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Current Question (ID: 21040)

Question:
$(\Delta H - \Delta U)$ \text{ for the formation of carbon monoxide (CO) from its elements at } 298 \text{ K is:} \ (R = 8.314 \text{ JK}^{-1} \text{ mol}^{-1})$
Options:
  • 1. $-1238.78 \text{ J mol}^{-1}$
  • 2. $1238.78 \text{ J mol}^{-1}$
  • 3. $-2477.57 \text{ J mol}^{-1}$
  • 4. $2477.57 \text{ J mol}^{-1}$
Solution:
$\text{Hint: Use, the formula that is, } \Delta H - \Delta U = \Delta n_g RT$ $\text{Step 1:}$ $\text{The relation between } \Delta H \text{ and } \Delta U \text{ is as follows:}$ $\Delta H - \Delta U = \Delta n_g RT$ $\text{First, calculate the value of } \Delta n_g \text{ as follows:}$ $\left[ \text{C(s) + } \frac{1}{2} \text{O}_2\text{(g)} \rightarrow \text{CO(g)} \right]$ $\Delta n_g = 1 - \frac{1}{2}$ $= \frac{1}{2}$ $\text{Step 2:}$ $\text{Calculate the value of } (\Delta H - \Delta U) \text{ as follows:}$ $\Delta H - \Delta U = \Delta n_g RT$ $= \frac{1}{2} \times 8.314 \times 298$ $= +1238.78 \text{ J mol}^{-1}$ $\text{Hence, option 2 is the correct answer.}$

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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
}