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Current Question (ID: 8395)

Question:
$\text{For the reaction } \text{N}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{NO}\text{(g)} \text{ the equilibrium constant is } K_1.$ $\text{The equilibrium constant is } K_2 \text{ for the reaction } 2\text{NO}\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}$ $\text{The value of K for the reaction given below will be:}$ $\text{NO}_2\text{(g)} \rightleftharpoons \frac{1}{2}\text{N}_2\text{(g)} + \text{O}_2\text{(g)}$
Options:
  • 1. $\frac{1}{4}(4)K_1K_2$
  • 2. $\left[\frac{1}{K_1K_2}\right]^{1/2}$
  • 3. $\frac{1}{(K_1K_2)}$
  • 4. $\frac{1}{(2K_1K_2)}$
Solution:
$\text{Hint: Use properties of Equilibrium constant.}$ $\text{Step 1:}$ $\text{N}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{NO}\text{(g)}; K_1 \ldots\ldots (i)$ $2\text{NO}\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}; K_2 \ldots\ldots (ii)$ $\text{Add equation i and ii}$ $\text{N}_2\text{(g)} + 2\text{O}_2\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}; K = K_1 \times K_2 \ldots\ldots (iii)$ $\text{Step 2:}$ $\text{Multiply equation iii with } \frac{1}{2} \text{ and reverse the reaction}$ $\text{therefore, } \text{NO}_2\text{(g)} \rightleftharpoons \frac{1}{2}\text{N}_2\text{(g)} + \text{O}_2\text{(g)};$ $K = \left[\frac{1}{K_1K_2}\right]^{1/2}$ $\text{Detailed Explanation of Operations:}$ $\text{Step 1: Adding reactions}$ $\text{When reactions are added, their equilibrium constants are multiplied:}$ $\text{Reaction (i) + Reaction (ii) → } K_{\text{net}} = K_1 \times K_2$ $\text{Step 2: Reversing a reaction}$ $\text{When a reaction is reversed, the equilibrium constant becomes its reciprocal:}$ $\text{If } A \rightleftharpoons B \text{ has } K, \text{ then } B \rightleftharpoons A \text{ has } \frac{1}{K}$ $\text{Step 3: Multiplying by a fraction}$ $\text{When a reaction is multiplied by } \frac{1}{2}, \text{ the equilibrium constant is raised to the power } \frac{1}{2}:$ $\text{If } A \rightleftharpoons B \text{ has } K, \text{ then } \frac{1}{2}A \rightleftharpoons \frac{1}{2}B \text{ has } K^{1/2}$ $\text{Combining operations:}$ $\text{From } \text{N}_2 + 2\text{O}_2 \rightleftharpoons 2\text{NO}_2 \text{ with } K = K_1K_2$ $\text{Reverse: } 2\text{NO}_2 \rightleftharpoons \text{N}_2 + 2\text{O}_2 \text{ with } K = \frac{1}{K_1K_2}$ $\text{Multiply by } \frac{1}{2}: \text{NO}_2 \rightleftharpoons \frac{1}{2}\text{N}_2 + \text{O}_2 \text{ with } K = \left(\frac{1}{K_1K_2}\right)^{1/2}$

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{
  "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
}