Answer the following questions (Do any two):
(a) Identify the order of the reaction from the following unit for its rate constant: L/mol/s
(b) The conversion of the molecules A to B follow second order kinetics. If the concentration of A is increased to three times, how will it affect the rate of concentration of B?
(c) Write the expression of integrated rate equation for zero order reaction?

(a) The unit of second order rate constant is L mol^–1 s^–1.

Hence, the order of the reaction, n=2.
$$\begin{gathered} \left(\mathrm{b}\right) \mathrm{A}\to \mathrm{B} \\ \begin{array}{l}\frac{-\mathrm{dA}}{\mathrm{dt}}=\frac{\mathrm{dB}}{\mathrm{dt}}=\mathrm{r}\\ \frac{-\mathrm{dA}}{\mathrm{dt}}=\text{ rate of consumption of 'A' }\\ \frac{\mathrm{dB}}{\mathrm{dt}}=\text{ rate of formation of ' }\mathrm{B}\text{ ' }\end{array} \end{gathered}$$
Since given conversion follows 2^nd order, rate law could be written as following
$\mathrm{r}=\mathrm{k}[\mathrm{A}{]}^2$
If concentration of ' A ' made 3 times
$\begin{array}{l}{\mathrm{r}}^{\mathrm{\prime }}=\mathrm{k}[3\mathrm{A}{]}^2\\ {\mathrm{r}}^{\mathrm{\prime }}=9\mathrm{k}[\mathrm{A}{]}^2\\ {\mathrm{r}}^{\mathrm{\prime }}=9\mathrm{r}\end{array}$
Hence rate of formation of B will become 9 times.
(C) For zero order reaction
$[\mathrm{a}{]}_{\mathrm{t}}=[\mathrm{a}{]}_0-\mathrm{kt}$
[a]_t = Concentration of reactant at time ‘t’
[a]_o = Initial concentration of the reactant
k = rate of constant
t = time taken