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Choose the Best Answer · Q18

Q.In a chemical equilibrium, the rate constant for the forward reaction is 2.5×1022.5 \times 10^2 and the equilibrium constant is 50. The rate constant for the reverse reaction is,

(a) 11.5
(b) 5
(c) 2×1022 \times 10^2
(d) 2×10−32 \times 10^{-3}
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Step 1. KC=kf/kbK_C = k_f/k_b, so kb=kf/KCk_b = k_f/K_C.

Step 2. Substitute kf=2.5×102k_f=2.5\times10^2 and KC=50K_C=50: …

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