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Current Question (ID: 21042)

Question:

$\text{Which of the following is the standard enthalpy of formation of carbon monoxide (CO), if the standard enthalpies of combustion of carbon (C) and carbon monoxide (CO) respectively are } -393.5 \text{ kJ mol}^{-1} \text{ and } -283 \text{ kJ mol}^{-1}?$

Options:

1. $110.5 \text{ kJ}$
2. $676.5 \text{ kJ}$
3. $-676.5 \text{ kJ}$
4. $-110.5 \text{ kJ}$

Solution:

$\text{Hint: Use Hess's Law of Constant Heat Summation}$ $\text{Step 1:}$ $\text{The combustion of carbon and carbon monoxide is as follows:}$ $\text{C(s) + O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \quad \Delta H_1 = -393.5 \text{ kJ mol}^{-1} \quad \text{(1)}$ $\text{CO(g) + } \frac{1}{2} \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \quad \Delta H_2 = -283 \text{ kJ mol}^{-1} \quad \text{(2)}$ $\text{The formation reaction of CO is as follows:}$ $\text{C(s) + } \frac{1}{2} \text{O}_2\text{(g)} \rightarrow \text{CO(g)} \quad \Delta H_3$ $\text{Step 2:}$ $\text{Reverse equation second and add with equation 1, the same scenario is done with enthalpy also.}$ $\Delta H_3 = \Delta H_2 + \Delta H_1$ $\Delta H_3 = 283 + (-393.5)$ $= -110.5 \text{ kJ mol}^{-1}$

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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
}