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Current Question (ID: 8425)

Question:

\text{For the reaction: } \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI}\text{(g)} ; K_c = 54.8 \text{ at } 700\text{K}. \text{If } 0.5 \text{ mol L}^{-1} \text{ of HI}\text{(g)} \text{ is present at equilibrium at } 700 \text{K, the concentration of } \text{H}_2\text{(g)} \text{ and } \text{I}_2\text{(g)} \text{ would be:}

Options:

1. $[\text{H}_2] = [\text{I}_2] = 0.05 \text{ mol L}^{-1}$
2. $[\text{H}_2] = 0.5 \text{ mol L}^{-1}, [\text{I}_2] = 0.05 \text{ mol L}^{-1}$
3. $[\text{H}_2] = 0.068 \text{ mol L}^{-1}, [\text{I}_2] = 0.55 \text{ mol L}^{-1}$
4. $[\text{H}_2] = [\text{I}_2] = 0.068 \text{ mol L}^{-1}$

Solution:

\text{Hint: Explanation:} \text{Step 1: Write down the given data.} \text{It is given that equilibrium constant } K_c \text{ for the reaction } \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)} \text{ is } 54.8 \text{Therefore, at equilibrium, the equilibrium constant } K_c' \text{ for the reaction } 2\text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \text{ will be } \frac{1}{54.8} [\text{HI}] = 0.5 \text{ mol L}^{-1} \text{Step 2: Calculate concentrations of } [\text{H}_2] \text{ and } [\text{I}_2] \text{Let the concentrations of hydrogen and iodine at equilibrium be } x \text{ mol L}^{-1} [\text{H}_2] = [\text{I}_2] = x \text{ mol L}^{-1} \text{Therefore, } \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} = K_c' \Rightarrow \frac{x \times x}{(0.5)^2} = \frac{1}{54.8} \Rightarrow x^2 = \frac{0.25}{54.8} \Rightarrow x = 0.068 \text{ mol L}^{-1} \text{ (approx.)} \text{Hence, at equilibrium, } [\text{H}_2] = [\text{I}_2] = 0.068 \text{ mol L}^{-1}

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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
}