Question 8 of 37intermediate🔧 ApplyNumerical3 marks

The rate of the chemical reaction doubles for an increase of 10K in absolute temperature from 298K. Calculate EaE_a.

Correct Answer

Given:

  • Initial temperature, T₁ = 298 K
  • Final temperature, T₂ = 298 + 10 = 308 K
  • Rate doubles, so k₂/k₁ = 2
  • Gas constant, R = 8.314 J K⁻¹ mol⁻¹

Formula: Using the Arrhenius equation: log(k₂/k₁) = Eₐ/(2.303R) × [(T₂ - T₁)/(T₁T₂)]

Calculation: log 2 = Eₐ/(2.303×8.3142.303 \times 8.314) × [(308 - 298)/(298×308298 \times 308)]

0.3010 = Eₐ/19.15 × [10/91784]

0.3010 = Eₐ/19.15×1.08919.15 \times 1.089 × 10⁻⁴

Eₐ = (0.3010×19.150.3010 \times 19.15)/(1.089×101.089 \times 10⁻⁴)

Eₐ = 52928.7 J mol⁻¹

Eₐ ≈\approx 52.93 kJ mol⁻¹

Exercise: Intext Questions | Q: 3.8 | (Page 24)
For More Understanding

Explanation

This is a standard NCERT problem applying the Arrhenius equation. When temperature increases by 10K from 298K, the rate doubles (k₂/k₁ = 2). The logarithmic form of Arrhenius equation relates rate constant ratio to activation energy and temperature. The calculated activation energy of approximately 53 kJ mol⁻¹ is a typical value for many chemical reactions.

Solution Steps

  1. Step 1: Identify given values - T₁ = 298K, T₂ = 308K, k₂/k₁ = 2

  2. Step 2: Write Arrhenius equation in logarithmic form: log(k₂/k₁) = Eₐ/(2.303R) × [(T₂-T₁)/(T₁T₂)]

  3. Step 3: Substitute values: log 2 = Eₐ/(2.303×8.3142.303 \times 8.314) × [10/(298×308298 \times 308)]

  4. Step 4: Solve for Eₐ: Eₐ = (0.3010×19.150.3010 \times 19.15)/(1.089×101.089 \times 10⁻⁴)

  5. Step 5: Calculate final answer: Eₐ = 52928.7 J mol⁻¹ ≈\approx 52.93 kJ mol⁻¹

PreviousNext