Import Question JSON

Current Question (ID: 8401)

Question:

$\text{Some reactions are written below in Column I and their equilibrium constants in terms of } K_c \text{ are written in Column II.}$ $\text{Match the following reactions with the corresponding equilibrium constant.}$ $\begin{array}{ll} \text{Column I (Reaction)} & \text{Column II (Equilibrium constant)} \\ \text{A. } 2\text{N}_2\text{(g)} + 6\text{H}_2\text{(g)} \rightleftharpoons 4\text{NH}_3\text{(g)} & \text{1. } 2K_c \\ \text{B. } 2\text{NH}_3\text{(g)} \rightleftharpoons \text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} & \text{2. } K_c^{1/2} \\ \text{C. } \frac{1}{2}\text{N}_2\text{(g)} + \frac{3}{2}\text{H}_2\text{(g)} \rightleftharpoons \text{NH}_3\text{(g)} & \text{3. } \frac{1}{K_c} \\ & \text{4. } K_c^2 \end{array}$

Options:

1. $\text{A-4, B-3, C-2}$
2. $\text{A-1, B-2, C-3}$
3. $\text{A-1, B-4, C-3}$
4. $\text{A-4, B-1, C-3}$

Solution:

$\text{Hint: Relations between Equilibrium Constants for a General Reaction and its Multiples.}$ $\text{For the reaction, } \text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ $\text{Equilibrium constant } K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$ $\text{A. The given reaction } 2\text{N}_2\text{(g)} + 6\text{H}_2\text{(g)} \rightleftharpoons 4\text{NH}_3\text{(g)} \text{ is twice the above reaction. Hence, } K = K_c^2$ $\text{B. The reaction } 2\text{NH}_3\text{(g)} \rightleftharpoons \text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \text{ is the reverse of the above reaction. Hence, } K = \frac{1}{K_c}$ $\text{C. The reaction } \frac{1}{2}\text{N}_2\text{(g)} + \frac{3}{2}\text{H}_2\text{(g)} \rightleftharpoons \text{NH}_3\text{(g)} \text{ is half of the above reaction. Hence, } K = \sqrt{K_c} = K_c^{1/2}$

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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
}