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# The equilibrium constant for${{N}_{2}}\left( g \right)+{{O}_{2}}\left( g \right)\rightleftharpoons 2NO\left( g \right)$is K, the equilibrium constant for$\dfrac{1}{2}{{N}_{2}}\left( g \right)+\dfrac{1}{2}{{O}_{2}}\left( g \right)\rightleftharpoons NO\left( g \right)$ will be:A. KB. ${{K}^{2}}$C. ${{K}^{\dfrac{1}{2}}}$D. $\dfrac{1}{2}K$

Last updated date: 20th Jun 2024
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Hint: Equilibrium constant constant is basically a proportionality constant, which indicates the relation between the molar concentration of reactants and the rate of reaction. And the expression for equilibrium constant is given by the equation:
$K=\dfrac{concentration\text{ }of\text{ }product}{concentration\text{ }of\text{ }reac\tan t}$

- The first reaction given is: ${{N}_{2}}\left( g \right)+{{O}_{2}}\left( g \right)\rightleftharpoons 2NO\left( g \right)$
$K=\dfrac{{{\left[ NO \right]}^{2}}}{\left[ {{N}_{2}} \right]\left[ {{O}_{2}} \right]}$
$\dfrac{1}{2}{{N}_{2}}\left( g \right)+\dfrac{1}{2}{{O}_{2}}\left( g \right)\rightleftharpoons NO\left( g \right)$
\begin{align}& K=\dfrac{\left[ NO \right]}{\sqrt{\left[ {{N}_{2}} \right]\left[ {{O}_{2}} \right]}} \\ & =\sqrt{\dfrac{{{\left[ NO \right]}^{2}}}{\left[ {{N}_{2}} \right]\left[ {{O}_{2}} \right]}} \\ & =\sqrt{K} \\ & ={{K}^{\dfrac{1}{2}}} \\ \end{align}
- Here, the value of equilibrium constant is ${{K}^{\dfrac{1}{2}}}$.
Hence, we can conclude that the correct option is (C), that is the value of equilibrium constant ${{K}^{\dfrac{1}{2}}}$.
Note: - We should not get confused in terms ${{k}_{C}}$and${{Q}_{C}}$ . As ${{k}_{C}}$is the equilibrium constant, that is the ratio of concentrations of products and reactants, when reaction is at equilibrium. While ${{Q}_{C}}$ is the reaction quotient, which is used to determine in which direction a reaction will proceed.