Question

Which of the following statement(s) is/are correct for zero order reaction?

• A. Quantity of the product formed is directly proportional to the time
• B. Larger the initial concentration of the reactant, greater the half-life period.
• C. If 50% of reaction takes place in 100 minutes, 75% reaction take place in 150 minutes
• D. Unit of rate constant is $s^{-1}$

Answer 1

Answer: (A), (B), (C)

If 50% of reaction takes place in 100 minutes, 75% reaction take place in 150 minutes

We know for zero order reaction,
$A\to X$
For (I):
$\text{Rate,}R=K[A{]}^0$
$-\frac{d\left[A\right]}{dt}=K[A{]}^0$
$-\frac{d\left[A\right]}{dt}=K\times 1$
$-d\left[A\right]=Kdt$
$\text{Product formed}\propto time$

For (II):
Half life is the time, at which the concentration becomes half of its initial concentration.
At $t=t_{\frac{1}{2}}$
$\left[A_t\right]=\frac{\left[A_o\right]}{2}$
we know for zero order reaction,
$[A{]}_t=[A_o]-Kt......eqn(1)$
Substituting $[A{]}_t$ value in equation (1), we get,
$\frac{[A{]}_o}{2}=[A{]}_o-K{t}_{\frac{1}{2}}.....eqn(2)t_{\frac{1}{2}}=\frac{[A{]}_o}{2K}{t}_{\frac{1}{2}}\propto [A_0]$

For (III): Using integrated law equation for zero order (equation 3)

At 50% of reaction,
$A_o-\frac{A_o}{2}=k\times \mathrm{100.....}\left(i\right)$

At 75% of reaction,
$A_o-\frac{A_o}{4}=k\times t.....\left(ii\right)$
Dividing $\left(i\right)by\left(ii\right)$
$\Rightarrow \frac{\frac{A_0}{2}}{\left(\frac{3A_0}{4}\right)}=\frac{k\times 100}{k\times t}$

$t=\frac{100\times 3}{2}=150minutes$

For (IV): Unit of rate constant
From eqaution (1)

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

Unit of K for zero order reaction is,
$mol{L}^{-1}{s}^{-1}$

Thus, statement a, b and c are correct.