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Current Question (ID: 21113)

Question:

$\text{A vessel at 1000 K contains CO}_2 \text{ with a pressure of 0.5 atm. Some of the CO}_2 \text{ is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is:}$

Options:

1. $1.8 \text{ atm}$
2. $3 \text{ atm}$
3. $0.3 \text{ atm}$
4. $0.18 \text{ atm}$

Solution:

$\text{Hint: } K = \frac{P_{\text{CO}}^2}{P_{\text{CO}_2}}$ $\text{Step 1:}$ $\text{The given reaction is as follows:}$ $\text{C(s) + CO}_2\text{(g)} \rightleftharpoons 2 \text{CO(g)}$ $\text{Initial: } 0.5 \text{ - }$ $\text{Final: } 0.5 - p \text{ } 2p$ $\text{The total pressure at equilibrium is 0.8.}$ $P(\text{CO}_2) + P(\text{CO}) = 0.8$ $0.5 - p + 2p = 0.8$ $p = 0.3$ $\text{At equilibrium, the pressure of CO}_2 \text{ is 0.2 atm, and the pressure of CO is 0.6.}$ $\text{Step 2:}$ $\text{Calculate the value of equilibrium constant K is as follows:}$ $K = \frac{P_{\text{CO}}^2}{P_{\text{CO}_2}}$ $= \frac{(0.6)^2}{0.2}$ $K = 1.8 \text{ atm}$ $\text{Hence, the correct option is the first option.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}