Question

In the reaction between $NO$ and $H_2$ the following data are obtained
Experiment 1 : $P_{H_2}$ = constant

(mm of Hg)	359	300	152
$-\frac{P_{H_2}}{dt}$	1.50	1.03	0.25

Experiment 2: $P_{NO}$= constant

$P_{H_2}$ (mm of Hg)	289	205	147
$-\frac{P_{H_2}}{dt}$	1.60	1.10	0.79

The orders with respect of $H_2$ and $NO$ are:

• A. 1 with respect to $NO$ and 2 with respect to $H_2$
• B. 2 with respect to $NO$ and 1 with respect to $H_2$
• C. 1 with respect to $NO$ and 3 with respect to $H_2$
• D. 2 with respect to $NO$ and 2 with respect to $H_2$

Answer 1

The correct option is B 2 with respect to $NO$ and 1 with respect to $H_2$

Let us consider the orders are $\alpha$ and $\beta$ with respect to $NO$ and $H_2$ respectively.

Expt 1.

$-\frac{P_{No}}{dt}=K_1{\left(P_{No}\right)}^{\alpha }$

$\Rightarrow \frac{359}{152}={\left(\frac{1.5}{0.25}\right)}^{\alpha }$

$\Rightarrow 2.36={\left(6\right)}^{\alpha }\Rightarrow 2.36={\left(2.36\right)}^2\Rightarrow \alpha =2$

Expt 2

$-\frac{P_{H_2}}{dt}=K_2.{\left(p_{H_2}\right)}^{\beta }$

now, $\frac{289}{147}={\left(\frac{1.6}{0.79}\right)}^{\beta }$

$\Rightarrow 1.97={\left(2.02\right)}^{\beta }\Rightarrow \beta \approx 1$

$\therefore$ The orders $1$ with $H_2$ and $2$ with respect to $NO$.