JEE Main 2022: Arrhenius Equation Calculation

chemca.in

Home Class XII Kinetics

Home › Class XII › Physical Chemistry › Chemical Kinetics › JEE Main 2022 Arrhenius Problem

Daily Challenge JEE Main 2022 (30 Jun, Shift 1)

Calculating Temperature from the Arrhenius Equation

Test your ability to carefully manipulate logarithms and units in the Arrhenius equation.

Question:

For the reaction $P \rightarrow B$, the values of frequency factor $A$ and activation energy $E_A$ are $4 \times 10^{13} \text{ s}^{-1}$ and $8.3 \text{ kJ mol}^{-1}$ respectively.

If the reaction is of first order, the temperature at which the rate constant is $2 \times 10^{-6} \text{ s}^{-1}$ is ________ $\times 10^{-1} \text{ K}$.

[Given: $\ln 10 = 2.3 \,;\, R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1} \,;\, \log 2 = 0.30$]

Enter your integer answer:

Detailed Solution

Step 1: Write down the Arrhenius Equation and Given Values

The Arrhenius equation is: $k = A e^{-\frac{E_a}{RT}}$

Taking the natural logarithm ($\ln$) on both sides gives:

$$ \ln k = \ln A - \frac{E_a}{RT} $$

Given values:

• $k = 2 \times 10^{-6} \text{ s}^{-1}$
• $A = 4 \times 10^{13} \text{ s}^{-1}$
• $E_a = 8.3 \text{ kJ mol}^{-1} = \mathbf{8.3 \times 10^3 \text{ J mol}^{-1}}$ (Must match R's units!)
• $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$

Unit Alert!

The most common mistake in Arrhenius problems is forgetting to convert Activation Energy ($E_a$) from $\text{kJ}$ to $\text{J}$. Since the gas constant $R$ is given in Joules ($8.3 \text{ J K}^{-1} \text{ mol}^{-1}$), $E_a$ MUST also be in Joules!

Step 2: Convert to Base-10 Logarithms

Since the problem provides values for $\log 2$ and $\ln 10$, we can convert the equation to base-10 log using the relationship $\ln x = 2.303 \log x$ (the problem approximates $2.303$ to $2.3$ via $\ln 10 = 2.3$).

$$ \ln k = \ln A - \frac{E_a}{RT} $$ $$ 2.3 \log k = 2.3 \log A - \frac{E_a}{RT} $$

Rearranging to solve for the temperature term:

$$ \frac{E_a}{RT} = 2.3 \log A - 2.3 \log k = 2.3 (\log A - \log k) $$ $$ \frac{E_a}{RT} = 2.3 \log \left( \frac{A}{k} \right) $$

Step 3: Substitute Values and Calculate

Substitute the given numbers into the rearranged equation:

$$ \frac{8.3 \times 10^3}{8.3 \times T} = 2.3 \log \left( \frac{4 \times 10^{13}}{2 \times 10^{-6}} \right) $$

Notice how perfectly $8.3$ cancels out on the left side! Now, simplify the log term on the right:

$$ \frac{10^3}{T} = 2.3 \log (2 \times 10^{19}) $$

Apply log rules: $\log(2 \times 10^{19}) = \log 2 + \log(10^{19}) = 0.30 + 19 = 19.3$

$$ \frac{1000}{T} = 2.3 \times 19.3 $$ $$ \frac{1000}{T} = 44.39 $$ $$ T = \frac{1000}{44.39} \approx 22.527 \text{ K} $$

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

$22.527 \text{ K} = 225.27 \times 10^{-1} \text{ K}$

Rounding to the nearest integer gives $225$.

Final Answer: 225

Review Arrhenius Equation

© 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