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Current Question (ID: 21030)

Question:

$\text{The incorrect expression among the following is -}$ $1. \text{In isothermal process, } W_{\text{reversible}} = -nRT \ln \frac{V_f}{V_i}$ $2. \ln K = \frac{\Delta H^\circ - T \Delta S^\circ}{RT}$ $3. K = e^{-\Delta G^\circ / RT}$ $4. \frac{\Delta G_{\text{system}}}{\Delta S_{\text{total}}} = -T$

Options:

1. $W_{\text{reversible}} = -nRT \ln \frac{V_f}{V_i}$
2. $\ln K = \frac{\Delta H^\circ - T \Delta S^\circ}{RT}$
3. $K = e^{-\Delta G^\circ / RT}$
4. $\frac{\Delta G_{\text{system}}}{\Delta S_{\text{total}}} = -T$

Solution:

$\text{Hint: Use the relation, that is, } -\frac{(\Delta H^\circ - T \Delta S^\circ)}{RT} = \ln K$ $1. \text{The expression of work done for isothermal reversible process is as follows:}$ $W_{\text{reversible}} = -nRT \ln \frac{V_f}{V_i}$ $2. \text{The relation between } \Delta G^\circ \text{ and } K \text{(equilibrium constant) is as follows:}$ $\Delta G^\circ = -RT \ln K \quad \text{.......(1)}$ $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \quad \text{.......(2)}$ $\text{From equation 1 and 2,}$ $-\frac{(\Delta H^\circ - T \Delta S^\circ)}{RT} = \ln K$ $\text{Hence, expression in option 2 is incorrect.}$ $3. \Delta G^\circ = -RT \ln K$ $\ln K = -\frac{\Delta G^\circ}{RT}$ $K = e^{-\Delta G^\circ / RT}$ $4. \text{For isothermal process, } \Delta H^\circ = 0$ $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$ $\frac{\Delta G_{\text{system}}}{\Delta S_{\text{total}}} = -T$

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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
}