Question

The rate law for a reaction between the substances A and B is given by:
Rate $=k\left[A{]}^n\right[B{]}^m$
On doubting 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:

• A. $(m+n)$
• B. $(n-m)$
• C. $2^{(n-m)}$
• D. $\frac{1}{2^{(m+n)}}$

Answer 1

Answer: (C)

$2^{(n-m)}$

$V_o=k\left[A{]}_o^n\right[B{]}_o^m$

using $\left[A_f\right]=\left(1/2\right)[A{]}_o$ and $\left[B_f\right]=\left(1/2\right)[B{]}_o$

$V_f=k{.2}^{(n-m)}\left[A{]}_o^n\right[B_o{]}^m=2^{(n-m)}{V}_o$