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Current Question (ID: 8424)

Question:

\text{The equilibrium pressure of C}_2\text{H}_6\text{ when it is placed in a flask at 4.0 atm pressure at 899 K would be:} \text{C}_2\text{H}_6\text{(g)} \rightleftharpoons \text{C}_2\text{H}_4\text{(g)} + \text{H}_2\text{(g)} K_p = 0.04 \text{ atm at } 899 \text{ K}

Options:

1. $4.12 \text{ atm}$
2. $3.62 \text{ atm}$
3. $1.54 \text{ atm}$
4. $2.16 \text{ atm}$

Solution:

\text{Hint: } K_p = \frac{P_{C_2H_4} \cdot P_{H_2}}{P_{C_2H_6}} \text{Step 1: Find equilibrium pressure p using } K_p \text{Let p = pressure of ethene and hydrogen at equilibrium} \text{Reaction: } C_2H_6(g) \leftrightarrow C_2H_4(g) + H_2(g) \text{Initial: 4.0 atm, 0, 0} \text{Equilibrium: } (4-p), p, p K_p = \frac{p \cdot p}{4-p} = 0.04 p^2 + 0.04p - 0.16 = 0 p = 0.38 \text{ atm} \text{Step 2: Equilibrium pressure of } C_2H_6 = 4 - 0.38 = 3.62 \text{ atm}

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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
}