$2 \mathrm{NO}+2 \mathrm{H}_{2} \rightarrow \mathrm{~N}_{2}+2 \mathrm{H}_{2} \mathrm{O}$ 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 $\mathrm{H}_{2}$ / kPa | Initial Pressure of NO/ kPa | Initial rate $\left(\frac{-\mathrm{dp}}{\mathrm{dt}}\right) /(\mathrm{kPa} / \mathrm{s})$ | |---|---|---|---| | 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 $\_\_\_\_$

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Chemical Kinetics

2 NO + 2 H_{2} \to N_{2} + 2 H_{2} O

The above reaction has been studied at

80 0^{\circ} 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 $(\frac{- dp}{dt}) / (kPa / s)$
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

To find the order with respect to NO, compare experiments where only the pressure of NO changes while

H X_{2}

pressure remains constant.

Compare experiments 1 and 2:

When

[NO]

is halved, the rate becomes

\frac{1}{4}

of the original.

This means:

(\frac{1}{2})^{n} = \frac{1}{4}

Solving:

n = 2

Therefore, the order with respect to NO is 2.

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Integer answer