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Current Question (ID: 21035)

Question:

$\text{The standard enthalpy of formation of NH}_3 \text{ is } -46.0 \text{ kJ mol}^{-1}. \text{ If the enthalpy of formation of H}_2 \text{ from its atoms is } -436 \text{ kJ mol}^{-1} \text{ and that of N}_2 \text{ is } -712 \text{ kJ mol}^{-1}, \text{ the average bond enthalpy of N - H bond in NH}_3 \text{ is:}$

Options:

1. $-1102 \text{ kJ mol}^{-1}$
2. $-964 \text{ kJ mol}^{-1}$
3. $+352 \text{ kJ mol}^{-1}$
4. $+1056 \text{ kJ mol}^{-1}$

Solution:

$\text{Hint: Follow the formula, that is,}$ $\Delta H_f = \frac{1}{2} BE_{\text{N}_2} + \frac{3}{2} BE_{\text{H}_2} - 3 BE_{\text{N-H}}$ $\text{Explanation:}$ $\text{The given reaction is as follows:}$ $\frac{1}{2} \text{N}_2(g) + \frac{3}{2} \text{H}_2(g) \rightarrow \text{NH}_3(g)$ $\text{Calculate the average bond enthalpy of N-H bond in NH}_3 \text{ is as follows:}$ $\Delta H = (B.E.)_{\text{reactants}} - (B.E.)_{\text{products}}$ $\Delta H_f = \frac{1}{2} BE_{\text{N}_2} + \frac{3}{2} BE_{\text{H}_2} - 3 BE_{\text{N-H}}$ $-46 = \frac{1}{2} \times (712) + \frac{3}{2} (436) - 3x$ $1056 = 3x$ $x = \frac{1056}{3} = 352 \text{ kJ/mol}$ $\text{Hence, option 3 is the correct answer.}$

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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
}