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Current Question (ID: 18298)

Question:
$ ext{For A + B} \rightarrow ext{C + D, } \Delta H = -20 \text{ kJ mol}^{-1},$ $\text{the activation energy of the forward reaction is } 85 \text{ kJ mol}^{-1}.$ $\text{The activation energy for the backward reaction is...}$
Options:
  • 1. $105$
  • 2. $85$
  • 3. $40$
  • 4. $65$
Solution:
$\Delta H = E_f - E_b$ $\text{Let 'A' is the } E_a \text{ for backward reaction.}$ $\text{For a reaction } E_a \text{ for forward reaction} = E_a \text{ for backward reaction} + \Delta H,$ $85 = A - 20$ $A = 105 \text{ KJ mol}^{-1}$

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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
}