ is stable at room temperature

Back to Equilibrium

JEE CHEMISTRY MCQ: Equilibrium

1 Easy

1 of 1

Analyzing +4 | -1

The equilibrium constant,

K

for the reaction

2\mathrm{HI}\left( \mathrm{g} \right)\;\leftrightharpoons\;{\mathrm{H}_2}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right)

At room temperature is 2.85 and that at

698\;K

, it is

1.4 \times {10^{ - 2}}

. This implies that

View answer and explanation

Correct option: D

\mathrm{HI}

is less stable than

{\mathrm{H}_2}

and

{\mathrm{I}_2}

Option D (HI is less stable than H₂ and I₂) is correct because the equilibrium constant decreases with increasing temperature, indicating that the reverse reaction (formation of H₂ and I₂) is favored at higher temperatures, meaning HI is less stable.