Question 16 of 37intermediate🔧 ApplyNumerical3 marks

The rate constant for the decomposition of hydrocarbons is 2.418×10−5s−12.418 \times 10^{-5} \text{s}^{-1} at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

Correct Answer

The value of pre-exponential factor (A) is 3.8×103.8 \times 10¹² s⁻¹.

Exercise: EXERCISES | Q: 3.23 | (Chapter: 27)
For More Understanding

Explanation

This question requires applying the Arrhenius equation to find the pre-exponential factor A. The Arrhenius equation relates rate constant k with activation energy Ea, temperature T, and pre-exponential factor A. Given k, Ea, and T, we can rearrange the equation to solve for A.

Solution Steps

  1. Step 1: Write the Arrhenius equation in logarithmic form: log k = log A - Ea/(2.303 RT)

  2. Step 2: Rearrange to find log A: log A = log k + Ea/(2.303 RT)

  3. Step 3: Substitute the given values: k = 2.418×102.418 \times 10⁻⁵ s⁻¹, Ea = 179.9 kJ/mol = 179.9×10179.9 \times 10³ J/mol, T = 546 K, R = 8.314 J mol⁻¹ K⁻¹

  4. Step 4: Calculate log k: log(2.418×102.418 \times 10⁻⁵) = log(2.418) + log(10⁻⁵) = 0.383 - 5 = -4.617

  5. Step 5: Calculate Ea/(2.303 RT): Ea/(2.303 RT) = 179900/(2.303×8.3142.303 \times 8.314 × 546) = 179900/10458.6 = 17.20

  6. Step 6: Calculate log A: log A = -4.617 + 17.20 = 12.58

  7. Step 7: Find A: A = 10^(12.58) = 10^(0.58) × 10¹² = 3.8×103.8 \times 10¹² s⁻¹

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