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Current Question (ID: 8412)

Question:

$\text{A sample of pure PCl}_5 \text{ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, a concentration of PCl}_5 \text{ was found to be } 0.5 \times 10^{-1} \text{ mol L}^{-1}. \text{ If the value of } K_c \text{ is } 8.3 \times 10^{-3} \text{ mol L}^{-1}, \text{ the concentrations of PCl}_3 \text{ and Cl}_2 \text{ at equilibrium would be:}$ $\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$

Options:

1. $[\text{PCl}_3] = 0.02 \text{ mol L}^{-1}, [\text{Cl}_2] = 0.04 \text{ mol L}^{-1}$
2. $[\text{PCl}_3] = [\text{Cl}_2] = 0.02 \text{ mol L}^{-1}$
3. $[\text{PCl}_3] = 0.04 \text{ mol L}^{-1}, [\text{Cl}_2] = 0.02 \text{ mol L}^{-1}$
4. $[\text{PCl}_3] = [\text{Cl}_2] = 0.04 \text{ mol L}^{-1}$

Solution:

$\text{Hint: } K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}$ $\text{Step 1: Use the expression of } K_c \text{ to find out the concentration.}$ $\text{Let the concentrations of both PCl}_3 \text{ and Cl}_2 \text{ at equilibrium be } x \text{ mol L}^{-1}. \text{ The given reaction is:}$ $\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$ $\text{Conc. at eq}^{\text{m}}: 0.5 \times 10^{-1}, \, x, \, x$ $\text{It is given that the value of the equilibrium constant, } K_c = 8.3 \times 10^{-3}$ $\text{Step 2:}$ $\text{Now, write the expression for the equilibrium constant as follows:}$ $K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}$ $\Rightarrow \frac{x \times x}{0.5 \times 10^{-1}} = 8.3 \times 10^{-3}$ $\text{On solving the above equation, we get}$ $\Rightarrow x = 0.02 \text{ (Approx)}$ $\text{Thus, at equilibrium; } [\text{PCl}_3] = [\text{Cl}_2] = 0.02 \text{ mol L}^{-1}$

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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
}