Q29 / 180 · KCET 2016 · Chemistry

Equilibrium constants \( K_{1} \), and \( K_{2} \) for the following equilibria
(a) \( \mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{NO}_{2}(g) \)
(b) \( 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \)
are related as :

  • A
    \( K_{1}=\sqrt{K_{2}} \)
  • Correct answer
    B
    \( K_{2}=\frac{1}{K_{1}} \)
  • C
    \( K_{1}=2 K_{2} \)
  • D
    \( O K_{2}=\frac{1}{K_{1}{ }^{2}} \)

Solution

For reaction (I),
\[ \begin{array}{l} N O(g)+\frac{1}{2} O_{2}(g) \rightleftharpoons N O_{2}(g) \text { or } 2 N O(g)+O_{2}(g) \rightleftharpoons 2 N O_{2}(g) \\ K_{1}=\frac{\left[N O_{2}\right]}{[N O]\left[O_{2}\right]^{1 / 2}} \text { or } K_{1}=\frac{\left[N O_{2}\right]^{2}}{[N O]^{2}\left[O_{2}\right]} \rightarrow(1) \end{array} \]
Now, for reaction (II),
\[ \begin{array}{l} 2 N O_{2}(g) \rightleftharpoons 2 N O(g)+O_{2}(g) \\ K_{1}=\frac{[N O]^{2}\left[O_{2}\right]}{\left[N O_{2}\right]^{2}} \rightarrow(2) \end{array} \]
Equating Eq. (1) and Eq. (2), we get
\[ K_{2}=\frac{1}{K_{1}} \]

Hence, the correct answer is option: 'B'.

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