Question

(a) For a reaction $A+B\to P$, the rate is given by

Rate$=k\left[A\right][B{]}^2$
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
(b) A first-order reaction takes 30 minutes for $50$ %completion. Calculate the time required for $90$% completion of this reaction.

Answer 1

a.

(i) For a reaction, $A+B\to P$

$R_1=k\left[A\right][B{]}^2$ ---------(i)

If the concentration of B is doubled,

$R_2=k\left[A\right][2B{]}^2$---------(ii)

On dividing (i) and (ii)

$\frac{R_1}{R_2}=\frac{k\left[A\right][B{]}^2}{k\left[A\right][2B{]}^2}$

$\frac{R_1}{R_2}=\frac{B^2}{4B^2}$

$R_2=4R_1$

The rate of reaction will be four times the initial rate.

(ii) If A is present in large excess, then the rate of the reaction will be independent of A and will depend only on the concentration of B. The overall rate of the reaction will be 2.

(b) $K$$=$$0.693/30$$=$$0.0231$

The time required to complete 90% reaction,

$t$$=$$\left(2.303/.0231\right)\times log\left(100/10\right)$$=$$100minutes$