Q87 · CSIR-NET Chemistry, December 2013

Paper: CSIR-NET December 2013 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Chemical Kinetics – General · Marks: 2 · Difficulty: Medium

In the reaction between NO and $\mathrm{H}_{2}$ the following data are obtained Experiment I: $\mathrm{P}_{\mathrm{H}_{2}}$ constant
$\mathrm{P}_{\mathrm{NO}}$ (mm of Hg)359300152
$\frac{-\mathrm{dP}_{\mathrm{NO}}}{\mathrm{dt}}$1.591.030.25
Experiment II : $\mathrm{P}_{\mathrm{NO}}=$ constant
$\mathrm{P}_{\mathrm{H}_{2}}$ (mm of Hg)289205147
$\frac{-\mathrm{dP}_{\mathrm{H}_{2}}}{\mathrm{dt}}$1.601.100.79
The orders with respect to $\mathrm{H}_{2}$ and NO are
(a)1 with respect to NO and 2 with respect to $\mathrm{H}_{2}$
(b)2 with respect to NO and 1 with respect to $\mathrm{H}_{2}$
(c)1 with respect to NO and 3 with respect to $\mathrm{H}_{2}$
(d)2 with respect to NO and 2 with respect to $\mathrm{H}_{2}$
Answer
Answer: B ✓ checked by 4AB · confidence high

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation
Expt I: (359/300)^n = 1.59/1.03 ⇒ n ≈ 2 (and (300/152)² ≈ 4.1 = 1.03/0.25). Expt II: (289/205)^m = 1.60/1.10 ⇒ m ≈ 1. Second order in NO, first in H₂.

Study loop for Chemical Kinetics

1. Practise the PYQsPrevious-year questions, with answers

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