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Current Question (ID: 21087)

Question:

$\text{For the following reactions, equilibrium constants are given below:}$ $\text{S}_{(s)} + \text{O}_2_{(g)} \rightleftharpoons \text{SO}_2_{(g)}; \ K_1 = 10^{52}$ $2\text{S}_{(s)} + 3\text{O}_2_{(g)} \rightleftharpoons 2\text{SO}_3_{(g)}; \ K_2 = 10^{129}$ $\text{The equilibrium constant for the reaction, } 2\text{SO}_2_{(g)} + \text{O}_2_{(g)} \rightleftharpoons 2\text{SO}_3_{(g)} \text{ is:}$

Options:

1. $10^{154}$
2. $10^{181}$
3. $10^{25}$
4. $10^{77}$

Solution:

$\text{Hint: The equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium constant for the forward reaction}$ $\text{Step 1:}$ $\text{S}_{(s)} + \text{O}_2_{(g)} \rightleftharpoons \text{SO}_2_{(g)}; \ K_1 = 10^{52}$ $2\text{S}_{(s)} + 3\text{O}_2_{(g)} \rightleftharpoons 2\text{SO}_3_{(g)}; \ K_2 = 10^{129}$ $\text{The desired equation is as follows:}$ $2\text{SO}_2_{(g)} + \text{O}_2_{(g)} \rightleftharpoons 2\text{SO}_3_{(g)} \quad K_4$ $\text{Reverse the first equation and multiply with 2.}$ $2\text{SO}_2_{(g)} \rightleftharpoons 2\text{S}_{(s)} + 2\text{O}_2_{(g)} \quad K_3 = \left(\frac{1}{K_1}\right)^2 \quad \ldots \ (3)$ $\text{Step 2:}$ $\text{Add equation 2 and 3, so that desired equation is obtained.}$ $K_4 = K_3 \times K_2$ $= 10^{129} \times \left(\frac{1}{10^{52}}\right)^2$ $K_4 = 10^{25}$

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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
}