Question

At a given temperature, the equilibrium constant for the reactions are ${{K}_{1}}$ and ${{K}_{2}}$ respectively. If ${{K}_{1}}$ is $4\times {{10}^{-3}}$ , then ${{K}_{2}}$ will be
\[\begin{array}{*{35}{l}}
NO\left( g \right)+\dfrac{1}{2}{{O}_{2}}\left( g \right)\rightleftarrows N{{O}_{2}}\left( g \right) \\
2N{{O}_{2}}\rightleftarrows 2NO\left( g \right)+{{O}_{2}}\left( g \right) \\
\end{array}\]
A. $8\times {{10}^{-3}}$
B. $16\times {{10}^{-3}}$
C. $6.25\times {{10}^{4}}$
D. $6.25\times {{10}^{6}}$

Answer 1

Hint: We have to find the relationship between ${{K}_{1}}$ and ${{K}_{2}}$ by using the given chemical reactions. Then we can easily identify the value of ${{K}_{2}}$ . There is a relationship between the given chemical reactions from which we can find the value of ${{K}_{2}}$ .
Equilibrium constant of reaction $\text{=}\dfrac{\text{ }\!\![\!\!\text{ concentration of the product }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ concentration of the reactants }\!\!]\!\!\text{ }}$

Complete step by step solution:
- The relationship between ${{K}_{1}}$ and ${{K}_{2}}$ is as follows.
- ${{K}_{1}}$ and ${{K}_{2}}$ are the equilibrium constant for the given reactions.
${{K}_{1}}$ is the equilibrium constant for $NO\left( g \right)+\dfrac{1}{2}{{O}_{2}}\left( g \right)\rightleftarrows N{{O}_{2}}\left( g \right)\to (a)$ and
${{K}_{2}}$ is the equilibrium constant for $2N{{O}_{2}}\rightleftarrows 2NO\left( g \right)+{{O}_{2}}\left( g \right)\to (b)$
- The equilibrium constants of the given chemical reaction (a) is as follows.
\[{{K}_{1}}=\dfrac{N{{O}_{2}}}{[NO]{{[{{O}_{2}}]}^{\dfrac{1}{2}}}}\]
- The equilibrium constants of the given chemical reaction (b) is as follows.
\[{{K}_{2}}=\dfrac{1}{\dfrac{{{[NO]}^{2}}[{{O}_{2}}]}{{{[N{{O}_{2}}]}^{2}}}}\]
- Therefore we can write the relationship between the equilibrium constants of the given chemical reactions as follows.
\[{{K}_{2}}=\dfrac{1}{K_{1}^{2}}\to (1)\]
- In the question it is given that the equilibrium constant of the reaction (a), ${{K}_{1}}$ is $4\times {{10}^{-3}}$ .
- By substituting the ${{K}_{1}}$ value in the above equation (1) we can find the value of equilibrium constant of the reaction (b), ${{K}_{2}}$ and it is as follows.
\[\begin{align}
& {{K}_{2}}=\dfrac{1}{K_{1}^{2}} \\
& {{K}_{2}}=\dfrac{1}{{{\left( 4\times {{10}^{-3}} \right)}^{2}}} \\
& {{K}_{2}}=\dfrac{1}{16\times {{10}^{-6}}} \\
& {{K}_{2}}=6.25\times {{10}^{4}} \\
\end{align}\]
- Therefore the value of equilibrium constant for the reaction (b) is $6.25\times {{10}^{4}}$ .

So, the correct option is C.

Note: By observing the given chemical reactions carefully we can write the relationship between the equilibrium constants. The chemical reactants in one chemical reaction are the products in the other chemical reaction and the products in one chemical reaction are the reactants in the other chemical reaction.