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Current Question (ID: 21045)

Question:

$\text{Assuming ideal behaviour, the magnitude of } \log K \text{ for the following reaction at } 25^\circ C \text{ is}$ $x \times 10^{-1}. \text{ The value of } x \text{ is:}$ $3\text{HC} \equiv \text{CH}_{(g)} \rightleftharpoons \text{C}_6\text{H}_6_{(l)}$ $\left[ \text{Given:}$ $\Delta_f G^\circ (\text{HC} \equiv \text{CH}) = -2.04 \times 10^5 \text{ J mol}^{-1}$ $\Delta_f G^\circ (\text{C}_6\text{H}_6) = -1.24 \times 10^5 \text{ J mol}^{-1}$ $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \right]$

Options:

1. 860
2. 875
3. 855
4. 895

Solution:

$\text{Hint: The relation between } \Delta G^\circ \text{ and } K \text{ is}$ $\Delta G^\circ = -RT \ln K$ $\text{Step 1:}$ $\text{Calculate the } \Delta G^\circ \text{ value for the given reaction}$ $3\text{HC} \equiv \text{CH}_{(g)} \rightarrow \text{C}_6\text{H}_6_{(l)}$ $\Delta G^\circ = \sum (\Delta G^\circ_f)_p - 3 \sum (\Delta G^\circ_f)_R$ $= -1.24 \times 10^5 \text{ J} - (3 \times -2.04 \times 10^5)$ $= 4.88 \times 10^5 \text{ J}$ $\text{Step 2:}$ $\text{The relation between } \Delta G^\circ \text{ and } K \text{ is as}$ $\Delta G^\circ = -RT \ln K$ $= -RT \ln K = 1 \times (-1.24 \times 10^5) - (-3 \times 2.04 \times 10^5)$ $4.88 \times 10^5 \text{ J} = -R \times T \times \ln K$ $\Rightarrow -2.303 \times R \times T \log K = 4.88 \times 10^5$ $\Rightarrow \log K = \frac{4.88 \times 10^5}{2.303 \times R \times T} = \frac{-488000}{5705.848} = 85.52$ $= 855 \times 10^{-1}$ $\Rightarrow x = 855$

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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
}