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Current Question (ID: 8431)

Question:

\text{Given the reaction: } 2\text{N}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{N}_2\text{O(g)} \text{If a mixture of } 0.482 \text{ mol N}_2 \text{ and } 0.933 \text{ mol of O}_2 \text{ is placed in a } 10 \text{ L vessel and allowed to form N}_2\text{O (K}_c = 2.0 \times 10^{-37} \text{ L mol}^{-1}\text{), the concentration of N}_2\text{O at equilibrium will be:}

Options:

1. $6.6 \times 10^{-21} \text{M}$ (Correct)
2. $0.6 \times 10^{-21} \text{M}$
3. $4.6 \times 10^{-11} \text{M}$
4. $3.6 \times 10^{-31} \text{M}$

Solution:

$\text{Hint: Use the basic equation of K}_c$ $\text{Explanation:}$ $\text{Step 1: Write down the given data}$ $\text{Let the concentration of N}_2\text{O at equilibrium be x.}$ $\text{The given reaction is:}$ $2\text{N}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{N}_2\text{O}\text{(g)}$ $\text{Initial conc.: 0.482 mol, 0.933 mol, 0}$ $\text{At equilibrium: (0.482 - x) mol, (0.933 - x/2) mol, x mol}$ $\text{Therefore, at equilibrium, in the 10 L vessel:}$ $[\text{N}_2] = \frac{0.482-x}{10}, [\text{O}_2] = \frac{0.933-x/2}{10}, [\text{N}_2\text{O}] = \frac{x}{10}$ $\text{Step 2: Calculate the concentration of N}_2\text{O}$ $\text{The value of the equilibrium constant i.e. K}_c = 2.0 \times 10^{-37} \text{ is very small. Therefore, the amount of N}_2 \text{ and O}_2 \text{ reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of N}_2 \text{ and O}_2\text{.}$ $\text{Then,}$ $[\text{N}_2] = \frac{0.482}{10} = 0.0482 \text{ mol L}^{-1} \text{ and } [\text{O}_2] = \frac{0.933}{10} = 0.0933 \text{ mol L}^{-1}$ $\text{Now,}$ $K_c = \frac{[\text{N}_2\text{O}\text{(g)}]^2}{[\text{N}_2\text{(g)}]^2[\text{O}_2\text{(g)}]}$ $\Rightarrow 2.0 \times 10^{-37} \text{ L/mol} = \frac{(\frac{x}{10})^2}{(0.0482)^2(0.0933)}$ $\Rightarrow \frac{x^2}{100} = 2.0 \times 10^{-37} \times (0.0482)^2 \times 0.0933$ $\Rightarrow x^2 = 43.35 \times 10^{-40}$ $\Rightarrow x = 6.6 \times 10^{-20}$ $[\text{N}_2\text{O}] = \frac{x}{10} = \frac{6.6 \times 10^{-20}}{10} = 6.6 \times 10^{-21} \text{ M}$

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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
}