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Current Question (ID: 21112)

Question:

$\text{If the equilibrium constant } (K_c) \text{ for the reaction}$ $\text{N}_2(g) + \text{O}_2(g) \rightarrow 2\text{NO}(g) \text{ at temperature } T \text{ is } 4 \times 10^{-4} \text{ then}$ $\text{what will be the value of } K_c \text{ for the reaction,}$ $\text{NO}(g) \rightarrow \frac{1}{2}\text{N}_2(g) + \frac{1}{2}\text{O}_2(g) \text{?}$

Options:

1. $2.5 \times 10^2$
2. $4 \times 10^{-4}$
3. $50.0$
4. $0.02$

Solution:

$\text{The given equation is as follows:}$ $\text{N}_2(g) + \text{O}_2(g) \rightarrow 2\text{NO}(g) \quad \text{(1)}$ $\text{The desired equation is}$ $\text{NO}(g) \rightarrow \frac{1}{2}\text{N}_2(g) + \frac{1}{2}\text{O}_2(g) \quad \text{(2)}$ $\text{Reverse equation 1 and multiply by } \frac{1}{2} \text{ as follows:}$ $\text{Calculate the equilibrium constant for second reaction from first reaction as follows:}$ $\therefore K'_C = \frac{1}{\sqrt{K_C}} = \frac{1}{\sqrt{4 \times 10^{-4}}}$ $= \frac{1}{2 \times 10^{-2}} = 50$

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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
}