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Current Question (ID: 18506)

Question:

$\text{At 1400 } ^\circ \text{C, the } \Delta G_r^\circ \text{ value for the reaction } 2\text{FeO} + 2\text{C} \rightarrow 2\text{Fe} + 2\text{CO} \text{ is}$

Options:

1. $-116 \text{ kJ mol}^{-1}$
2. $-106 \text{ kJ mol}^{-1}$
3. $100 \text{ kJ mol}^{-1}$
4. $-100 \text{ kJ mol}^{-1}$

Solution:

$\text{Consider the two reactions.}$ $2\text{FeO} \rightarrow 2\text{Fe} + \text{O}_2 \quad \Delta G_r^\circ = +341 \text{ kJ mol}^{-1} \quad \ldots (1)$ $2\text{C} + \text{O}_2 \rightarrow 2\text{CO} \quad \Delta G_r^\circ = -447 \text{ kJ mol}^{-1} \quad \ldots (2)$ $\text{When reaction 1 is combined with reaction 2 then the desired reaction, that is, } 2\text{FeO} + 2\text{C} \rightarrow 2\text{Fe} + 2\text{CO} \text{ is obtained.}$ $\text{Hence, the } \Delta G_r^\circ \text{ value for the reaction } 2\text{FeO} + 2\text{C} \rightarrow 2\text{Fe} + 2\text{CO} \text{ is}$ $\Delta G_r^\circ = \Delta G_r^\circ (1) + \Delta G_r^\circ (2)$ $\Delta G_r^\circ = 341 + (-447)$ $\Delta G_r^\circ = -106 \text{ kJ mol}^{-1}$

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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
}