2NO + 2 H _ 2 → N _ 2 + 2 H _ 2 O The above reaction has been studied at 800 ° C . The related data are given in the table below Reaction serial number Initial pressure of H _ 2 / kPa Initial Pressure of NO / kPa

Correct answer: 0 The rate law for the reaction 2NO + 2 H _ 2 → N _ 2 + 2 H _ 2 O is r = k [ H _ 2 ] ^ m [NO] ^ n The powers of the concentration terms m, n are the order with respect to hydrogen and nitric oxide respectively. r _ 1 = [65 . 6] ^ m [40 . 0] ^ n = 0 . 135 r _ 2 = [65 . 6] ^ m [20 . 1] ^ n = 0 . 033 r _ 3 = [38 . 6] ^ m [65 . 6] ^ n = 0 . 214 r _ 4 = [19 . 2] ^ m [65 . 6] ^ n = 0 . 106 r _ 1 r _ 2 = [65 . 6] ^ m [40 . 0] ^ n [65 . 6] ^ m [20 . 1] ^ n = (0 . 135)/(0 . 033) = 4 ⇒ 2 ^ n = 4 ⇒ n = 2 Hence, the order with respect to NO is 2 .

JEE Main 202227 JulyMorning Shift

Exam: JEE Main
Year: 2022
Date: 27 July
Shift: Morning Shift
Subject: Chemistry
Chapter: Chemical Kinetics

JEE Main 2022 · 27 July · Morning Shift · Chemistry · Chemical Kinetics

2NO + 2 H _ 2 → N _ 2 + 2 H _ 2 O The above reaction has been...

2NO + 2{H}_{2} \to {N}_{2} + 2{H}_{2}O $</p><p>The above reaction has been studied at$ 800 ° C $. The related data are given in the table below</p><table border="1" cellpadding="1" cellspacing="1" style="width: 500px;"> <tbody> <tr> <td>Reaction serial number</td> <td>Initial pressure of$ {H}_{2} / kPa $</td> <td>Initial Pressure<br /> of$ NO / kPa $</td> <td>Initial rate \left(\dfrac{- dp}{dt}\right) / \left(kPa / s\right)</td> </tr> <tr> <td>$ 1 $</td> <td>$ 65 . 6 $</td> <td>$ 40 . 0 $</td> <td>$ 0 . 135 $</td> </tr> <tr> <td>$ 2 $</td> <td>$ 65 . 6 $</td> <td>$ 20 . 1 $</td> <td>$ 0 . 033 $</td> </tr> <tr> <td>$ 3 $</td> <td>$ 38 . 6 $</td> <td>$ 65 . 6 $</td> <td>$ 0 . 214 $</td> </tr> <tr> <td>$ 4 $</td> <td>$ 19 . 2 $</td> <td>$ 65 . 6 $</td> <td>$ 0 . 106 $</td> </tr> </tbody></table><p>The order of the reaction with respect to NO is____</p>$

Correct answer

Step-by-step solution

The rate law for the reaction $2NO + 2{H}_{2} \to {N}_{2} + 2{H}_{2}O$ is
$r = k{\left[{H}_{2}\right]}^{m}{\left[NO\right]}^{n}$
The powers of the concentration terms m, n are the order with respect to hydrogen and nitric oxide respectively.
${r}_{1} = {\left[65 . 6\right]}^{m}{\left[40 . 0\right]}^{n} = 0 . 135{r}_{2} = {\left[65 . 6\right]}^{m}{\left[20 . 1\right]}^{n} = 0 . 033{r}_{3} = {\left[38 . 6\right]}^{m}{\left[65 . 6\right]}^{n} = 0 . 214{r}_{4} = {\left[19 . 2\right]}^{m}{\left[65 . 6\right]}^{n} = 0 . 106$
$\dfrac{{r}_{1}}{{r}_{2}} = \dfrac{{\left[65 . 6\right]}^{m}{\left[40 . 0\right]}^{n}}{{\left[65 . 6\right]}^{m}{\left[20 . 1\right]}^{n}} = \dfrac{0 . 135}{0 . 033} = 4 ⇒ {2}^{n} = 4 ⇒ n = 2$
Hence, the order with respect to $NO$ is $2 .$

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