JEE Main 2021: First Order Rate Constant

chemca.in

Home Class XII Kinetics

Home › Class XII › Physical Chemistry › Chemical Kinetics › JEE Main 2021 Problem

Daily Challenge JEE Main 2021 (22 Jul, Shift 2)

First Order Rate Constant Calculation

Apply the integrated rate law for a first-order reaction and manage unit conversions.

Question:

$$ N_2O_5(g) \rightarrow 2NO_2(g) + \frac{1}{2}O_2(g) $$

In the above first order reaction the initial concentration of $N_2O_5$ is $2.40 \times 10^{-2} \text{ mol L}^{-1}$ at $318 \text{ K}$.

The concentration of $N_2O_5$ after $1 \text{ hour}$ was $1.60 \times 10^{-2} \text{ mol L}^{-1}$.

The rate constant of the reaction at $318 \text{ K}$ is ________ $\times 10^{-3} \text{ min}^{-1}$. (Nearest integer)

[Given: $\log 3 = 0.477 \,,\, \log 5 = 0.699$]

Enter your integer answer:

Detailed Solution

Unit Check: The problem gives the time in "hours" but asks for the rate constant in "$\text{min}^{-1}$". Always ensure your time units match the requested output before calculating!
$t = 1 \text{ hour} = \mathbf{60 \text{ min}}$

Step 1: Write the Integrated Rate Equation

For a first-order reaction, the integrated rate law is:

$$ k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} \right) $$

Given values:

• $[A]_0$ (Initial conc.) = $2.40 \times 10^{-2} \text{ mol L}^{-1}$
• $[A]_t$ (Final conc.) = $1.60 \times 10^{-2} \text{ mol L}^{-1}$
• $t$ = $60 \text{ min}$

Step 2: Substitute and Simplify

Substitute the given values into the equation:

$$ k = \frac{2.303}{60} \log \left( \frac{2.40 \times 10^{-2}}{1.60 \times 10^{-2}} \right) $$

The $10^{-2}$ terms cancel out. Simplify the fraction:

$$ k = \frac{2.303}{60} \log \left( \frac{2.4}{1.6} \right) = \frac{2.303}{60} \log \left( \frac{3}{2} \right) $$

Step 3: Calculate using Log Properties

Use the property $\log(a/b) = \log a - \log b$:

$$ k = \frac{2.303}{60} (\log 3 - \log 2) $$

We are given $\log 3 = 0.477$. We also know $\log 2 \approx 0.301$ (or $0.30$ as often approximated in these problems, though using $0.301$ is slightly more precise if you remember it).

$$ k = \frac{2.303}{60} (0.477 - 0.301) $$ $$ k = \frac{2.303}{60} (0.176) $$ $$ k = \frac{0.4053}{60} $$ $$ k \approx 0.006755 \text{ min}^{-1} $$

The question asks for the answer in the format $Y \times 10^{-3} \text{ min}^{-1}$.

$0.006755 = \mathbf{6.755} \times 10^{-3} \text{ min}^{-1}$

Rounding to the nearest integer gives $7$.

Final Answer: 7

Exam Tip: Log Values

Always try to memorize the basic log values for competitive exams, even if some are given:

• $\log 2 \approx 0.301$
• $\log 3 \approx 0.477$
• $\log 5 \approx 0.699$ (which is $1 - \log 2$)

Review First Order Kinetics

© 2026 chemca.in. Empowering JEE & NEET Aspirants.

🏠 Home 📝 Quiz 📘 Revision Notes ⬇ Downloads 📚 Class Notes ▶ Video Lectures

Get new posts by email

Powered by

📚 Also Read

Lecture Notes