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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 concentrati n of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

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

Grade:12

2 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
7 years ago
The rate law for a reaction between the substances A and B is given by Rate = k[A]^n [B]^m
initial rate law is given by =k[A]^n[B]^m
final rate ,when concentration of A is doubled and concentration of B is halfved.
R2=K[2A]^m[B/2]^n
=kA^mB^n *2^m/2^n
Now R2/R1 =2^(m-n)
Gaurav
askIITians Faculty 164 Points
7 years ago
R = k[A]n[B]m
Doubling the concentration of A we get

R’ =k[2A]n[B/2]m = (2n-m)R
Now,
\frac{R'}{R} = \frac{2^n^-^m}{1}

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