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Current Question (ID: 21083)

Question:

$\text{Given that the equilibrium constant } K_C \text{ at 800 K for the reaction}$ $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \text{ is } 64. \text{ What is the equilibrium constant } K_C \text{ at}$ $800 \text{ K for the given-below reaction is:}$ $\text{NH}_3(g) \rightleftharpoons \frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g)$

Options:

1. $\frac{1}{4}$
2. $\frac{1}{8}$
3. $8$
4. $\frac{1}{64}$

Solution:

$\text{Hint: Use the formula, that is, } K'_c = \left(\frac{1}{K_c}\right)^{\frac{1}{2}}$ $\text{The given reaction is as follows:}$ $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad \ldots(1)$ $\text{Reverse the first reaction and multiply with } \frac{1}{2} \text{ so that desired equation is obtained.}$ $\text{NH}_3(g) \rightleftharpoons \frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g)$ $K'_c = \left(\frac{1}{K_c}\right)^{\frac{1}{2}}$ $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 \rightarrow K_C = 64$ $K'_c = \left(\frac{1}{64}\right)^{\frac{1}{2}}$ $= \frac{1}{8}$

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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
}