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Current Question (ID: 18330)

Question:

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

Options:

1. $4.0 \times 10^{12} \ \text{s}^{-1}$
2. $7.8 \times 10^{-13} \ \text{s}^{-1}$
3. $3.8 \times 10^{-12} \ \text{s}^{-1}$
4. $4.7 \times 10^{12} \ \text{s}^{-1}$

Solution:

$\text{Hint: Use Arrhenius Equation}$ $\text{Step 1:}$ $\text{Given data:}$ $k = 2.418 \times 10^{-5} \ \text{s}^{-1}; \ T = 546 \ \text{K}; \ E_a = 179.9 \ \text{kJ mol}^{-1}$ $\text{Step 2:}$ $\text{According to the Arrhenius equation}$ $k = A \times e^{-E_a/RT}$ $\log k = \log A - \frac{E_a}{RT}$ $\log A = \log (2.418 \times 10^{-5}) + \frac{179.9}{10^3 \ \text{J mol}^{-1}} \times 2.303 \times 8.314 \ \text{JK}^{-1} \times 546 \ \text{K}$ $\log A = 12.60$ $A = \text{antilog}(12.60)$ $A = 3.98 \times 10^{12} \ \text{s}^{-1}$ $\text{Hence, the value of pre-exponential factor = } 4.0 \times 10^{12} \ \text{s}^{-1}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}