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Current Question (ID: 18316)

Question:
$\text{Given the following reaction:}$ $\text{N}_2\text{O}_5 \; \text{as} \; \text{N}_2\text{O}_5 \rightleftharpoons 2\text{NO}_2 + (1/2)\text{O}_2$ $\text{The values of rate constants for the above reaction are} \; 3.45 \times 10^{-5} \; \text{and} \; 6.9 \times 10^{-3} \; \text{at} \; 27^\circ \text{C} \; \text{and} \; 67^\circ \text{C} \; \text{respectively. The activation energy for the above reaction is:}$
Options:
  • 1. $102 \times 10^2 \; J$
  • 2. $488.5 \; kJ$
  • 3. $112 \; J$
  • 4. $112.5 \; kJ$
Solution:
$\text{Use Arrhenius equation at different temperatures.}$ $\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right]$ $\log \frac{6.9 \times 10^{-3}}{3.45 \times 10^{-5}} = \frac{E_a}{2.303 \times 8.31} \left[ \frac{40}{300 \times 340} \right]$ $E_a = 112.5 \; kJ$

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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
}