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 23.1 minutes for 50% completion. Calculate the time required for 75% completion of this reaction. (log 2 =0.301, log 3 =0.4771, log 4 =0.6021)

Answer 1

(a) (i) Rate will become four times because rate $\propto [B{]}^2$.
(i) 2

(b) Time taken for completion of 50% of a reaction is half life period.
$\therefore t_{1/2}=23.1$ minutes

Now, $t_{1/2}=\frac{0.693}{k}$

or $k=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{23.1}=0.03mi{n}^{-1}$

For 75% completion of reaction
x =0.75 a

$\therefore t=\frac{2.303}{k}log\frac{a}{a-x}$

$=\frac{2.303}{0.03}log\frac{a}{a-0.75a}$

$=\frac{2.303}{0.03}log\frac{a}{0.25a}$

$=\frac{2.303}{0.03}log\frac{100}{25}$

$=\frac{2.303}{0.03}log4$

$=\frac{2.303}{0.03}\times 0.6021=46.2$ minutes