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Current Question (ID: 8434)

Question:

$\text{Given the reaction:}$ $2\text{BrCl}_{(g)} \rightleftharpoons \text{Br}_2\text{(g)} + \text{Cl}_2\text{(g)}; K_c = 32 \text{ at } 500 \text{ K. If the initial concentration of BrCl is } 3.3 \times 10^{-3} \text{ mol L}^{-1}\text{, the molar concentration of BrCl in the mixture at equilibrium would be:}$

Options:

1. $3.0 \times 10^{-2} \text{ mol L}^{-1}$
2. $2.0 \times 10^{-4} \text{ mol L}^{-1}$
3. $2.5 \times 10^{-6} \text{ mol L}^{-1}$
4. $3.0 \times 10^{-4} \text{ mol L}^{-1}$

Solution:

$\text{Hint: } K_c = \frac{[\text{Br}_2][\text{Cl}_2]}{[\text{BrCl}]^2}$ $\text{Let the amount of Br}_2 \text{ and Cl}_2 \text{ formed at equilibrium be } c.$ $\text{Given reaction: } 2\text{BrCl}_{(g)} \rightleftharpoons \text{Br}_2\text{(g)} + \text{Cl}_2\text{(g)}$ $\begin{array}{lccc} & 2\text{BrCl}_{(g)} & \text{Br}_2\text{(g)} & \text{Cl}_2\text{(g)} \\ \text{Initial conc.} & 3.3 \times 10^{-3} & 0 & 0 \\ \text{At eq.} & (3.3 \times 10^{-3}) - 2c & c & c \end{array}$ $\text{Now, } K_c = \frac{[\text{Br}_2][\text{Cl}_2]}{[\text{BrCl}]^2}$ $32 = \frac{c \times c}{((3.3 \times 10^{-3}) - 2c)^2} \Rightarrow c = 1.5 \times 10^{-3}$ $\text{Thus, } [\text{BrCl}] = 3.3 \times 10^{-3} - (2 \times 1.5 \times 10^{-3})$ $= 3.0 \times 10^{-4} \text{ mol L}^{-1}$

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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
}