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Current Question (ID: 8276)

Question:
$\Delta_f\text{U}^\circ \text{ of formation of } \text{CH}_4\text{(g) at certain temperature is } -393 \text{ kJ mol}^{-1}\text{. The value of } \Delta_f\text{H}^\circ \text{ is-}$
Options:
  • 1. $\text{Zero}$
  • 2. $< \Delta_f\text{U}^\circ$
  • 3. $> \Delta_f\text{U}^\circ$
  • 4. $\text{Equal to } \Delta_f\text{U}^\circ$
Solution:
$\text{Hint: } \Delta_f\text{H}^\circ = \Delta_f\text{U}^\circ + \Delta n_g \text{RT}$\n\n$\text{The reaction is}$\n\n$\text{The formula is as follows:}$\n\n$\Delta_f\text{H}^\circ = \Delta_f\text{U}^\circ + \Delta n_g \text{RT}$\n\n$\text{For the given reaction, } \text{C(graphite) + 2H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}$\n\n$\text{calculate the value of } \Delta n_g$\n\n$\Delta n_g = (n_p - n_r)_g = 1 - 2 = -1$\n\n$\Delta_f\text{H}^\circ = \Delta_f\text{U}^\circ + \Delta n_g \text{RT}$\n\n$\Delta n_g = -1$\n\n$\Delta_f\text{H}^\circ = \Delta_f\text{U}^\circ - 2\text{RT}$\n\n$\therefore \Delta_f\text{H}^\circ < \Delta_f\text{U}^\circ$

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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
}