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Current Question (ID: 18192)

Question:

$\Delta G^\circ \text{ for the reaction given below is:}$ $\frac{1}{2} \text{A(g)} + \frac{3}{2} \text{B(g)} \rightleftharpoons \text{C(g)}$ $(K_{\text{eq}} = 826 \ \text{atm}^{-1} \ \text{at} \ 298 \ \text{K})$

Options:

1. $-8.32 \ \text{kJ}$
2. $8.32 \ \text{kJ}$
3. $16.64 \ \text{kJ}$
4. $-16.64 \ \text{kJ}$

Solution:

$\text{HINT: } \Delta G^\circ = -2.303 \ RT \ \log_{10} \ K_{\text{eq}}$ $\text{Explanation:}$ $\text{Given: } K_{\text{eq}} = 826 \ \text{atm}^{-1} \ \text{and} \ T = 298 \ \text{K}$ $\Delta G^\circ = -2.303 \times 8.314 \times 10^{-3} \times 298 \times \log \ 826$ $= -16.64 \ \text{KJ}$

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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
}