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Current Question (ID: 8364)

Question:

\text{Calculate } \Delta_r H° \text{ (in kJ/mol) for Cr}_2\text{O}_3 \text{ from the } \Delta_r G° \text{ and the } S° \text{ values provided at 27°C} 4\text{Cr(s)} + 3\text{O}_2\text{(g)} → 2\text{Cr}_2\text{O}_3\text{(s)} \Delta_r G° = -2093.4 \text{ kJ/mol} \text{S° (J/K·mol): Cr(s) = 24} \text{O}_2\text{(g) = 205; Cr}_2\text{O}_3\text{(s) = 81}

Options:

1. $-2258.1 \text{ kJ/mol}$
2. $-1129.05 \text{ kJ/mol}$
3. $-964.35 \text{ kJ/mol}$
4. $\text{None of the above}$

Solution:

$\text{Hint: } \Delta_r G^o = \Delta_r H^o - T \times \Delta_r S^o$\n\n$\Delta_r G^o = \Delta_r H^o - T \times \Delta_r S^o$\n\n$\Delta_r S^o = 2 \times 81 - 4 \times 24 - 3 \times 205 \text{ J/mol}$\n$\Delta_r S^o = 162 - 96 - 615 = -549 \text{ J/mol}$\n\n$\text{Converting to J/mol K: } \Delta_r S^o = -549 \text{ J/mol K}$\n$\text{T} = 27^\circ\text{C} = 300 \text{ K}$\n\n$\Delta_r H^o = \Delta_r G^o + T \times \Delta_r S^o$\n$\Delta_r H^o = -2093.4 \times 10^3 + 300 \times (-549)$\n$\Delta_r H^o = -2093400 - 164700 = -2258100 \text{ J/mol}$\n$\Delta_r H^o = -2258.1 \text{ kJ/mol}$\n\n$\text{Since } \Delta_r H^o = 2 \times \Delta_f H^o(\text{Cr}_2\text{O}_3)$\n\n$\therefore \Delta_f H^o(\text{Cr}_2\text{O}_3) = -\frac{2258.1}{2} = -1129.05 \text{ kJ/mol}$

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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
}