For a reversible first-order reaction R ⇌ S, the forward rate constant is 0.30 min -1 and the backward rate constant is 0.10 min -1 . The initial concentrations are [R] = 0.20 M and [S] = 1.00 M. The

Question

For a reversible first-order reaction R ⇌ S, the forward rate constant is 0.30 min -1 and the backward rate constant is 0.10 min -1 . The initial concentrations are [R] = 0.20 M and [S] = 1.00 M. The concentration of S that is converted into R before equilibrium is reached is

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K = k f /k b = 0.30/0.10 = 3 At the start [S]/[R] = 5, which is above K, so the net change converts S into R Let y M of S react: [R] = 0.20 + y and [S] = 1.00 - y k f (0.20 + y) = k b (1.00 - y) 0.30(0.20 + y) = 0.10(1.00 - y) 0.06 + 0.30y = 0.10 - 0.10y 0.40y = 0.04 so y = 0.10 M This is option (c).

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