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Current Question (ID: 8273)

Question:
$\text{The } \Delta\text{H for vaporization of a liquid is 20 kJ/mol. Assuming ideal behaviour, the change in internal energy for the vaporization of 1 mol of the liquid at } 60^{\circ}\text{C and 1 bar is close to:}$
Options:
  • 1. $13.2 \text{ kJ/mol}$
  • 2. $17.2 \text{ kJ/mol}$
  • 3. $19.5 \text{ kJ/mol}$
  • 4. $20.0 \text{ kJ/mol}$
Solution:
$\text{Hint: } \Delta\text{H} = \Delta\text{E} + \Delta\text{n}_g \text{RT}$ $\text{For the vaporization process (liquid } \rightarrow \text{ gas):}$ $\Delta\text{H} = \Delta\text{E} + \Delta\text{n}_g \text{RT}$ $\text{Where } \Delta\text{n}_g = 1 \text{ mol (as 1 mol of gas is formed from 0 mol of gas)}$ $\text{Temperature T} = 60^{\circ}\text{C} = 333 \text{ K}$ $\text{Gas constant R} = 8.314 \times 10^{-3} \text{ kJ mol}^{-1}\text{ K}^{-1}$ $\text{Rearranging the equation to solve for } \Delta\text{E:}$ $\Delta\text{E} = \Delta\text{H} - \Delta\text{n}_g \text{RT}$ $\Delta\text{E} = 20 - (1)(8.314 \times 10^{-3})(333)$ $\Delta\text{E} = 20 - 2.8$ $\Delta\text{E} = 17.2 \text{ kJ/mol}$

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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
}