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Current Question (ID: 8359)

Question:

$\text{The equilibrium constant for a reaction is 10. The value of } \Delta G^o \text{ will be:}$\n\n$(R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}; T = 300 \text{ K})$

Options:

1. $-5.74 \text{ kJ mol}^{-1}$
2. $-5.74 \text{ J mol}^{-1}$
3. $+4.57 \text{ kJ mol}^{-1}$
4. $-57.4 \text{ kJ mol}^{-1}$

Solution:

$\text{Step 1: Write down the given data}$\n\n$K_{\text{eq}} = 10; T = 300 \text{ K and } R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$\n\n$\text{Step 2: Calculate } \Delta G^o$\n\n$\text{We know that; } \Delta G^o = -2.303 RT \log K_{\text{eq}}$\n$\Rightarrow \Delta G^o = -2.303 \times 8.314 \times 300 \times \log 10$\n$\Rightarrow \Delta G^o = -5744.14 \text{ J mol}^{-1}$\n$\Rightarrow \Delta G^o = -5.74 \text{ kJ mol}^{-1}$

Import JSON File

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
}