For an elementary chemical reaction, \(\mathrm{A_2} \underset{K_{-1}}{\overset{K_1}{\rightleftharpoons}} 2\mathrm{A}\), the expression for \(\frac{d[A]}{dt}\) is
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- \(2K_1[A_2] - K_{-1}[A]^2\)
- \(K_1[A_2] - K_{-1}[A]^2\)
- \(K_1[A_2] + K_{-1}[A]^2\)
- \(2K_1[A_2] - 2K_{-1}[A]^2\)
The Correct Option is D
Solution and Explanation
Backward (elementary): \(2\mathrm{A} \to \mathrm{A_2}\), rate \(= K_{-1}[A]^2\); 2 mol of A are consumed per event, so A is used up at \(2K_{-1}[A]^2\).
\(\frac{d[A]}{dt} = 2K_1[A_2] - 2K_{-1}[A]^2\).
Answer: (4) \(2K_1[A_2] - 2K_{-1}[A]^2\)
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