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Current Question (ID: 21169)

Question:

$\text{Consider the following reaction at equilibrium at a temperature of T Kelvin, with a given equilibrium constant } K_c = 3 \times 10^{-13}:$ $\text{SO}_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightleftharpoons \text{SO}_3 (g)$ $\text{Now, consider the reverse reaction:}$ $2\text{SO}_3 (g) \rightleftharpoons 2\text{SO}_2 (g) + \text{O}_2 (g)$ $\text{The equilibrium constant for this reaction is denoted as } K'_c \text{ which}$ $\text{can be expressed as } a \times 10^{+b} \text{ in scientific notation.}$ $\text{What is the value of } a + b?$

Options:

1. 26
2. 29
3. 20
4. 19

Solution:

$\text{Hint: } K'_c = \left( \frac{1}{K_c} \right)^2$ $K'_c = \left( \frac{1}{3 \times 10^{-13}} \right)^2 = 1.11 \times 10^{25}$ $a = 1.11 \text{ and } b = 25$ $a + b = 26.11$

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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
}