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Q.64 Chemistry ⭐ 360° Option Elimination Active
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NEET 2026 Chemistry Question 64 Solution

Topic: Chemical Kinetics · Official NTA Booklet Code 50
NEET 2026 Chemistry Question 64: Arrhenius Plot Intercept & Calculation of Target Temperature

Question

For an elementary chemical reaction, the Arrhenius plot (ln k vs 1/T) is given below. If the energy of activation is 6.64 kJ mol⁻¹ and R = 8.3 J K⁻¹ mol⁻¹, the temperature at which the rate constant becomes e² min⁻¹, is

✓
Verified Against NTA Final Answer Key
Option (3) : 200 K
📚 Chemical Kinetics Medium
📋 Given Parameters & Problem Setup
Arrhenius Plot ln k vs 1/T with y-intercept = 6
Activation Energy (E_a) 6.64 kJ mol⁻¹ = 6640 J mol⁻¹
Gas Constant (R) 8.3 J K⁻¹ mol⁻¹
Target Rate Constant (k) e² min⁻¹ (ln k = 2)
⚡

The 30-Second Exam Hall Shortcut: ⚡ The Arrhenius Straight Line Trick

⏱️ Save ~60 seconds

ln k = ln A - (E_a / RT). Intercept is ln A = 6 -> When ln k = 2, E_a / RT MUST equal 4!

1 From the graph, at 1/T = 0, y = ln A = 6.
2 Target: k = e² ⇒ ln k = ln(e²) = 2.
3 Substitute: 2 = 6 - (E_a / RT) ⇒ E_a / RT = 4.
4 Solve for T: T = E_a / (4R) = 6640 / (4 × 8.3) = 6640 / 33.2 = 200 K!
5 Option (3) is locked in 15 seconds!
🛡️
Why You Can 100% Trust This Shortcut: Direct mathematical deduction from the logarithmic Arrhenius equation ln k = ln A - E_a/(RT).
📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Formulate the Linear Arrhenius Equation

The Arrhenius equation describes how the rate constant depends on absolute temperature: k = A e^{-E_a / RT} Taking natural logarithm on both sides: ln k = ln A - (E_a / R)(1/T)

ln⁡k=ln⁡A−EaR(1T)\ln k = \ln A - \frac{E_a}{R}\left(\frac{1}{T}\right)
  • • y = mx + c with y = ln k, x = 1/T, slope m = -E_a/R, and intercept c = ln A.
  • • From the graph, the y-intercept is clearly 6 ⇒ ln A = 6.
Step 2

Evaluate the Target Condition

We need the temperature at which the rate constant becomes k = e² min⁻¹.

ln⁡k=ln⁡(e2)=2\ln k = \ln(e^2) = 2
  • • Substitute ln k = 2 and ln A = 6 into the linear equation:
  • • 2 = 6 - \frac{E_a}{RT} \implies \frac{E_a}{RT} = 4
Step 3

Calculate the Absolute Temperature T

Convert E_a to joules: E_a = 6.64 kJ mol⁻¹ = 6640 J mol⁻¹. Substitute R = 8.3 J K⁻¹ mol⁻¹.

T=Ea4R=6640 J mol−14×8.3 J K−1 mol−1=664033.2=200 KT = \frac{E_a}{4R} = \frac{6640\text{ J mol}^{-1}}{4 \times 8.3\text{ J K}^{-1}\text{ mol}^{-1}} = \frac{6640}{33.2} = 200\text{ K}
  • • Temperature = 200 K.
  • • Matches Option (3).
⭐

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 A
✗ Incorrect
Verdict & Scientific Reason:

Incorrect. Result of dividing by incorrect integer denominator.

⚠️
Student Fallacy (Why candidates pick this trap): Miscalculating the difference between 6 and 2.
2 B
✗ Incorrect
Verdict & Scientific Reason:

Incorrect. Arithmetic slip in division.

⚠️
Student Fallacy (Why candidates pick this trap): Estimating without working out 6640 / 33.2.
3 C
✓ 100% Correct
Verdict & Scientific Reason:

Exact mathematical solution: T = 6640 / 33.2 = 200 K.

💡
Why this is airtight: None; clean algebraic and graphical execution.
4 D
✗ Incorrect
Verdict & Scientific Reason:

Incorrect. Arises from using E_a / (RT) = 2 instead of 4.

⚠️
Student Fallacy (Why candidates pick this trap): Setting E_a / RT equal to ln k instead of (ln A - ln k).
⚠️

NTA Trap Alert: Forgetting to Convert kJ to J

Students frequently write 6.64 / (4 × 8.3) = 0.2 K because they forget to multiply 6.64 kJ by 1000 to match R in J/(mol K).

🛡️ How to avoid the -1 mark: Always match units: R is in Joules, so E_a MUST be converted from kJ to Joules (6640 J).
📖

Core NEET Syllabus Standard

NEET Chemistry Core Syllabus · Chemical Kinetics Standard Benchmark
Core Scientific Principle Tested:

"In a plot of ln k vs 1/T, the y-intercept is ln A and the slope is -E_a/R."

Syllabus Focus: Temperature dependence of rate of reaction - Arrhenius equation
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗
ln A is 6, ln k is 2, the drop is 4: T = 6640 / 33.2 = 200 K for sure!

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