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# The decomposition of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ to ${\text{N}}{{\text{O}}_{\text{2}}}$ is carried out at 280K in chloroform. When equilibrium is reached, 0.2 mol of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ and ${{2 \times 1}}{{\text{0}}^{{\text{ - 3}}}}$mol of ${\text{N}}{{\text{O}}_{\text{2}}}$ are present in a 2 litre solution. The equilibrium constant for the below reaction is${{\text{N}}_{\text{2}}}{{\text{O}}_{{\text{4(g)}}}} \rightleftharpoons 2{\text{N}}{{\text{O}}_{\text{2}}}_{{\text{(g)}}}$(A) ${\text{1 x 1}}{{\text{0}}^{{\text{ - 2}}}}$(B) ${\text{2 x 1}}{{\text{0}}^{{\text{ - 3}}}}$(C) ${\text{1 x 1}}{{\text{0}}^{{\text{ - 5}}}}$(D) ${\text{2 x 1}}{{\text{0}}^{{\text{ - 5}}}}$

Last updated date: 12th Sep 2024
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Hint: The equilibrium constant for a reaction is defined as the number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.

Complete step by step answer: It is given that the decomposition of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ to ${\text{N}}{{\text{O}}_{\text{2}}}$ is carried out
at 280K in chloroform. At equilibrium, number of moles of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ is 0.2 mol and number of moles of ${\text{N}}{{\text{O}}_{\text{2}}}$ is ${{2 \times 1}}{{\text{0}}^{{\text{ - 3}}}}$mol. The total volume of the solution is 2 litres.
We can find the concentration of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$&${\text{N}}{{\text{O}}_{\text{2}}}$ by the following formula:
${\text{Concentration = }}\dfrac{{{\text{Number of moles}}}}{{{\text{Volume}}}}$
Concentration can be expressed in moles per litre or mol/L. it is also denoted by M.
The concentration of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ is found by substituting the given values, we get,
$\Rightarrow {\text{[}}{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}{\text{] = }}\dfrac{{0.2{\text{ mol}}}}{{2{\text{ L}}}} = 0.1{\text{ mol/L}}$
The concentration of ${\text{N}}{{\text{O}}_{\text{2}}}$ is found by substituting the given values, we get,
$\Rightarrow {\text{[N}}{{\text{O}}_{\text{2}}}{\text{] = }}\dfrac{{{{2 \times 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{mol}}}}{{{\text{2 L}}}}{{ = 1 \times 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{ mol/L}}$
The equilibrium constant (K or Keq or KC) is the ratio of the mathematical product of the concentrations of the products of a reaction to the mathematical product of the concentrations of the reactants of the reaction. Each concentration is raised to the power of its coefficient in the balanced chemical equation.
The equilibrium reaction for the decomposition of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ to ${\text{N}}{{\text{O}}_{\text{2}}}$ is given as:
${{\text{N}}_{\text{2}}}{{\text{O}}_{{\text{4(g)}}}} \rightleftharpoons 2{\text{N}}{{\text{O}}_{\text{2}}}_{{\text{(g)}}}$
The equilibrium constant, K can be given as
${\text{K = }}\dfrac{{{{{\text{[N}}{{\text{O}}_{\text{2}}}{\text{]}}}^{\text{2}}}}}{{{\text{[}}{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}{\text{]}}}} \\ \Rightarrow {\text{K = }}\dfrac{{{{{{(1 \times 1}}{{\text{0}}^{{\text{ - 3}}}})}^2}}}{{{\text{0}}{\text{.1}}}} \\ \Rightarrow {\text{K = }}\dfrac{{{{1 \times 1}}{{\text{0}}^{{{ - 6}}}}}}{{{\text{0}}{\text{.1}}}} \\ \Rightarrow {{K = 1 \times 1}}{{\text{0}}^{{\text{ - 5}}}} \\$
The equilibrium constant for the decomposition of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ to ${\text{N}}{{\text{O}}_{\text{2}}}$ is ${{1 \times 1}}{{\text{0}}^{{\text{ - 5}}}}$

So, the correct option is C.

Note: Equilibrium constant has no unit since it is the ratio of two concentrations, namely products and reactants respectively. The equilibrium constant can help us understand whether the reaction tends to have a higher concentration of products or reactants at equilibrium.