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Current Question (ID: 18311)

Question:

$\text{The rate of reaction triples when the temperature changes from } 20^\circ\text{C to } 50^\circ\text{C. The energy of activation for the reaction will be:}$

Options:

1. $28.81 \text{ kJ mol}^{-1}$
2. $38.51 \text{ kJ mol}^{-1}$
3. $18.81 \text{ kJ mol}^{-1}$
4. $8.31 \text{ kJ mol}^{-1}$

Solution:

$\text{Step 1:}$ $\ln \frac{k_2}{k_1} = \frac{E_a}{R} \left( \frac{T_2 - T_1}{T_1 \times T_2} \right)$ $\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 \times T_2} \right)$ $\text{Step 2:}$ $\log(3) = \frac{E_a}{2.303 \times 8.314} \left( \frac{30}{323 \times 293} \right)$ $E_a = 28.81 \text{ kJ mol}^{-1}$ $\text{Hence, option 1 is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}