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Current Question (ID: 8402)

Question:

$\text{For the reaction } \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI}\text{(g)}, \text{ the standard free energy is}$ $\Delta G^\Theta > 0. \text{ The equilibrium constant (}K_c\text{) would be :}$

Options:

1. $K_c = 0$
2. $K_c > 1$
3. $K_c = 1$
4. $K_c < 1$

Solution:

$\text{Hint: } \Delta G^\Theta = -RT \ln K_c$ $\text{Step 1:}$ $\text{The relation between } \Delta G^\Theta \text{ and } K_c \text{ is as follows:}$ $\Delta G^\Theta = -RT \ln K_c$ $\text{Step 2:}$ $\text{When } \Delta G^\Theta > 0 \text{ means } \Delta G^\Theta \text{ is positive. This can be so only if ln } K_c \text{ is negative i.e., } K_c < 1$ $\text{Therefore, when the standard free energy change is positive, the equilibrium constant is less than 1, indicating that the reaction does not favor product formation at equilibrium.}$

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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
}