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Current Question (ID: 21132)

Question:

$\text{The equilibrium constant for the reaction}$ $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g)$ $\text{at temperature T is } 4 \times 10^{-4}. \text{ The value of } K_c \text{ for the reaction,}$ $\text{NO}(g) \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \text{ at the same temperature is:}$

Options:

1. $2.5 \times 10^2$
2. $5 \times 10^1$
3. $4 \times 10^{-4}$
4. $2 \times 10^{-2}$

Solution:

$\text{Hint: Equilibrium constant for the reverse reaction is the inverse of}$ $\text{the equilibrium constant for the reaction in the forward direction.}$ $\text{The given equation is as follows:}$ $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g) \quad k = 4 \times 10^{-4}$ $\text{Convert the equation and multiply by } 1/2.$ $\text{NO}(g) \rightleftharpoons \frac{1}{2}\text{N}_2(g) + \frac{1}{2}\text{O}_2(g) \quad k''$ $\text{The relation between } k \text{ and } k'' \text{ is as follows:}$ $k'' = \left(\frac{1}{k}\right)^{\frac{1}{2}}$ $k'' = \left(\frac{1}{4 \times 10^{-4}}\right)^{\frac{1}{2}}$ $k'' = 50$

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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
}