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Q.96 Chemistry · Section B ⭐ 360° Option Elimination Active
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NEET 2024 Chemistry Question 96 Solution

Topic: Chemical Kinetics · Official NTA Booklet Code T1
NEET 2024 Chemistry Question 96: Calculation of Activation Energy from Rate Quadrupling Across a Temperature Interval

Question

The rate of a reaction quadruples when temperature changes from 27°C to 57°C. Calculate the energy of activation. Given R = 8.314 J K⁻¹ mol⁻¹, log 4 = 0.6021

✓
Verified Against NTA Final Answer Key
Option (3) : 38.04 kJ/mol
📚 Chemical Kinetics Medium
📋 Given Parameters & Problem Setup
Initial temperature (T1) 27°C = 300.15 K ≈ 300 K
Final temperature (T2) 57°C = 330.15 K ≈ 330 K
Rate factor (k2 / k1) 4 (quadruples)
Gas constant (R) 8.314 J K^-1 mol^-1
Given log 4 0.6021
⚡

The 30-Second Exam Hall Shortcut: ⚡ Arrhenius Two-Temperature Factor Shortcut

⏱️ Save ~45 seconds

log(k2/k1) = [E_a / (2.303 R)] · [ΔT / (T1 · T2)]. 0.6021 = [E_a / 19.147] · [30 / (300 × 330)]. E_a = 0.6021 × 19.147 × 3300 = 38040 J = 38.04 kJ/mol!

1 Arrhenius formula: log10(k2 / k1) = (E_a / 2.303 R) × [ (T2 - T1) / (T1 · T2) ].
2 Convert temperatures to Kelvin: T1 = 27 + 273 = 300 K; T2 = 57 + 273 = 330 K.
3 Temperature difference: ΔT = 330 - 300 = 30 K. Denominator product: T1 · T2 = 300 × 330 = 99,000.
4 Bracket factor: 30 / 99,000 = 1 / 3300.
5 Constant: 2.303 × R = 2.303 × 8.314 = 19.147 J K⁻¹ mol⁻¹.
6 Rearrange for E_a: E_a = log(4) × 19.147 × 3300 = 0.6021 × 19.147 × 3300 = 38,043 J/mol = 38.04 kJ/mol.
7 Option (3) in 25 seconds!
🛡️
Why You Can 100% Trust This Shortcut: NCERT Chemistry · Chemical Kinetics (Arrhenius equation and quantitative calculation of activation energy).
📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Two-Point Logarithmic Form of Arrhenius Equation

The relationship between rate constants k1, k2 at absolute temperatures T1, T2 is given by:

log⁡10(k2k1)=Ea2.303 R(T2−T1T1T2)\log_{10}\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303\, R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)
Step 2

Substitution of Given Physical Quantities

Given: - k2 / k1 = 4 ⟹ log10(4) = 0.6021 - T1 = 27°C = 300 K - T2 = 57°C = 330 K - R = 8.314 J K⁻¹ mol⁻¹ - 2.303 × R = 19.147 J K⁻¹ mol⁻¹

0.6021=Ea19.147(330−300300×330)=Ea19.147(3099000)=Ea19.147(13300)0.6021 = \frac{E_a}{19.147} \left(\frac{330 - 300}{300 \times 330}\right) = \frac{E_a}{19.147} \left(\frac{30}{99000}\right) = \frac{E_a}{19.147} \left(\frac{1}{3300}\right)
Step 3

Solving for Activation Energy (Ea)

Cross-multiplying to isolate Ea:

Ea=0.6021×19.147×3300=38043.6  J/molE_a = 0.6021 \times 19.147 \times 3300 = 38043.6\;\text{J/mol}
Step 4

Conversion to Kilojoules per Mole (kJ/mol)

Dividing by 1000 to convert Joules to Kilojoules:

Ea=38043.61000  kJ/mol=38.04  kJ/molE_a = \frac{38043.6}{1000}\;\text{kJ/mol} = 38.04\;\text{kJ/mol}
⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan
1 (1)
✗ Incorrect
Verdict & Scientific Reason:

Off by a factor of 10 due to decimal error.

⚠️
Student Fallacy (Why candidates pick this trap): Dividing by 10,000 instead of 1,000.
2 (2)
✗ Incorrect
Verdict & Scientific Reason:

Represents 38040 divided by 10 instead of 1000.

⚠️
Student Fallacy (Why candidates pick this trap): Units confusion.
3 (3)
✓ 100% Correct
Verdict & Scientific Reason:

Exact mathematical calculation: 38,044 J/mol = 38.04 kJ/mol.

💡
Why this is airtight: None.
4 (4)
✗ Incorrect
Verdict & Scientific Reason:

Off by a factor of 10.

⚠️
Student Fallacy (Why candidates pick this trap): Dividing by 100 instead of 1000.
⚠️

NTA Trap Alert: Power of 10 Decimal Shift Trap

The options all share the digits "3804" with varying powers of 10: 3.80, 38.04, 380.4, 3804. Watch the division: 30 / 99000 = 1 / 3300, and converting J to kJ divides by 1000!

🛡️ How to avoid the -1 mark: Always check: 0.6 × 19 ≈ 11.5; 11.5 × 3.3 ≈ 38 kJ/mol.
📖

Core NEET Syllabus Standard

Chemistry Core Physical Chemistry · Chemical Kinetics Standard Benchmark
Core Scientific Principle Tested:

"Activation energy is determined from the two-temperature Arrhenius equation: log(k2/k1) = Ea / (2.303 R) · [ΔT / (T1 T2)]."

Syllabus Focus: Temperature Dependence of Reaction Rates: Arrhenius Activation Energy
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗
Multiply point-six by nineteen and three-point-three: thirty-eight point zero-four is the key!

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