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Current Question (ID: 21014)

Question:

$\text{The reaction of cyanamide, } \text{NH}_2\text{CN}_{(s)} \text{ with oxygen was run in a bomb calorimeter and } \Delta U \text{ was found to be } -742.24 \text{ kJ mol}^{-1}. \text{ The magnitude of } \Delta H_{298} \text{(KJ) for the given-below reaction is:}$ $\text{NH}_2\text{CN}_{(s)} + \frac{3}{2} \text{O}_2_{(g)} \rightarrow \text{N}_2_{(g)} + \text{O}_2_{(g)} + \text{H}_2\text{O}_{(l)}$ $\text{[Assume ideal gases and } R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}]}$

Options:

1. $741 \text{ KJ}$
2. $745 \text{ KJ}$
3. $720 \text{ KJ}$
4. $734 \text{ KJ}$

Solution:

$\text{Hint: Use the formula } \Delta H = \Delta U + \Delta n_g RT$ $\text{Step 1:}$ $\text{The relation between } \Delta H \text{ and } \Delta U \text{ is as follows:}$ $\Delta H = \Delta U + \Delta n_g RT$ $\text{The given values are as follows:}$ $\Delta U = -742.24 \text{ kJ mol}^{-1}$ $R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \text{ in kJ } 8.314 \times 10^{-3} \text{ J mol}^{-1} \text{ K}^{-1}$ $T = 298 \text{ K}$ $\text{Calculate the } \Delta n_g \text{ value as follows:}$ $\Delta n_g = (n_{\text{N}_2} + n_{\text{O}_2}) - \left(\frac{3}{2} n_{\text{O}_2}\right)$ $\Delta n_g = 1 + 1 - \left(\frac{3}{2}\right)$ $\Delta n_g = \frac{1}{2}$ $\text{Step 2:}$ $\text{Calculate the value of } \Delta H \text{ is as follows:}$ $\Delta H = \Delta U + \Delta n_g RT$ $= -742.24 + \frac{1}{2} \times \frac{8.314}{1000} \times 298$ $= -741 \text{ kJ}$ $\text{Hence answer is option one.}$

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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
}