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Current Question (ID: 8358)

Question:

$\text{For a given reaction, } \Delta H = 35.5 \text{ kJ mol}^{-1} \text{ and } \Delta S = 83.6 \text{ J K}^{-1} \text{mol}^{-1}\text{. The reaction is spontaneous at:}$\n\n$\text{(Assume that } \Delta H \text{ and } \Delta S \text{ do not vary with temperature)}$

Options:

1. $T < 425 \text{ K}$
2. $T > 425 \text{ K}$
3. $\text{All temperatures}$
4. $T > 298 \text{ K}$

Solution:

\text{Hint: use Gibbs-free energy equation, Gibbs energy } (\Delta G) = \Delta H - T\Delta S \text{where, } \Delta H = \text{Enthalpy change} \Delta S = \text{Entropy change} T = \text{Temperature} \text{For a reaction to be spontaneous } \Delta G < 0 \therefore \text{ Gibbs energy equation becomes:} \Delta G = \Delta H - T\Delta S < 0 \text{or, } \Delta H < T\Delta S T > \frac{\Delta H}{\Delta S} = \frac{35.5 \text{ kJ mol}^{-1}}{83.6 \text{ J K}^{-1}\text{mol}^{-1}} = \frac{35.5 \times 1000 \text{ J mol}^{-1}}{83.6 \text{ J K}^{-1}\text{mol}^{-1}} = 425 \text{ K} T > 425\text{ K}

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}