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Current Question (ID: 21151)

Question:

$\text{For a hypothetical reaction}$ $\text{A} \rightleftharpoons \text{B} ; \text{K}_{\text{eq}} = 10^2$ $\text{(Use T = 27 °C, R = 8.3 JK}^{-1}\text{mol}^{-1}, \ln 10 = 2.3)}$ $\text{If the value of } \Delta G^\circ \text{ for the above reaction is } -x \text{ kJ, the value of } 2x \text{ will be:}$ $\text{(Round off to the nearest integer)}$

Options:

1. 11
2. 19
3. 23
4. 28

Solution:

$\text{Step 1:}$ $\text{The relation between } \Delta G^\circ \text{ and equilibrium constant (} K_{\text{equ}} \text{) is as follows:}$ $\Delta G^\circ = -2.3RT \log K_{\text{eq}}$ $\text{The given values are as follows:}$ $K_{\text{equ}} = 10^2$ $R = 8.3 \text{ JK}^{-1}\text{mol}^{-1}$ $T = 27 \text{ °C} = 300 \text{ K}$ $\text{Step 2:}$ $\text{Calculate the value of } \Delta G^\circ \text{ as follows:}$ $\Delta G^\circ = -2.3RT \log K_{\text{eq}}$ $= -8.3 \times 300 \times 2.3 \log(10^2)$ $\Delta G^\circ = -11454 \text{ J}$ $\Delta G^\circ = -11.454 \text{ kJ}$ $x = 11.454$ $2x = 22.908$ $\text{Thus the value of } 2x \text{ is 23. Hence, option third is the correct answer.}$

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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
}