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The rate law for a reaction between the substances A and B is given by Rate = k [ A ]^ n [ B ]^ m On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

Options

  1. A(m+n)
  2. B(n-m)
  3. C2^ ( n - m )
  4. D1 2^ ( m + n )

Correct answer

C. 2^ ( n - m )

Step-by-step solution

Rate ₁= k [ A ]^ n [ B ]^ m ; Rate ₂= k [2 ~A ]^ n [1 / 2 ~B ]^ m Rate ₂ Rate ₁ = k [2 ~A ]^n[1 / 2 ~B ]^m k [ A ]^n[ ~B ]^m =[2]^n[1 / 2]^m=2^ n 2^ -m =2^ n - m

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