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Current Question (ID: 18319)

Question:

$\text{The rate constant for the first-order decomposition of } \text{H}_2\text{O}_2 \text{ is given by the equation:}$ $\log k = 14.34 - 1.25 \times 10^4 \frac{K}{T}. \text{ The value of } E_a \text{ for the reaction would be:}$

Options:

1. $249.34 \text{ kJ mol}^{-1}$
2. $242.64 \text{ J mol}^{-1}$
3. $-275.68 \text{ kJ mol}^{-1}$
4. $239.34 \text{ kJ mol}^{-1}$

Solution:

$\text{HINT: Use the Arrhenius equation.}$ $\text{Explanation:}$ $\text{Step 1: According to Arrhenius equation,}$ $\log k = \log A - \frac{E_a}{RT} \quad \text{.......(i)}$ $\text{Given equation } = \log k = 14.34 - 1.25 \times 10^4 \frac{K}{T} \quad \text{....(ii)}$ $\text{Step 2:}$ $\text{By comparing (i) and (ii), we get}$ $\frac{E_a}{2.303 \ RT} = 1.25 \times 10^4 KT$ $E_a = 1.25 \times 10^4 K \times 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \times 2.303$ $E_a = 239339.3 \text{ J mol}^{-1}$ $E_a = 239.34 \text{ kJ mol}^{-1}$

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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
}