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Current Question (ID: 8436)

Question:

$\text{The equilibrium constant for the following reaction is } 1.6 \times 10^5 \text{ at } 1024\text{K}$ $\text{H}_2\text{(g)} + \text{Br}_2\text{(g)} \rightleftharpoons 2\text{HBr}\text{(g)}$ $\text{If HBr at pressure 10.0 bar is introduced into a sealed container at 1024 K, the equilibrium pressure of HBr will be:}$

Options:

1. $11.20 \text{ bar}$
2. $5.56 \text{ bar}$
3. $7.30 \text{ bar}$
4. $9.95 \text{ bar}$

Solution:

$\text{Hint: } K_p' = \frac{P_{\text{H}_2}}{P_{\text{Br}_2} (P_{\text{HBr}})^2}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{H}_2\text{(g)} + \text{Br}_2\text{(g)} \rightleftharpoons 2\text{HBr}\text{(g)}, K_p = 1.6 \times 10^5$ $\text{Let p be the pressure of both H}_2 \text{ and Br}_2 \text{ at the equilibrium}$ $\text{For the reverse reaction,}$ $2\text{HBr}\text{(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{Br}_2\text{(g)}, K_p' = 6.25 \times 10^{-6}$ $\text{At } t = 0: \quad 10 \quad 0 \quad 0$ $\text{At eq}^m: \quad 10 - 2p \quad p \quad p$ $\text{Step 2:}$ $\text{Now, } K_p' = \frac{P_{\text{H}_2} \times P_{\text{Br}_2}}{(P_{\text{HBr}})^2}$ $\Rightarrow 6.25 \times 10^{-6} = \frac{p}{p} \times (10 - 2p)^2$ $\Rightarrow p = 2.49 \times 10^{-2}$ $\text{Thus, at eq}^m P_{\text{HBr}} = 10 - (2 \times 2.49 \times 10^{-2}) = 9.95 \text{ bar}$

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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
}