Consider the first-order conversion of reactant A to product B. If k 1 is equal

Question

Consider the first-order conversion of reactant A to product B. If k1 is equal to 0.013 s-1, and if k2 is equal to 0.003 s-1, then what will be the velocity of the forward reaction (vf) when equilibrium is reached? Recall that vf (at equilibrium) is equal to k1 times the equilibrium concentration of reactant A. Also, assume that the initial concentration of A, is 0.99 M. Report your answer in terms of millimolar concentration (mM) per second to the nearest tenth. Thought question: what will be the value of velocity of the reverse reaction (vr) when equilibrium is reached?

Explanation / Answer

For forward reaction, k1 = 0.013 s-1

For backward reaction, k2 = 0.003 s-1

so Keq = K1 / K2 = 0.013 / 0.003 = 4.33

                                      A ---> B

Initial concentration 0.99 0

Change                          -x +x

equilibrium 0.99-x x

Keq = [B] / [A] = x / 0.99-x = 4.33

4.28 - 4.33x = x

x = 0.797

We will use integrated rate law expression for first order

t = 1/K ln (A0 / At)

A0= 0.99

At = 0.797

K = 0.013

t = 1/0.013 ln(0.99/0.797) =   16.68 seconds

So velocity = change in concentration / time = (0.99-0.797)/16.68 = 0.0115 M / second

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