$$2 \mathrm{NO}+2 \mathrm{H}_{2} \rightarrow \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}$$
The above reaction has been studied at $800^{\circ} \mathrm{C}$. The related data are given in the table below
| Reaction serial number | Initial Pressure of ${H_2}/kPa$ | Initial Pressure of $NO/kPa$ | Initial rate $\left( {{{ - dp} \over {dt}}} \right)/(kPa/s)$ |
|---|---|---|---|
| 1 | 65.6 | 40.0 | 0.135 |
| 2 | 65.6 | 20.1 | 0.033 |
| 3 | 38.6 | 65.6 | 0.214 |
| 4 | 19.2 | 65.6 | 0.106 |
The order of the reaction with respect to NO is ___________.
Answer (integer)
2
Solution
Let the rate of reaction ( $r$ ) is as
<br/><br/>
$\mathrm{r}=\mathrm{K}[\mathrm{NO}]^{n}\left[\mathrm{H}_{2}\right]^{\mathrm{m}}$
<br/><br/>
From $1^{\text {st }}$ data
<br/><br/>
$0.135=\mathrm{K}[40]^{\mathrm{n}} \cdot(65.6)^{\mathrm{m}}\quad\quad...(1)$
<br/><br/>
From $2^{\text {nd }}$ data
<br/><br/>
$0.033=\mathrm{K}(20.1)^{\mathrm{n}} \cdot(65.6)^{\mathrm{m}}\quad\quad...(2)$
<br/><br/>
On dividing equation (1) by equation (2)
<br/><br/>
$$
\begin{aligned}
&\frac{0.135}{0.033}=\left(\frac{40}{20.1}\right)^{n} \\\\
&4=(2)^{n} \\\\
&\therefore n=2 \\\\
&\therefore \text { Order of reaction w.r.t. NO is } 2 .
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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