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Current Question (ID: 21029)

Question:

$\text{Given}$ $\text{C}_{(\text{graphite})} + \text{O}_2 (g) \rightarrow \text{CO}_2 (g)$ $\Delta_r H^\circ = -393.5 \text{ kJ mol}^{-1}$ $\text{H}_2(g) + \frac{1}{2} \text{O}_2 (g) \rightarrow \text{H}_2\text{O}(l)$ $\Delta_r H^\circ = -285.8 \text{ kJ mol}^{-1}$ $\text{CO}_2( g) + 2\text{H}_2\text{O}(l) \rightarrow \text{CH}_4( g) + 2\text{O}_2( g)$ $\Delta_r H^\circ = +890.3 \text{ kJ mol}^{-1}$ $\text{Based on the above thermochemical equations, the value of } \Delta_r H^\circ \text{ at } 298 \text{ K for the reaction}$ $\text{C}_{(\text{graphite})} + 2\text{H}_2(g) \rightarrow \text{CH}_4(g) \text{ will be:}$

Options:

1. $-74.8 \text{ kJ mol}^{-1}$
2. $-144.0 \text{ kJ mol}^{-1}$
3. $+74.8 \text{ kJ mol}^{-1}$
4. $+144.0 \text{ kJ mol}^{-1}$

Solution:

$\text{Hint: Hess's Law of Constant Heat Summation concept}$ $\text{Calculate the } \Delta_r H^\circ \text{ as follows:}$ $\text{C}_{(\text{graphite})} + \text{O}_2 (g) \rightarrow \text{CO}_2 (g); \Delta_r H^\circ = -393.5 \text{ kJ} \ldots (i)$ $\text{H}_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightarrow \text{H}_2\text{O}(l); \Delta_r H^\circ = -285.8 \text{ kJ} \ldots (ii)$ $\text{CO}_2 (g) + 2\text{H}_2\text{O}(l) \rightarrow \text{CH}_4 (g) + 2\text{O}_2 (g); \Delta_r H^\circ = +890.3 \text{ kJ}$ $\text{C}_{(\text{graphite})} + 2\text{H}_2 (g) \rightarrow \text{CH}_4 (g); \Delta_r H^\circ = ?$ $(iv) = \text{eq. (i)} + 2 \times \text{eq. (ii)} + \text{eq. (iii)}$ $= -393.5 + 2(-285.8) + 890.3$ $= -74.8 \text{ kJ/mol}$

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{
  "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
}