Question

The rate law of a reaction between the substance A and B is given by, rate=k[A]^n[B]^m. On doubling the concentration of A and making the volume of B half the ratio of new rate to the earlier rate of reaction will be?

Answer 1

Answer: Ratio is $2^{n-m}$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$A+B\rightarrow products$

$rate =k[A]^n[B]^m$

$rate =k\frac{C_a}{V}^n\times \frac{C_b}{V}^m$

On doubling the concentration of A, making the volume of B half.

$rate'=k\frac{2C_a}{V}^n\times \frac{C_b}{2V}^m$

$rate'=k\frac{2^nC_a^n}{V^n}\times \frac{C_b^m}{2^mV^m}$

Thus $rate'=2^{n-m}rate$

$\frac{rate'}{rate}=2^{n-m}$

Answer 2

Answer:

Explanation:rate α k[Ca/Va]^n [Cb/Vb]^m

If Ca is doubled and Vb is half then

Rate α k[2Ca/Va]^n [Cb/(Vb/2)]^m

So answer is 2^n+m