The complex [Co(NH3)5F]^2+ reacts with water as: [Co(NH3)5F]^2+ + H2O → [Co(NH3)5(H2O)]^3+ + F^-. The rate of reaction may be given as, r = K[complex]^a[H^+]^b. The reaction is acid catalysed and henc

Question

The complex [Co(NH3)5F]^2+ reacts with water as: [Co(NH3)5F]^2+ + H2O → [Co(NH3)5(H2O)]^3+ + F^-. The rate of reaction may be given as, r = K[complex]^a[H^+]^b. The reaction is acid catalysed and hence [H^+] does not change during the reaction. Thus, the rate may be given as r = K'[complex]^a where K' = K[H^+]^b. The value of (a + b) is
[Complex]=0.1M, [H^+]=0.01M, t1/2=1.0h, t3/4=2.0h.
[Complex]=0.2M, [H^+]=0.02M, t1/2=0.5h, t3/4=1.0h.

Prashant Kotian · Accepted Answer

Correct answer: 2. Since t3/4 = 2 × t1/2 in both cases, the reaction is first-order with respect to the complex, so a = 1. For a pseudo-first-order reaction, t1/2 = 0.693 / K'. Comparing the two sets: doubling both [Complex] and [H^+] halves the t1/2, which means K' is doubled. Since K' = K[H^+]^b, doubling [H^+] doubles K', implying b = 1. (a + b) = 1 + 1 = 2. Therefore, correct answer is 2.

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