For each of the questions, four choices have been provided. Select the correct alternative. The equilibrium constant for the reactions \(\quad N_{2_{(g)}}+O_{2_{(g)}} \rightleftarrows 2 N O_{(g)}\) and \(\mathrm{NO}_{(\mathrm{g})} \rightleftarrows \frac{1}{2} \mathrm{~N}_{2_{(\mathrm{B})}}+\frac{1}{2} \mathrm{O}_{2_{(\mathrm{g})}}\) 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) \(\mathrm{k}^{1}=\left(\frac{1}{\mathrm{k}}\right)^{2}\)

We have two reactions as follows: Reaction 1: \(N_{2_{(g)}}+O_{2_{(g)}} \rightleftarrows 2 NO_{(g)}\) with the equilibrium constant k. Reaction 2: \(NO_{(g)} \rightleftarrows \frac{1}{2} N_{2_{(g)}}+\frac{1}{2} O_{2_{(g)}}\) with the equilibrium constant k¹. It is evident that if we multiply the coefficients of reaction 2 by 2, the reactions are the inverse of each other.

Since the reactions are the inverse of each other, we can write the relationship between their equilibrium constants: For the modified reaction: \(2 NO_{(g)} \rightleftarrows N_{2_{(g)}}+O_{2_{(g)}}\) The equilibrium constant for this modified reaction (let's call it k') is: \(k'=\frac{1}{k}\), following the standard rule that the equilibrium constant of an inverse reaction is the reciprocal of the original equilibrium constant. Now, let's compare the modified reaction with the second reaction: Modified Reaction: \(2 NO_{(g)} \rightleftarrows N_{2_{(g)}}+O_{2_{(g)}}\) with the equilibrium constant k'. Reaction 2: \(NO_{(g)} \rightleftarrows \frac{1}{2} N_{2_{(g)}}+\frac{1}{2} O_{2_{(g)}}\) with the equilibrium constant k¹. The equilibrium constant for the modified reaction (k') is related to the equilibrium constant k¹ of the second reaction since they involve the same species but with double the coefficients: \(k' = (k¹)²\) We know that \(k' = \frac{1}{k}\), so we have: \(\frac{1}{k} = (k¹)²\) With some rearrangement, we have: \(k = \left(\frac{1}{k¹}\right)^2\)

Comparing our result with the given options, we see that our result corresponds to option (b): Relation between k and k¹: \(k = \left(\frac{1}{k¹}\right)^2\)