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Current Question (ID: 21071)

Question:

$\text{At 1990 K and 1 atm pressure, there are an equal number of Cl}_2 \text{ molecules and Cl atoms in the reaction mixture.}$ $\text{The value of } K_P \text{ for the reaction } \text{Cl}_2{(g)} \rightleftharpoons 2 \text{Cl}{(g)} \text{ under the above conditions is } x \times 10^{-1}. \text{ The value of } x \text{ is:}$

Options:

1. $4$
2. $8$
3. $5$
4. $10$

Solution:

$\text{Hint: } K_P = \frac{(P_{\text{Cl}})^2}{P_{\text{Cl}_2}}$ $\text{Step 1:}$ $\text{The number of molecules of Cl}_2 \text{ and the number of atoms of Cl is equal hence, the number of moles of both must be equal.}$ $\text{Cl}_2 \rightleftharpoons 2 \text{Cl}$ $\text{Let mol of both of Cl}_2 \text{ and Cl is } x. \text{ Calculate the partial pressure of Cl}_2 \text{ and Cl as follows:}$ $\text{Partial pressure} = P_T X_A$ $P_{\text{Cl}} = \frac{x}{2x} \times 1 = \frac{1}{2}$ $P_{\text{Cl}_2} = \frac{x}{2x} \times 1 = \frac{1}{2}$ $\text{Step 2:}$ $K_P = \frac{P_{\text{Cl}}^2}{P_{\text{Cl}_2}}$ $K_P = \frac{\left(\frac{1}{2}\right)^2}{\frac{1}{2}} = \frac{1}{2} = 0.5 \Rightarrow 5 \times 10^{-1}$ $\text{Hence, option third is the correct answer.}$

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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
}