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Question:
$\text{The equilibrium reaction that doesn't have equal values for K}_C \text{ and K}_P \text{ is:}$
Options:
  • 1. $\text{2NO(g)} \rightleftharpoons \text{N}_2\text{(g)} + \text{O}_2\text{(g)}$
  • 2. $\text{SO}_2\text{(g)} + \text{NO}_2\text{(g)} \rightleftharpoons \text{SO}_3\text{(g)} + \text{NO(g)}$
  • 3. $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons \text{2HI(g)}$
  • 4. $\text{2C(s)} + \text{O}_2\text{(g)} \rightleftharpoons \text{2CO}_2\text{(g)}$
Solution:
$\text{HINT: Use K}_P = \text{K}_C(\text{RT})^{\Delta n}$ $\text{The relationship between K}_P \text{ and K}_C \text{ for gas phase reactions is given by:}$ $\text{K}_P = \text{K}_C(\text{RT})^{\Delta n}$ $\text{where } \Delta n = n_P - n_R \text{ (difference between moles of gaseous products and gaseous reactants)}$ $\text{If } \Delta n = 0 \text{, then K}_P = \text{K}_C$ $\text{Let's analyze each reaction:}$ $\text{1. 2NO(g)} \rightleftharpoons \text{N}_2\text{(g)} + \text{O}_2\text{(g)}$ $n_P = 2 \text{ moles of gas (N}_2 \text{ and O}_2\text{)}$ $n_R = 2 \text{ moles of gas (2NO)}$ $\Delta n = n_P - n_R = 2 - 2 = 0$ $\text{Therefore, K}_P = \text{K}_C$ $\text{2. SO}_2\text{(g)} + \text{NO}_2\text{(g)} \rightleftharpoons \text{SO}_3\text{(g)} + \text{NO(g)}$ $n_P = 2 \text{ moles of gas (SO}_3 \text{ and NO)}$ $n_R = 2 \text{ moles of gas (SO}_2 \text{ and NO}_2\text{)}$ $\Delta n = n_P - n_R = 2 - 2 = 0$ $\text{Therefore, K}_P = \text{K}_C$ $\text{3. H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons \text{2HI(g)}$ $n_P = 2 \text{ moles of gas (2HI)}$ $n_R = 2 \text{ moles of gas (H}_2 \text{ and I}_2\text{)}$ $\Delta n = n_P - n_R = 2 - 2 = 0$ $\text{Therefore, K}_P = \text{K}_C$ $\text{4. 2C(s)} + \text{O}_2\text{(g)} \rightleftharpoons \text{2CO}_2\text{(g)}$ $\text{Note: Solids like C(s) do not contribute to the gas phase equilibrium and are not counted in } n_P \text{ or } n_R$ $n_P = 2 \text{ moles of gas (2CO}_2\text{)}$ $n_R = 1 \text{ mole of gas (O}_2\text{)}$ $\Delta n = n_P - n_R = 2 - 1 = 1$ $\text{Since } \Delta n \neq 0, \text{ K}_P = \text{K}_C(\text{RT})^1 = \text{K}_C(\text{RT})$ $\text{Therefore, K}_P \neq \text{K}_C$ $\text{The correct answer is option 4: 2C(s)} + \text{O}_2\text{(g)} \rightleftharpoons \text{2CO}_2\text{(g)}$

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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
}