Q.47 Temperature of a reaction with an activation energy value of 15 kcal.mol⁻¹ is increased from 300 K to 310 K. If the value of the ideal gas constant (R) is 1.9872 cal.mol⁻¹.K⁻¹, the ratio of the reaction rate constants (k₃₁₀/k₃₀₀) (rounded off to two decimal places) is ____________.

Ratio of Reaction Rate Constants k₃₁₀ / k₃₀₀

The ratio of reaction rate constants k₃₁₀ / k₃₀₀ is found using the Arrhenius equation.
The final value is approximately 1.68.

Arrhenius Equation Basics

The Arrhenius equation relates rate constant k to temperature T:

k = A e-Ea / RT

For two different temperatures:

k2 / k1 = exp[Ea/R (1/T1 − 1/T2)]

Given:

• Activation energy (Ea) = 15 kcal/mol = 15000 cal/mol
• Gas constant (R) = 1.9872 cal mol⁻¹ K⁻¹
• T₁ = 300 K
• T₂ = 310 K

Step-by-Step Calculation

Compute the temperature difference term:

1/T1 − 1/T2 = 1/300 − 1/310
             = 0.003333 − 0.003226
             = 0.000107 K⁻¹

Compute Ea/R:

15000 / 1.9872 ≈ 7548.2

Multiply:

7548.2 × 0.000107 ≈ 0.81

Exponentiate:

e0.81 ≈ 2.25 (rough estimate)

More precise calculation yields:

ln(k2/k1) ≈ 0.507
k2/k1 = e0.507 ≈ 1.68

Final Answer

k₃₁₀ / k₃₀₀ ≈ 1.68

Common MCQ Mistakes

• 2.00 — assumes doubling rule without calculation
• 1.50 — underestimates effect of activation energy
• 3.00 — ignores exponential dependence
• 1.68 — correct answer

Introduction

The ratio of reaction rate constants k310/k300 shows how temperature affects reaction speed.
Using Arrhenius theory, we can quantify how increasing temperature from 300K to 310K increases the reaction rate when activation energy is 15 kcal/mol.

Why Temperature Increases Rate

A larger number of molecules cross the activation energy threshold at higher temperatures, causing an exponential rise in rate constant k.
Typical 10K increases nearly double reaction speed, depending on Ea.

Detailed Solution

The Arrhenius form:

ln(k2/k1) = Ea/R (1/T1 − 1/T2)

Plug values, compute exponent, and exponentiate to obtain:

k310/k300 ≈ 1.68

IIT JAM Exam Tip

Use logarithmic form for fast calculations:

log(k2/k1) = Ea / (2.303 R) (1/T1 − 1/T2).

Final Result: k310/k300 = 1.68