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Current Question (ID: 8429)

Question:

\text{When the initial concentration of ICl is 0.78 M, the equilibrium concentration of ICl will be:} 2\text{ICl}_{(g)} \rightleftharpoons \text{I}_{2(g)} + \text{Cl}_{2(g)}; \quad K_c = 0.14

Options:

1. $0.446 \text{ M}$
2. $0.876 \text{ M}$
3. $0.345 \text{ M}$
4. $0.986 \text{ M}$

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{.} \text{The given reaction is:} 2\text{N}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{N}_2\text{O(g)} \begin{array}{|l|c|c|c|} \hline \text{} & \text{N}_2 & \text{O}_2 & \text{N}_2\text{O} \\ \hline \text{Initial conc.} & 0.482 \text{ mol} & 0.933 \text{ mol} & 0 \\ \hline \text{At equilibrium} & (0.482 - x)\text{ mol} & (0.933 - x/2)\text{ mol} & x \text{ mol} \\ \hline \end{array} \text{Therefore, at equilibrium, in the 10 L vessel:} [N_2] = \frac{0.482-x}{10}, \quad [O_2] = \frac{0.933-x/2}{10}, \quad [N_2O] = \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.} \text{Therefore, the amount of N}_2 \text{ and O}_2 \text{ reacted is also very small. Thus, } x \text{ can be neglected from the expressions of molar concentrations of N}_2 \text{ and O}_2\text{.} \text{Then,} [N_2] = \frac{0.482}{10} = 0.0482 \text{ mol L}^{-1} \text{ and } [O_2] = \frac{0.933}{10} = 0.0933 \text{ mol L}^{-1} \text{Now,} K_c = \frac{[N_2O]^2}{[N_2]^2 [O_2]} \Rightarrow 2.0 \times 10^{-37} \text{ L/mol} = \frac{\left(\frac{x}{10}\right)^2}{(0.0482)^2 \times (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} [N_2O] = \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
}