Question

The equilibrium constant for the reactions ${\mathrm{N}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftarrows 2\mathrm{NO}\left(\mathrm{g}\right)$ and $\mathrm{NO}\left(\mathrm{g}\right)\rightleftarrows \frac{1}{2}{\mathrm{N}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2$are $\mathrm{k}$ and ${\mathrm{k}}^1$, respectively, the relation between $\mathrm{k}$ and ${\mathrm{k}}^1$ is

• A. $\mathrm{k}={\left({\mathrm{k}}^1\right)}^2$
• B. $\mathrm{k}={\left(\frac{1}{{\mathrm{k}}^1}\right)}^2$
• C. ${\mathrm{k}}^2={\mathrm{k}}^1$
• D. ${\mathrm{k}}_1={\left(\frac{1}{\mathrm{k}}\right)}^2$

Answer 1

The correct option is D

${\mathrm{k}}_1={\left(\frac{1}{\mathrm{k}}\right)}^2$

The explanation for the correct option D):

1. When a chemical equation is reversed, its equilibrium constant becomes the reciprocal of the equilibrium equation of the forward direction.
2. Also, whenever a balanced equation is multiplied by a coefficient n, then its equilibrium constant is raised to the power n.
3. Thus, the second equation is obtained by reversing the first equation and then multiplying by a factor of$\frac{1}{2}$.
4. $$\begin{gathered} {\mathrm{N}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftarrows 2\mathrm{NO}\left(\mathrm{g}\right)\mathrm{Equation}1=\mathrm{k} \\ \mathrm{Nitrogen}\mathrm{Oxygen}\mathrm{Nitogen}\mathrm{oxide} \end{gathered}$$
5. Reverse equation 1: $$\begin{gathered} 2\mathrm{NO}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{N}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\mathrm{Equation}2=\frac{1}{\mathrm{k}} \\ \mathrm{Nitrogen}\mathrm{oxide}\mathrm{Nitrogen}\mathrm{Oxygen} \end{gathered}$$
6. Finally, divide equation 2 by a factor of $\frac{1}{2}$: $$\begin{gathered} \mathrm{NO}\left(\mathrm{g}\right)\rightleftharpoons \frac{1}{2}{\mathrm{N}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right){\mathrm{k}}_1={\left(\frac{1}{\mathrm{k}}\right)}^{1/2} \\ \mathrm{Nitrogen}\mathrm{oxide}\mathrm{Nitrogen}\mathrm{Oxygen} \end{gathered}$$
7. Hence, the equilibrium constant of the second equation is: $\mathrm{k}={\left(\frac{1}{{\mathrm{k}}_1}\right)}^{1/2}\mathrm{or}{\mathrm{k}}_1={\left(\frac{1}{\mathrm{k}}\right)}^2$

conclusion: The correct option is (D).