Consider the following data for the reaction: 2\, HI(g) rightarrow H 2(g) + I 2(g) Experiment [ HI ] (mol L -1 ) Rate (mol L -1 s -1 ) --- --- --- 1 0.005 7.5 × 10 -4 2 0.010 3.0 × 10 -3 3 0.020 1.2 × 10 -2 Determine the order of the reaction with respect to HI .

Chemistry · Chemical Kinetics

JEE Main 2024 — 27 January, Shift 1 — Question 74

Consider the following data for the reaction:

2\,\mathrm{HI(g)} \rightarrow \mathrm{H_2(g)} + \mathrm{I_2(g)}

Experiment	$\left[\mathrm{HI}\right]$ (mol L $^{-1}$ )	Rate (mol L $^{-1}$ s $^{-1}$ )
1	0.005	$7.5 \times 10^{-4}$
2	0.010	$3.0 \times 10^{-3}$
3	0.020	$1.2 \times 10^{-2}$

Determine the order of the reaction with respect to $\mathrm{HI}$ .

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

For the reaction: $2\,\mathrm{HI(g)} \rightarrow \mathrm{H_2(g)} + \mathrm{I_2(g)}$ .

When the concentration of HI is doubled from $0.005$ to $0.01\ \text{mol L}^{-1}$ , the rate increases from $7.5 \times 10^{-4}$ to $3.0 \times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}$ , i.e. four times. When $\left[\mathrm{HI}\right]$ is doubled again from $0.01$ to $0.02\ \text{mol L}^{-1}$ , the rate increases from $3.0 \times 10^{-3}$ to $1.2 \times 10^{-2}\ \text{mol L}^{-1}\text{s}^{-1}$ , again four times.

Using the rate law, $\text{Rate} \propto [\mathrm{HI}]^n$ , we get $2^n = 4 \Rightarrow n = 2$ .

Hence, the order of the reaction is 2.

Answer key and solution verified before publishing.

Practise Chemical Kinetics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 27 January, Shift 1
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Molecularity and Practical Methods to Determine Order of Reaction