Import Question JSON

Current Question (ID: 8304)

Question:

$\text{Consider the following reaction,}$ $\text{S} + \text{O}_2 \rightarrow \text{SO}_2, \Delta\text{H} = - 298.2 \text{ kJ mol}^{-1}$ $\text{SO}_2 + 1/2 \text{ O}_2 \rightarrow \text{SO}_3, \Delta\text{H} = - 98.7\text{ kJ mol}^{-1}$ $\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4, \Delta\text{H} = - 130.2 \text{ kJ mol}^{-1}$ $\text{H}_2 + 1/2 \text{ O}_2 \rightarrow \text{H}_2\text{O}, \Delta\text{H} = - 287.3 \text{ kJ mol}^{-1}$ $\text{the enthalpy of formation of H}_2\text{SO}_4 \text{ at 298 K will be}−$

Options:

1. $− 814.4 \text{ kJ mol}^{-1}$
2. $+ 814.4 \text{ kJ mol}^{-1}$
3. $− 650.3 \text{ kJ mol}^{-1}$
4. $− 433.7 \text{ kJ mol}^{-1}$

Solution:

$\text{Hint: Hess's Law of Constant Heat Summation}$ $\text{S}+ \text{O}_2 \rightarrow \text{SO}_2, \Delta\text{H} = − 298.2 \text{ kJ mol}^{-1}$ $\text{SO}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{SO}_3, \Delta\text{H} = − 98.7 \text{ kJ mol}^{-1}$ $\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4, \Delta\text{H} = − 130.2 \text{ kJ mol}^{-1}$ $\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O}, \Delta\text{H} = − 287.3 \text{ kJ mol}^{-1}$ $\text{Net reaction S + H}_2 + 2\text{O}_2 \rightarrow \text{H}_2\text{SO}_4 ; \Delta\text{H}$ $= − 814.4 \text{ kJ mol}^{-1}$

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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
}