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Current Question (ID: 8415)

Question:

\text{Consider the following reaction: CO(g) + 3H}_2\text{(g) } \leftrightarrow \text{ CH}_4\text{(g) + H}_2\text{O(g). If the reaction mixture contains 0.30 mol of CO, 0.10 mol of H}_2\text{, and 0.02 mol of H}_2\text{O, and an unknown amount of CH}_4\text{ at equilibrium at 1300 K in a 1L flask, the concentration of CH}_4\text{ in the mixture will be (Kc = 3.90 at 1300 K)}

Options:

1. $5.85 \times 10^2 \text{ M}$
2. $5.85 \times 10^{-1} \text{ M}$
3. $5.85 \times 10^3 \text{ M}$
4. $5.85 \times 10^{-2} \text{ M}$

Solution:

\text{Hint: } K_c = \frac{[H_2O][CH_4]}{[CO][H_2]^3} \text{Explanation:} \text{Step 1: Write down the given data} \text{Let the concentration of methane at equilibrium be } x \text{ and } K_c = 3.90 \text{Given Reaction: } \text{CO(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_4\text{(g)} + \text{H}_2\text{O(g)} \text{At equilibrium conc.: CO = 0.3M, H}_2 \text{ = 0.1M, CH}_4 \text{ = }x\text{M, H}_2\text{O = 0.02M} \text{Step 2: Find the concentration of CH}_4 K_c = \frac{[H_2O][CH_4]}{[CO][H_2]^3} 3.90 = \frac{0.02 \times x}{0.3 \times (0.1)^3} x = \frac{3.90 \times 0.3 \times (0.1)^3}{0.02} x = 5.85 \times 10^{-2} \text{ M} \text{Hence, the concentration of CH}_4 \text{ at equilibrium is } 5.85 \times 10^{-2} \text{ M}

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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
}