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Q.

A reaction has a forward rate constant of 2.3×106s−1 and an equilibrium constant of 4.0×108. What is the rate constant for the reverse reaction?

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a

1.7×1028−1

b

9.2×1014s−1

c

5.8×10−3s−1

d

1.1×10−15s−1

answer is B.

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Detailed Solution

We know that,

K=KfKb

Here the rate constant;
⇒Kb=2.3×1064×108

K=5.8×10−3s−1

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