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Current Question (ID: 8437)

Question:

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO}_2 \text{ in equilibrium with solid carbon has 90.55% CO by mass.}$ $\text{C}_{(s)} + \text{CO}_2_{(g)} \rightleftharpoons 2\text{CO}_{(g)}$ $\text{At the specified temperature, } K_c \text{ for this reaction would be:}$

Options:

1. $0.25 \text{ mol L}^{-1}$
2. $0.34 \text{ mol L}^{-1}$
3. $0.15 \text{ mol L}^{-1}$
4. $1.25 \text{ mol L}^{-1}$

Solution:

$\text{Hint: Use, } K_p = K_c((RT))^{\Delta n}$ $\text{Explanation:}$ $\text{Step 1: Find out the partial pressure of CO and CO}_2$ $\text{Let the total mass of the gaseous mixture = 100 g}$ $\text{Mass of CO = 90.55 g ; mass of CO}_2 = 100 - 90.55 = 9.45\text{g}$ $\text{Now, the number of moles of CO, } n_{CO} = \frac{90.55}{28} = 3.234\text{mol}$ $\text{Number of moles of CO}_2, n_{(CO)_2} = \frac{9.45}{44} = 0.215\text{mol}$ $\text{Thus, partial pressure of CO,}$ $p_{CO} = \frac{n_{CO}}{n_{Total}} \times p_{total} = \frac{3.234}{3.234+0.215} \times 1 = 0.938\text{atm}$ $\text{The partial pressure of CO}_2,$ $p_{(CO)_2} = \frac{n_{(CO)_2}}{n_{Total}} \times p_{total} = \frac{0.215}{3.234+0.215} \times 1 = 0.062\text{atm}$ $\text{Step 2: Calculate } K_p \text{ and } K_c$ $K_p = \frac{(p_{CO})^2}{(p_{(CO)_2})} = \frac{(0.938)^2}{0.062} = 14.19 \text{ atm}$ $\text{For the given reaction, } \Delta n = 2 - 1 = 1$ $\text{We know that, } K_p = K_c((RT))^{\Delta n}$ $K_c = \frac{14.19}{0.0821 \times 1127} = 0.15\text{mol L}^{-1}$

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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
}