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Current Question (ID: 21021)

Question:

$\text{If, for a dimerization reaction, } 2\text{A}_{(g)} \rightarrow \text{A}_2{(g)} \text{ at } 298 \text{ K, } \Delta U^\circ = -20 \text{ kJ mol}^{-1} \text{, } \Delta S^\circ = -30 \text{ J K}^{-1}\text{mol}^{-1}\text{, then } \Delta G^\circ \text{ will be:}$

Options:

1. $-10.4 \text{ kJ}$
2. $18.9 \text{ kJ}$
3. $-13.5 \text{ kJ}$
4. $17.4 \text{ kJ}$

Solution:

$\text{Hint: Follow the formula, } \Delta G^\circ = (\Delta U^\circ + \Delta n_g RT) - T \Delta S^\circ$ $\text{Step 1:}$ $\text{The relation between } \Delta H^\circ \text{ and } \Delta U^\circ \text{ is as follows:}$ $\Delta H^\circ = \Delta U^\circ + \Delta n_g RT \quad \text{.......(1)}$ $\text{The relation between } \Delta H^\circ \text{ and } \Delta S^\circ \text{ is as follows:}$ $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \quad \text{....... (2)}$ $\text{Step 2:}$ $\text{From equation 1 and 2, the new formula is as follows:}$ $\Delta G^\circ = (\Delta U^\circ + \Delta n_g RT) - T \Delta S^\circ$ $= \left\{ \left( -20 + (-1) \frac{8.314}{1000} \times 298 \right) - \frac{298}{1000} \times (-30) \right\} \text{ kJ}$ $= -13.5 \text{ kJ}$

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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
}