KCET · Chemistry · Chemical Equilibrium
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 \( K_{1}=\sqrt{K_{2}} \)
- B \( K_{2}=\frac{1}{K_{1}} \)
- C \( K_{1}=2 K_{2} \)
- D \( O K_{2}=\frac{1}{K_{1}{ }^{2}} \)
Answer & Solution
Correct Answer
(B) \( K_{2}=\frac{1}{K_{1}} \)
Step-by-step Solution
Detailed explanation
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
\[
K_{2}=\frac{1}{K_{1}}
\]
\[
\begin{array}{l}
N O(g)+\frac{1}{2} O_{2}(g) \rightleftharpoons N O_{2}(g) \text { or } 2 N O(g)+O_{2}(g) \rightleftharpoons 2 N O_{2}(g) \\
K_{1}=\frac{\left[N O_{2}\right]}{[N O]\left[O_{2}\right]^{1 / 2}} \text { or } K_{1}=\frac{\left[N O_{2}\right]^{2}}{[N O]^{2}\left[O_{2}\right]} \rightarrow(1)
\end{array}
\]
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
\[
K_{2}=\frac{1}{K_{1}}
\]
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