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Current Question (ID: 18350)

Question:

$\text{If the rate of reaction doubles when the temperature is raised from } 20 \, ^\circ\text{C to } 35 \, ^\circ\text{C, then the activation energy for the reaction will be:}$ $\text{(R = 8.314 J mol}^{-1} \text{ K}^{-1})$

Options:

1. $342 \text{ kJ mol}^{-1}$
2. $269 \text{ kJ mol}^{-1}$
3. $34.7 \text{ kJ mol}^{-1}$
4. $15.1 \text{ kJ mol}^{-1}$

Solution:

$\text{HINT: Use Arrhenius equation}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{The formula of Arrhenius equation is as follows:}$ $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 \times R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$ $k_2 = 2k_1, \; T_1 = 20 + 273 = 293 \text{ K}$ $T_2 = 35 + 273 = 308 \text{ K}$ $R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}$ $\text{Step 2:}$ $\text{Calculate the value of activation energy as follows:}$ $\log 2 = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{293} - \frac{1}{308} \right)$ $0.3010 = \frac{E_a}{19.147} \times \frac{15}{293 \times 308}$ $E_a = 34673 \text{ J mol}^{-1} = 34.7 \text{ kJ mol}^{-1}$

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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
}