Question

For the decomposition of 1 mol of $\text{N}{{\text{H}}_{\text{3}}}\left( \text{g} \right)$ into ${{\text{N}}_{2}}\left( \text{g} \right)$ and ${{\text{H}}_{2}}\left( \text{g} \right)$, $\text{K=2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{M}$. If the equilibrium concentration of ${{\text{N}}_{2}}\left( \text{g} \right)$ is $\text{0}\text{.09M}$ and ${{\text{H}}_{2}}\left( \text{g} \right)$ is $\text{0}\text{.04M}$, then the equilibrium concentration of $\text{N}{{\text{H}}_{\text{3}}}\left( \text{g} \right)$ is:
(A)- $\text{1}\text{.2M}$
(B)- $\text{2}\text{.88 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{M}$
(C)- $\text{5}\text{.36 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-2}}}\text{M}$
(D)- None of these

Answer 1

Hint: Equilibrium constant of any reaction is defined as the ratio of the equilibrium concentration of the formed products in the chemical reaction to the equilibrium concentration of reactants present in the chemical reaction.

Complete step by step solution:
First we write balanced chemical equation for the given reaction:
\[\text{2N}{{\text{H}}_{\text{3}}}\to {{\text{N}}_{\text{2}}}\text{+3}{{\text{H}}_{\text{2}}}\]
Given that, Equilibrium concentration of nitrogen gas i.e. ${{\text{N}}_{2}}\left( \text{g} \right)$ = $\text{0}\text{.09M}$,
Equilibrium concentration of hydrogen gas i.e. ${{\text{H}}_{2}}\left( \text{g} \right)$ = $\text{0}\text{.04M}$,
Equilibrium constant of given reaction $\text{K=2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{M}$.
We know that equilibrium constant for given reaction will be calculated as follow:
\[\text{K=}\dfrac{\left[ {{\text{N}}_{\text{2}}} \right]{{\left[ {{\text{H}}_{\text{2}}} \right]}^{\text{3}}}}{{{\left[ \text{N}{{\text{H}}_{\text{3}}} \right]}^{\text{2}}}}\]
On putting the values of equilibrium concentration of nitrogen gas, hydrogen gas & equilibrium constant in the above equation we get,
\[\text{2}\times \text{1}{{\text{0}}^{-3}}\text{=}\dfrac{\left[ 0.09 \right]{{\left[ 0.04 \right]}^{\text{3}}}}{{{\left[ \text{N}{{\text{H}}_{\text{3}}} \right]}^{\text{2}}}}\]
\[{{\left[ \text{N}{{\text{H}}_{\text{3}}} \right]}^{\text{2}}}=\dfrac{\left[ 0.09 \right]{{\left[ 0.04 \right]}^{\text{3}}}}{\text{2}\times \text{1}{{\text{0}}^{-3}}}\]
\[{{\left[ \text{N}{{\text{H}}_{\text{3}}} \right]}^{\text{2}}}=2.8\times {{10}^{-3}}\]
\[\left[ \text{N}{{\text{H}}_{\text{3}}} \right]=\sqrt{2.8\times {{10}^{-3}}}=1.6\times {{10}^{-3}}\]
From the above discussion it is clear that the equilibrium concentration of $\text{N}{{\text{H}}_{\text{3}}}\left( \text{g} \right)$ is $\text{1}\text{.6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}\text{M}$.

Hence, the option (D) is correct because none of the above-given options matches the answer.

Additional information: Equilibrium constant of any chemical reaction defines the amount of prepared product & reactant in the equilibrium condition of a reversible reaction, where the rate of forwarding reaction is equal to the rate of backward reaction.
Note: Always keep in mind that during the calculation of equilibrium constant of any reaction first, you have to balance the given chemical reaction, because if you do not do so then you will get the wrong answer.