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 $300\;{\rm{K}}$ is ${\rm{2}}{\rm{.85}}$ and at $698\;{\rm{K}}\,\;1.84 \times {10^{ - 2}}.$ Hence the reason that ${\rm{HI}}$ exists as a stable compound at room temperature is because:

The heat of reaction at room temperature is $ - 5.31\;{\rm{kcal}}$

It decomposes so slowly that equilibrium is not readily achieved

The ${\rm{HI}}$ bond has a large covalent contribution

NEET UG CHEMISTRY MCQ: Equilibrium

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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

300\;\mathrm{K}

\mathrm{2}\mathrm{.85}

and at

698\;\mathrm{K}\,\;1.84 \times {10^{ - 2}}.

Hence the reason that

\mathrm{HI}

exists as a stable compound at room temperature is because:

View answer and explanation

Correct option: C

The heat of reaction at room temperature is

- 5.31\;\mathrm{kcal}

Option A is correct because the slow decomposition rate of HI at room temperature prevents the attainment of equilibrium, making it appear stable.