Q.89 Chemistry ⭐ 360° Option Elimination Active

NEET 2021 Chemistry Question 89 Solution

Topic: Chemical Kinetics · Official NTA Booklet Code M1

Question

The slope of Arrhenius Plot (ln k v/s 1/T) of first order reaction is −5×10³ K. The value of Ea of the reaction is. Choose the correct option for your answer. [Given R = 8.314 J K⁻¹ mol⁻¹]

✓

Verified Against NTA Final Answer Key

Option (1) : 41.5 kJ mol^-1

📚 Chemical Kinetics Easy

📋 Given Parameters & Problem Setup

Plot ln k vs 1/T

Slope -5 × 10^3 K

Gas Constant (R) 8.314 J K^-1 mol^-1

⚡

The 30-Second Exam Hall Shortcut: ⚡ Arrhenius Slope: Ea = -Slope × R

⏱️ Save ~20 seconds

For $\ln k$ vs $1/T$ , the slope is $-\frac{E_a}{R}$ . Therefore, $E_a = -\text{Slope} \times R = -(-5 \times 10^3\text{ K}) \times 8.314\text{ J K}^{-1}\text{mol}^{-1} = 41,570\text{ J mol}^{-1} = 41.57\text{ kJ mol}^{-1} \approx 41.5\text{ kJ mol}^{-1}$ ! Direct 10-second multiplication matching Option (1)!

1 Arrhenius equation in natural log:

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

2 Plot of

\ln k

\frac{1}{T}

has slope

m = -\frac{E_a}{R}

3 Equate:

-\frac{E_a}{R} = -5 \times 10^3\text{ K} \implies E_a = 5 \times 10^3 \times 8.314

4 Compute:

E_a = 41,570\text{ J/mol} = 41.57\text{ kJ/mol} \approx 41.5\text{ kJ/mol}

5 Match Option (1).

🛡️

Why You Can 100% Trust This Shortcut: Canonical Arrhenius equation temperature dependence from NCERT Chemical Kinetics.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Linear Form of Arrhenius Equation

The Arrhenius equation expresses the dependence of the rate constant $k$ on temperature $T$ : $k = A e^{-E_a / RT}$ Taking the natural logarithm ( $\ln$ ) of both sides:

\ln k = \ln A - \frac{E_a}{R} \left(\frac{1}{T}\right)

Step 2

Relating Slope to Activation Energy

Comparing this expression with the standard equation of a straight line $y = c + mx$ , where $y = \ln k$ and $x = \frac{1}{T}$ : - $y$ -intercept $= \ln A$ - $\text{Slope } m = -\frac{E_a}{R}$

Step 3

Calculating Ea in kJ/mol

Given slope $= -5 \times 10^3\text{ K}$ and $R = 8.314\text{ J K}^{-1}\text{mol}^{-1}$ : $-\frac{E_a}{R} = -5 \times 10^3\text{ K} \implies E_a = (5 \times 10^3\text{ K}) \times (8.314\text{ J K}^{-1}\text{mol}^{-1})$

E_a = 41570\text{ J mol}^{-1} = \frac{41570}{1000}\text{ kJ mol}^{-1} = 41.57\text{ kJ mol}^{-1} \approx 41.5\text{ kJ mol}^{-1}

⭐

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

✓ 100% Correct

Verdict & Scientific Reason:

Exact activation energy: $5 \times 10^3 \times 8.314 = 41.57\text{ kJ mol}^{-1} \approx 41.5\text{ kJ mol}^{-1}$ .

💡

Why this is airtight: None; precise kinetic derivation.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Result of multiplying by a factor of 2 ( $2 \times 41.5 = 83$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Accidental doubling of slope or rate constant.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Factor of 4 error.

⚠️

Student Fallacy (Why candidates pick this trap): Multiple multiplication mistakes.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Negative value (activation energy is strictly positive) with doubling error.

⚠️

Student Fallacy (Why candidates pick this trap): Carrying negative sign to activation energy.

⚠️

NTA Trap Alert: ln k vs. log10 k Slope Formula Trap

If the plot were $\log_{10} k$ vs $1/T$ , the slope would be $-\frac{E_a}{2.303 R}$ . The question explicitly specifies $\ln k$ (natural log), so the slope is simply $-\frac{E_a}{R}$ . Candidates who insert an unnecessary $2.303$ get erroneous values.

🛡️ How to avoid the -1 mark:

\ln k

1/T \implies

Slope

= -E_a/R

\log_{10} k

1/T \implies

Slope

= -E_a/(2.303 R)

📖

Core NEET Syllabus Standard

NCERT Chemistry · Chemical Kinetics Standard Benchmark

Core Scientific Principle Tested:

"A plot of ln k vs 1/T gives a straight line with slope equal to -Ea/R."

Syllabus Focus: Temperature Dependence of the Rate of a Reaction (Arrhenius Equation)

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Chemistry Question 89 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Arrhenius Slope: Ea = -Slope × R

Step-by-Step Rigorous Workflow & Mathematical Proof

Linear Form of Arrhenius Equation

Relating Slope to Activation Energy

Calculating Ea in kJ/mol

360° Option Elimination Engine Key Feature

NTA Trap Alert: ln k vs. log10 k Slope Formula Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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