COMEDK · Chemistry · 7. Chemical Equilibrium
Consider the following equilibrium,
\(\begin{aligned}
& 2 \mathrm{No}(g) \rightleftharpoons \mathrm{N}_2+\mathrm{O}_2 ; K_{\mathrm{G}_1}=2.4 \times 10^{20} \\
& \mathrm{No}(g)+\frac{1}{2} \mathrm{Br}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NoBr}(g) ; K_{C_2}=1.4
\end{aligned}\)
Calculate \(K_C\) for the reaction,
\(\frac{1}{2} \mathrm{~N}_2(g)+\frac{1}{2} \mathrm{O}_2(g)+\frac{1}{2} \mathrm{Br}_2(g) \rightleftharpoons \mathrm{NOBr}(g)\)
- A \(8.96 \times 10^{-11}\)
- B \(9.48 \times 10^{-\theta}\)
- C \(8.08 \times 10^{-12}\)
- D \(8.96 \times 10^{11}\)
Answer & Solution
Correct Answer
(A) \(8.96 \times 10^{-11}\)
Step-by-step Solution
Detailed explanation
For, \(2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_2+\mathrm{O}_2\) \(\mathrm{K}_{\mathrm{C}_1}=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]}{[\mathrm{NO}]^2}=2.4 \times 10^{20}\) For,…
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