Question

Initial concentration of the reactant is $1.0M$. The concentration becomes $0.9M,0.8M$ and $0.7M$ in $2$ hours, $4$ hours and $6$ hours respectively. Then the order of reaction is:

• A. $2$
• B. $1$
• C. $Zero$
• D. $3$

Answer 1

The correct option is C $Zero$

Here initial concentration is $0.1M=A_1$

Other time and corresponding concentration are given as

Time	Cocentration
$t_1=0H$	$A_1=1.0M$
$t_2=2H$	$A_2=0.9M$
$t_2=4H$	$A_3=0.8M$
$t_4=6H$	$A_4=0.7M$

Using the relation as:

$\frac{t_{final}-t_{initial}}{A_{initial}-A_{final}}=K$

Where $k=\text{constant}$

$\ast \frac{t_2-t_1}{A_1-A_2}=\frac{2-0}{1-0.9}=\frac{2}{0.1}=\frac{2\times 10}{1}=20=K_1$

$\ast \frac{t_3-t_1}{A_1-A_3}=\frac{4-0}{1-0.8}=\frac{4}{0.2}=\frac{4\times 10}{2}=20=K_2$

$\ast \frac{t_4-t_1}{A_1-A_4}=\frac{6-0}{1-0.7}=\frac{6}{0.3}=\frac{6\times 10}{3}=20=K_3$

Since $K_1=K_2=K_3=20$ all are constant

Hence, by the relations:

$A_0=A+Kt\to \text{for zero order}$

$A_0=A_2+K{t}_2.....\left(i\right)$

$A_0=A_1+K{t}_1.....\left(ii\right)$

Subtracting $\left(ii\right)$ from $\left(i\right)$

$A_0=A_2+K\cdot t_2$

$-A_0=-A_1+\left(-k{t}_1\right)$

$A_2-A_1=K\left[t_1-t_2\right]$

Or $K=\frac{t_2-t_1}{A_1-A_2}=\text{constant}$

So, it is type of zero order.