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

(a) The rate of order is n=2
(b) $\begin{array}{l}\mathrm{A}\to \mathrm{B}\\ \frac{-\mathrm{dA}}{\mathrm{dt}}=\frac{\mathrm{dB}}{\mathrm{dt}}=\mathrm{r}\end{array}$
$\frac{-\mathrm{dA}}{\mathrm{dt}}=$rate of consumption of ' A '
$\frac{\mathrm{dB}}{\mathrm{dt}}=$rate of formation of ' B '
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
$\begin{array}{r}{\mathrm{r}}^{\mathrm{\prime }}=\mathrm{k}[3\mathrm{A}{]}^2\\ {\mathrm{r}}^{\mathrm{\prime }}=9\mathrm{k}[\mathrm{A}{]}^2\\ {\mathrm{r}}^{\mathrm{\prime }}=9\mathrm{r}\end{array}$
Hence rate of formation of B will become 9 times.
(C) For zero order reaction
$[\mathrm{a}{]}_{\mathrm{t}}=[\mathrm{a}{]}_0-\mathrm{kt}$
$[\mathrm{a}{]}_{\mathrm{t}}=$Concentration of reactant at time ‘t’
$[\mathrm{a}{]}_0=$Initial concentration of the reactant
k = rate of constant
t = time taken