Q.59 Chemistry · Section A ⭐ 360° Option Elimination Active

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NEET 2024 Chemistry Question 59 Solution

Topic: Chemical Kinetics · Official NTA Booklet Code T1

Question

Activation energy of any chemical reaction can be calculated if one knows the value of

(1)orientation of reactant molecules during collision.
(2)rate constant at two different temperatures.
(3)rate constant at standard temperature.
(4)probability of collision.

✓

Verified Against NTA Final Answer Key

Option (2) : rate constant at two different temperatures.

📚 Chemical Kinetics Easy

📋 Given Parameters & Problem Setup

Target quantity Activation energy (E_a) of a chemical reaction

Governing equation Arrhenius equation: k = A e^(-E_a / RT)

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The 30-Second Exam Hall Shortcut: ⚡ Two-Point Arrhenius Formula Rule

⏱️ Save ~30 seconds

ln(k2 / k1) = (E_a / R) · (1/T1 - 1/T2) -> Knowing rate constant at two temperatures gives E_a directly!

1 Recall the logarithmic two-temperature Arrhenius equation:

2 ln(k2 / k1) = (E_a / R) [ (T2 - T1) / (T1 · T2) ].

3 Gas constant R is a universal physical constant.

4 If rate constants k1 and k2 are determined at temperatures T1 and T2, E_a is the ONLY unknown variable.

5 Thus, E_a can be computed precisely from the rate constant at two different temperatures.

6 Option (2) in under 5 seconds!

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Why You Can 100% Trust This Shortcut: NCERT Chemistry · Chemical Kinetics (Temperature dependence of rate of a reaction and two-point Arrhenius calculation).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

The Differential and Logarithmic Arrhenius Form

The Arrhenius equation describes how the rate constant k of a reaction depends exponentially on absolute temperature T and activation energy Ea:

k = A e^{-\frac{E_a}{RT}} \implies \ln k = \ln A - \frac{E_a}{RT}

Step 2

Two-Point Formulation Eliminating Pre-Exponential Factor A

Writing the equation at two distinct absolute temperatures T1 and T2 with rate constants k1 and k2:

\ln k_1 = \ln A - \frac{E_a}{R T_1}, \quad \ln k_2 = \ln A - \frac{E_a}{R T_2}

Step 3

Subtraction and Direct Solution for Activation Energy

Subtracting the first equation from the second eliminates the unknown frequency factor A entirely:

\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) = \frac{E_a}{R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)

Step 4

Conclusion

Because R is known, experimental determination of the rate constant at two distinct temperatures permits exact calculation of Ea without prior knowledge of orientation factors or pre-exponential constant A.

E_a = \frac{R \cdot T_1 T_2}{T_2 - T_1} \ln\left(\frac{k_2}{k_1}\right)

⭐

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:

Relates to the steric probability factor P in collision theory, not the calculation of Ea.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing the collision steric factor with activation energy.

2 (2)

✓ 100% Correct

Verdict & Scientific Reason:

Standard Arrhenius two-point method eliminates A and yields Ea directly.

💡

Why this is airtight: None.

3 (3)

✗ Incorrect

Verdict & Scientific Reason:

At a single temperature, Ea cannot be solved without knowing pre-exponential factor A.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming single rate constant suffices without knowing A.

4 (4)

✗ Incorrect

Verdict & Scientific Reason:

Determines collision frequency Z, not the energetic barrier Ea.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing collision rate with kinetic activation barrier.

⚠️

NTA Trap Alert: Attempting to Determine Ea from a Single Temperature

At a single temperature T, the equation k = A exp(-Ea/RT) has two unknowns: A and Ea. You must have two temperatures to eliminate A!

🛡️ How to avoid the -1 mark: Remember: 2 unknowns (A, Ea) require 2 data points (k1 at T1, k2 at T2).

📖

Core NEET Syllabus Standard

Chemistry Core Physical Chemistry · Chemical Kinetics Standard Benchmark

Core Scientific Principle Tested:

"Activation energy Ea is quantitatively determined by evaluating the rate constants at two different temperatures: ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)."

Syllabus Focus: Temperature Dependence of Reaction Rates: Arrhenius Equation

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Two temperatures, two k’s in sight: Pre-exponential A drops out of the fight!

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NEET 2024 Chemistry Question 59 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Two-Point Arrhenius Formula Rule

Step-by-Step Rigorous Workflow & Mathematical Proof

The Differential and Logarithmic Arrhenius Form

Two-Point Formulation Eliminating Pre-Exponential Factor A

Subtraction and Direct Solution for Activation Energy

Conclusion

360° Option Elimination Engine Key Feature

NTA Trap Alert: Attempting to Determine Ea from a Single Temperature

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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