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Current Question (ID: 8362)

Question:

$\text{The standard free energies of formation (in kJ/mol) at 298 K are -237.2, -394.4, and -8.2 for H}_2\text{O}_{\text{(l)}}\text{, CO}_{2\text{(g)}}\text{, and pentane (g), respectively. The value of E}_{\text{cell}} \text{ for the pentane-oxygen fuel cell is:}$

Options:

1. $1.968 \text{ V}$
2. $2.0968 \text{ V}$
3. $1.0968 \text{ V}$
4. $0.0968 \text{ V}$

Solution:

$\text{Hint: } \Delta G = \sum \Delta G_P - \sum G_R$\n\n$\text{The given reaction is as follows:}$\n\n$\text{C}_5\text{H}_{12} + 8\text{O}_2 \rightarrow 5\text{CO}_2 + 6\text{H}_2\text{O}$\n\n$\text{Calculate } \Delta G \text{ is as follows:}$\n$\Delta G = \sum \Delta G_P - \sum G_R$\n$= 5 \times [-394.4] + 6[-237.2] - (-8.2)$\n$= -3387 \text{ kJ} = -3387 \times 10^3 \text{ J}$\n\n$\text{Calculate E}_{\text{cell}}$\n$\Delta G = -nE_{\text{cell}}F$\n\n$\text{loss of electron for each C} = 6.4 \times 5 = 32$\n\n$E = \frac{3387 \times 10^3}{96500 \times 32} = 1.0968 \text{ V}$

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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
}