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Question:
$\text{The reaction of cyanamide, NH}_2 \text{CN (s) with dioxygen, was carried out in a bomb calorimeter, and } \Delta\text{U was found to be } -742.7 \text{ kJ mol}^{-1} \text{ at 298 K.}$ $\text{NH}_2\text{CN(s)} + \frac{3}{2}\text{O}_2\text{(g)} \rightarrow \text{N}_2\text{(g)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}$ $\text{The enthalpy change for the reaction at 298 K would be:}$
Options:
  • 1. $-741.3 \text{ kJ mol}^{-1}$
  • 2. $+ 753.9 \text{ kJ mol}^{-1}$
  • 3. $+ 772.7 \text{ kJ mol}^{-1}$
  • 4. $-845.1 \text{ kJ mol}^{-1}$
Solution:
$\text{Hint: Use relation between } \Delta\text{H and } \Delta\text{U}$ $\text{Step 1: Write down the relation between } \Delta\text{H and } \Delta\text{U}$ $\text{Enthalpy change for a reaction (}\Delta\text{H) is given by the expression,}$ $\Delta\text{H} = \Delta\text{U} + \Delta\text{n}_g \text{ RT} \ldots\text{(I)}$ $\text{Where,}$ $\Delta\text{U = change in internal energy}$ $\Delta\text{n}_g \text{ = change in number of moles}$ $\text{Step 2: Find out } \Delta\text{n}_g \text{ for the given reaction}$ $\text{For the given reaction,}$ $\Delta\text{n}_g = \sum \text{n}_g\text{(products)} - \sum \text{n}_g\text{(reactants) = (2 - 1.5) moles}$ $\Delta\text{n}_g = 0.5 \text{ moles}$ $\text{Step 3: Substitute the given values in (I) to calculate } \Delta\text{H}$ $\Delta\text{U} = -742.7 \text{ kJ mol}^{-1}$ $\text{T} = 298 \text{ K}$ $\text{R} = 8.314 \times 10^{-3} \text{ kJ mol}^{-1} \text{ K}^{-1}$ $\text{Substituting the values in the expression of } \Delta\text{H:}$ $\Delta\text{H} = (-742.7 \text{ kJ mol}^{-1}) + (0.5 \text{ mol})(298 \text{ K})(8.314 \times 10^{-3} \text{ kJ mol}^{-1} \text{ K}^{-1})$ $\Delta\text{H} = -741.3 \text{ kJ mol}^{-1}$

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{
  "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
}