Question

In a certain reaction, $10$% of the reactant decomposes in one hour, $20$% in one two, $30$% in three hour, and so on. The dimensions of the rate constant is:
A. ${\text{hou}}{{\text{r}}^{ - 1}}$
B. ${\text{mole}}\,\,\,{\text{lite}}{{\text{r}}^{ - 1}}{\text{se}}{{\text{c}}^{ - 1}}$
C.$\,{\text{liter}}\,\,{\text{mol}}{{\text{e}}^{ - 1}}\,{\text{se}}{{\text{c}}^{ - 1}}$
D.$\,{\text{mol}}\,\,{\text{se}}{{\text{c}}^{ - 1}}$

Answer 1

We will determine the order of the decomposition. For this we will use rate constant formula for zero, then first and second order reaction and fins in which order reaction the rate constant comes constant. The order does not change with concentration of the reactant. After determining order we can determine the unit of rate constant by using rate constant formula.

Complete answer:
The rate constant formula for zero order reaction is as follows:
${\text{x}}\,{\text{ = }}\,{\text{kt}}$
Where,
${\text{x}}$is the concentration of reactant at time t.
${\text{k}}$is rate constant.
${\text{t}}$is the time.
We will rearrange the equation for rate constant and determine the constant.
$\Rightarrow{\text{k}} = \,\dfrac{{\text{x}}}{{\text{t}}}\,\,$
$10$% of the reactant decomposes in one hour so, rate constant is,
$\Rightarrow {\text{k}} = \,\dfrac{{10}}{{100 \times 1}}\,\,$
$\Rightarrow {\text{k}} = \,0.10\,$
$20$% of the reactant decomposes in two hour so, rate constant is,
$\Rightarrow {\text{k}} = \,\dfrac{{20}}{{100 \times 2}}\,\,$
$\Rightarrow {\text{k}} = \,0.10\,$
$30$% of the reactant decomposes in three hour so, rate constant is,
$\Rightarrow {\text{k}} = \,\dfrac{{30}}{{100 \times 3}}\,\,$
$\Rightarrow {\text{k}} = \,0.10\,$

So, the value of rate constant remains constant with respect to the time so the given decomposition follows zero order.
For zero order reaction the unit of rate constant is as follows:
The unit of concentration is mol/L. The time can be taken as minute or sec.
${\text{k = }}\,\dfrac{{{\text{mole/L}}}}{{{\text{sec}}}}\,\,$
$\,{\text{k = }}\,{\text{mole}}\,\,\,{\text{lite}}{{\text{r}}^{ - 1}}{\text{se}}{{\text{c}}^{ - 1}}$
So, the dimensions of the rate constant is${\text{mole}}\,\,\,{\text{lite}}{{\text{r}}^{ - 1}}{\text{se}}{{\text{c}}^{ - 1}}$.

Therefore, option (C) ${\text{mole}}\,\,\,{\text{lite}}{{\text{r}}^{ - 1}}{\text{se}}{{\text{c}}^{ - 1}}$ is correct.

Note: Primary batteries are used only for once. They become dead after one use. Example of the primary battery is leclanche cell and mercury cell. The primary batteries are used in clocks. Secondary batteries are used in inverters. In a leclanche battery, zinc works as an anode, and carbon wrapped with magnesium oxide works as a cathode. In mercury batteries, zinc-mercury amalgam works as anode carbon with a paste of mercury oxide working as a cathode.