JEE MAIN - Chemistry (2022 - 27th July Morning Shift - No. 15)
$$2 \mathrm{NO}+2 \mathrm{H}_{2} \rightarrow \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}$$
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
2
Explanation
Let the rate of reaction ( $r$ ) is as
$$ \mathrm{r}=\mathrm{K}[\mathrm{NO}]^{n}\left[\mathrm{H}_{2}\right]^{\mathrm{m}} $$
From $1^{\text {st }}$ data
$$ 0.135=\mathrm{K}[40]^{\mathrm{n}} \cdot(65.6)^{\mathrm{m}}\quad\quad...(1) $$
From $2^{\text {nd }}$ data
$$ 0.033=\mathrm{K}(20.1)^{\mathrm{n}} \cdot(65.6)^{\mathrm{m}}\quad\quad...(2) $$
On dividing equation (1) by equation (2)
$$ \begin{aligned} &\frac{0.135}{0.033}=\left(\frac{40}{20.1}\right)^{n} \\\\ &4=(2)^{n} \\\\ &\therefore n=2 \\\\ &\therefore \text { Order of reaction w.r.t. NO is } 2 . \end{aligned} $$
$$ \mathrm{r}=\mathrm{K}[\mathrm{NO}]^{n}\left[\mathrm{H}_{2}\right]^{\mathrm{m}} $$
From $1^{\text {st }}$ data
$$ 0.135=\mathrm{K}[40]^{\mathrm{n}} \cdot(65.6)^{\mathrm{m}}\quad\quad...(1) $$
From $2^{\text {nd }}$ data
$$ 0.033=\mathrm{K}(20.1)^{\mathrm{n}} \cdot(65.6)^{\mathrm{m}}\quad\quad...(2) $$
On dividing equation (1) by equation (2)
$$ \begin{aligned} &\frac{0.135}{0.033}=\left(\frac{40}{20.1}\right)^{n} \\\\ &4=(2)^{n} \\\\ &\therefore n=2 \\\\ &\therefore \text { Order of reaction w.r.t. NO is } 2 . \end{aligned} $$
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