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Current Question (ID: 8280)

Question:

$\text{The difference in } \Delta H \text{ and } \Delta U \text{ for combustion of benzoic acid at 300 K is equal to-}$

Options:

1. $-1.247 \text{ kJ}$
2. $+1.247 \text{ kJ}$
3. $-1.247 \text{ J}$
4. $+1.247 \text{ J}$

Solution:

$\text{Hint: } \Delta H = \Delta U + \Delta n_g RT$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{The combustion reaction of benzoic acid is as follows:}$ $\text{C}_6\text{H}_6 - \text{COOH}_{(s)} + \frac{15}{2}\text{O}_{2(g)} \rightarrow 7\text{CO}_{2(g)} + 3\text{H}_2\text{O}_{(l)}$ $\text{Calculate the } \Delta n_p \text{ value as follows:}$ $\Delta n_p = n_p - n_R = 7 - \frac{15}{2} = -\frac{1}{2}$ $\text{Step 2:}$ $\text{Calculate the difference in } \Delta H \text{ and } \Delta U \text{ as follows:}$ $\Delta H = \Delta U + \Delta n_g RT$ $\Delta H - \Delta U = \left(-\frac{1}{2}\right) \times 8.314 \times 300$ $\Delta H - \Delta U = -1247.1 \text{ J} = -1.247 \text{ kJ}$ $\text{Hence, the difference in } \Delta H \text{ and } \Delta U \text{ for the combustion of solid benzoic acid at 300 K is -1.247 kJ.}$

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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
}