Question 37 of 37intermediate🔧 ApplyNumerical3 marks

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Correct Answer

The activation energy of the reaction is 52.8 kJ mol⁻¹.

Exercise: EXERCISES | Q: 3.30 | (Page 28)
For More Understanding

Explanation

This numerical problem applies the Arrhenius equation to calculate activation energy when the rate constant changes with temperature. The context provides the logarithmic form of the Arrhenius equation: ln k = -Ea/RT + ln A. By writing this equation for two different temperatures and subtracting, we can eliminate the pre-exponential factor A and solve for Ea. The key relationship is ln(k₂/k₁) = Ea/R × (1/T₁ - 1/T₂).

Solution Steps

  1. Step 1: Given data — Rate quadruples, so k₂/k₁ = 4; T₁ = 293 K; T₂ = 313 K; R = 8.314 J mol⁻¹ K⁻¹

  2. Step 2: Using the Arrhenius equation in logarithmic form for two temperatures: ln(k₂/k₁) = Ea/R × (1/T₁ - 1/T₂)

  3. Step 3: Substituting the values: ln(4) = Ea/8.314 × (1/293 - 1/313)

  4. Step 4: Calculating: ln(4) = 1.386; (1/293 - 1/313) = 0.003413 - 0.003195 = 2.18×102.18 \times 10⁻⁴ K⁻¹

  5. Step 5: Solving for Ea: 1.386 = Ea/8.314×2.188.314 \times 2.18 × 10⁻⁴; Ea = (1.386×8.3141.386 \times 8.314)/(2.18×102.18 \times 10⁻⁴) = 52812 J mol⁻¹ = 52.8 kJ mol⁻¹

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