Question

How long does it take for reactant concentration to decrease from $4\times {10}^{-4}mol{L}^{-1}$ to $2\times {10}^{-4}mol{L}^{-1}$ in a zero order reaction.
$\text{Rate constant of the reaction,}k=5\times {10}^{-8}mol{L}^{-1}{s}^{-1}$

• A. $7500s$
• B. $2500s$
• C. $4000s$
• D. $5000s$

Answer 1

The correct option is C $4000s$

Let us consider a zero order reaction:

$A\to B$
Rate Law for Zero Order Reaction is given as:
$\text{Rate,}R=k[A{]}^0$
$-\frac{d\left[A\right]}{dt}=k[A{]}^0$
$-\frac{d\left[A\right]}{dt}=k\times 1$
$-d\left[A\right]=kdt$
Taking limit on both side
$-\underset{[a{]}_0}{\overset{[A{]}_t}{\int }}d\left[A\right]=k\underset{0}{\overset{t}{\int }}dt$

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

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

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

$[A{]}_t=[A{]}_0-kt.......eqn\left(1\right)$
where $[A{]}_0$ is the initial concentration of Reactant A.

$K=\frac{[A{]}_0-[A{]}_t}{t}$

$t=\frac{A_0-A_t}{k}$

$t=\frac{(4\times {10}^{-4})-(2\times {10}^{-4})}{5\times {10}^{-8}}$

$\text{Time,}t=4000s$

Note:
Rate constant $k$ for any reaction depends on temperature only and not on concentration.