Question

In certain reaction, 10% of the reactant decomposes in one hour, 20% in two hours, 30% in three hours and so on. The dimensions of the rate constant is

• A. $hou{r}^{-}{}^1$
• B. $mol{L}^{-}{}^1{s}^{-}{}^1$
• C. $Lmo{l}^{-1}{s}^{-1}$
• D. $mol{s}^{-}{}^1$

Answer 1

The correct option is B $mol{L}^{-}{}^1{s}^{-}{}^1$

The given reaction is a zero order reaction because on plotting the percent concentration of reactant left after decomposition of reactants vs time, we will obtain a straight line.

For a zero order reaction, concentration versus time graph is a straight line with a nagative slope.

Consider a zero order reaction,
$\text{Rate,}R=K[A{]}^0$
$-\frac{d\left[A\right]}{dt}=K[A{]}^0$
$-\frac{d\left[A\right]}{dt}=K\times 1$
$-d\left[A\right]=Kdt$
Taking limit on both side
$-\underset{[a{]}_0}{\overset{[A{]}_t}{\int }}d\left[A\right]=K\underset{0}{\overset{t}{\int }}dt$

when,
$time=0,A=[A{]}_o$
$time=t,A=[A{]}_t$

$-[A{]}_{[A{]}_o}^{[A{]}_t}=K[t{]}_0^t$

$[A{]}_0-[A{]}_t=Kt$

$[A{]}_t=[A{]}_0-Kt$

Its in $y=-mx+c$ form.

Hence,
Order of the given reaction is zero
$\text{Rate}=\text{Rate constant}=\frac{dx}{dt}$
$\text{units of K}=mol{L}^{-1}{s}^1$

Generally, decomposition of gases on metal surface at high concentration follows zero order kinetics.
1. $2N{H}_3\left(g\right)\stackrel{Pt}{⟶}{N}_2\left(g\right)+3H_2\left(g\right)$
2. $2HI\left(g\right)\stackrel{Au}{⟶}{H}_2\left(g\right)+I_2\left(g\right)$
3. $2P{H}_3\left(g\right)\stackrel{Ni}{⟶}2P\left(g\right)+3H_2\left(g\right)$
4. $H_2\left(g\right)+C{l}_2\left(g\right)\stackrel{hv}{⟶}2HCl\left(g\right)$