Given the reaction between 2 gases represented by ${A_2}$ and ${B_2}$ to give the compound $A{B_{\left( {\rm{g}} \right)}}{\rm{.}}$ ${A_{2\left( {\rm{g}} \right)}} + {B_{2\left( {\rm{g}} \right)}}\;\leftrightharpoons\;2A{B_{\left( {\rm{g}} \right)}}$ At equilibrium, the concentration of ${A_2} = 3.0 \times {10^{ - 3}}\;{\rm{M}}$ of ${B_2} = 4.2 \times {10^{ - 3}}\;{\rm{M}}$ of $AB = 2.8 \times {10^{ - 3}}\;{\rm{M}}$ If the reaction takes place in a sealed vessel at $527^\circ C$, then the value of ${K_c}$ will be :

JEE CHEMISTRY MCQ: Equilibrium

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Given the reaction between 2 gases represented by

{A_2}

and

{B_2}

to give the compound

A{B_{\left( \mathrm{g} \right)}}\mathrm{.}

{A_{2\left( \mathrm{g} \right)}} + {B_{2\left( \mathrm{g} \right)}}\;\leftrightharpoons\;2A{B_{\left( \mathrm{g} \right)}}

At equilibrium, the concentration
of

{A_2} = 3.0 \times {10^{ - 3}}\;\mathrm{M}

{B_2} = 4.2 \times {10^{ - 3}}\;\mathrm{M}

AB = 2.8 \times {10^{ - 3}}\;\mathrm{M}

If the reaction takes place in a sealed vessel at

527^\circ C

, then the value of

{K_c}

will be :

View answer and explanation

Correct option: C

0.62

The correct answer is 0.62 (Option C) because it is calculated by substituting the given equilibrium concentrations into the equilibrium constant expression,

{K_c} = \frac{{{{\left[ {AB} \right]}^2}}}{{\left[ {{A_2}} \right]\left[ {{B_2}} \right]}}

JEE CHEMISTRY MCQ: Equilibrium

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