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Current Question (ID: 18536)

Question:

$\text{Consider the following reactions at } 1000 \ ^\circ\text{C.}$ $\text{(I) ZnS + } \frac{1}{2}\text{O}_2\text{(g)} \xrightarrow{\Delta} \text{ZnO(g)}; \ \Delta G^\circ = -360 \ \text{kJ mol}^{-1}$ $\text{(II) Cs + } \frac{1}{2}\text{O}_2\text{(g)} \xrightarrow{\Delta} \text{CO(g)}; \ \Delta G^\circ = -460 \ \text{kJ mol}^{-1}$ $\text{Choose the correct statement about the above reactions at } 1000 \ ^\circ\text{C:}$

Options:

1. $\text{ZnO is more stable than CO}$
2. $\text{ZnO can be reduced to Zn by C}$
3. $\text{ZnO, and CO are formed at an equal rate}$
4. $\text{ZnO can not be reduced to Zn by C}$

Solution:

$\text{(I) Zn(s) + } \frac{1}{2}\text{O}_2\text{(g)} \xrightarrow{\Delta} \text{ZnO(g)};$ $\Delta G^\circ = -360 \ \text{kJ mol}^{-1}$ $\text{(II) C(s) + } \frac{1}{2}\text{O}_2\text{(g)} \xrightarrow{\Delta} \text{CO(g)};$ $\Delta G^\circ = -460 \ \text{kJ mol}^{-1}$ $\text{Thus, ZnO(g) + C(s)} \xrightarrow{\Delta}_{1000 \ ^\circ\text{C}} \text{Zn(g) + CO(g)}, \Delta G^\circ = -100 \ \text{kJ mol}^{-1}$ $\text{The free energy of the reaction is negative.}$ $\text{Hence, ZnO can be reduced to Zn by C at } 1000 \ ^\circ\text{C.}$

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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
}