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Current Question (ID: 8399)

Question:

$\text{For the reaction } 2\text{NOCl}_{(g)} \rightleftharpoons 2\text{NO}_{(g)} + \text{Cl}_2\text{(g)}, K_C \text{ at } 427°\text{C is } 3 \times 10^{-6} \text{ mol L}^{-1}\text{. The value of } K_P \text{ will be:}$

Options:

1. $1.72 \times 10^{-4}$ (Correct)
2. $7.50 \times 10^5$
3. $2.50 \times 10^{-5}$
4. $2.50 \times 10^{-4}$

Solution:

$\text{HINT: } K_P = K_C(RT)^{\Delta n}$ $\text{Explanation:}$ $\text{For a given reaction: } 2\text{NOCl}_{(g)} \rightleftharpoons 2\text{NO}_{(g)} + \text{Cl}_2\text{(g)}$ $\Delta n = 3 - 2 = 1$ $K_P = K_C(RT)^1$ $= 3 \times 10^{-6} \times 0.0821 \times 700\text{K}$ $= 1.72 \times 10^{-4}$

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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
}