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Current Question (ID: 8433)

Question:

\text{At a certain temperature and pressure of } 10^5 \text{ Pa, iodine vapour contains 40% by volume of I atoms. The } K_p \text{ for the equilibrium of the reaction would be:} \text{I}_2\text{(g)} \leftrightarrow 2\text{I}\text{(g)}

Options:

1. $2.67 \times 10^4 \text{ Pa}$
2. $1.00 \times 10^5 \text{ Pa}$
3. $3.63 \times 10^4 \text{ Pa}$
4. $2.18 \times 10^5 \text{ Pa}$

Solution:

$\text{Hint: Use general expression of } K_p = \frac{[P_I]^2}{[P_{I_2}]}$ $\text{Step 1:}$ $\text{Calculate the partial pressure of I and I}_2 \text{ is as follows:}$ $\text{The formula of partial pressure is as follows:}$ $P = P_T X$ $\text{The partial pressure for I atoms is, } P_I = \frac{40}{100} \times 10^5 = 4 \times 10^4 \text{ Pa}$ $\text{The partial pressure of I}_2 \text{ molecules is , } P_{I_2} = \frac{60}{100} \times 10^5 = 6 \times 10^4 \text{ Pa}$ $\text{Step 2:}$ $\text{Calculate the value of } K_p \text{ as follows:}$ $K_p = \frac{(P_I)^2}{P_{I_2}}$ $K_p = \left( \frac{(4\times10^4)^2}{6\times10^4} \right)$ $K_p = 2.67 \times 10^4 \text{ Pa}$

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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
}