At a given temperature the ${K_c}$ for the reaction, ${\rm{PC}}{{\rm{l}}_5}\left( {\rm{g}} \right)\;\leftrightharpoons\;{\rm{PC}}{{\rm{l}}_3}\left( {\rm{g}} \right) + {\rm{C}}{{\rm{l}}_2}\left( {\rm{g}} \right)$ is $2.4 \times {10^{ - 3}}$. At the same temperature, the ${K_c}$ for the reaction ${\rm{PC}}{{\rm{l}}_3}\left( {\rm{g}} \right) + {\rm{C}}{{\rm{l}}_2}\left( {\rm{g}} \right)\;\leftrightharpoons\;{\rm{PC}}{{\rm{l}}_5}\left( {\rm{g}} \right)$ is :

NEET UG CHEMISTRY MCQ: Equilibrium

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At a given temperature the

{K_c}

for the reaction,

\mathrm{PC}{\mathrm{l}_5}\left( \mathrm{g} \right)\;\leftrightharpoons\;\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)

2.4 \times {10^{ - 3}}

. At the same temperature, the

{K_c}

for the reaction

\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)\;\leftrightharpoons\;\mathrm{PC}{\mathrm{l}_5}\left( \mathrm{g} \right)

is :

View answer and explanation

Correct option: C

4.2 \times {10^{ - 2}}

Option C is correct because the second reaction is the reverse of the first, so its equilibrium constant is the inverse of the first: Kc' = 1/Kc = 1/(2.4 × 10^-3) = 416.67. However, the provided answer is incorrect. The correct answer should be approximately 416.67, which is not among the options.

NEET UG CHEMISTRY MCQ: Equilibrium

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