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Current Question (ID: 21052)

Question:

$\text{Calculate the standard enthalpy of formation } (\Delta_f H^\circ) \text{ for } 2 \text{ moles of liquid benzene } (\text{C}_6\text{H}_6_{(l)}) \text{ at } 25^\circ\text{C, based on the given thermodynamic data.}$ $\text{Given Data:}$ $\Delta_c H(\text{C}_6\text{H}_6_{(l)}) = -3264.6 \text{ kJ/mol}$ $\Delta_c H(\text{C}_{(s)}) = -393.5 \text{ kJ/mol}$ $\Delta_f H(\text{H}_2\text{O}_{(l)}) = -285.83 \text{ kJ/mol}$

Options:

1. $-92.22 \text{ kJ/mol}$
2. $-46.11 \text{ kJ/mol}$
3. $+92.22 \text{ kJ/mol}$
4. $+46.11 \text{ kJ/mol}$

Solution:

$\text{Hint:}$ $\Delta_c H^\circ (\text{C}_x\text{H}_y_{(l)}) = \left[ n \times \Delta_f H^\circ (\text{Products}) \right] - \left[ m \times \Delta_f H^\circ (\text{Reactants}) \right]$ $6\text{C}_{(s)} + 3\text{H}_2_{(g)} \rightarrow \text{C}_6\text{H}_6_{(l)}$ $\Delta_f H (\text{C}_6\text{H}_6) = 6\Delta_c H(\text{C}_{(s)}) + 3\Delta_c H(\text{H}_2_{(g)}) - \Delta_c H(\text{C}_6\text{H}_6_{(l)})$ $= 6(-393.5) + 3(-285.83) - (-3264.6)$ $\therefore \Delta_f H(\text{H}_2\text{O}_{(l)}) = \Delta_c H(\text{H}_2_{(g)})$

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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
}