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Current Question (ID: 8369)

Question:

$\text{I}_2\text{(s)} + 5\text{F}_2\text{(g)} \rightarrow 2\text{IF}_5\text{(g)}$ $\text{The equilibrium constant } K_C \text{ expression for the above mentioned reaction is:}$

Options:

1. $K_C = \frac{[\text{IF}_5]^2}{[\text{F}_2]^5}$
2. $K_C = \frac{[\text{IF}_5]^2}{[\text{F}_2]^5 [\text{I}_2]}$
3. $K_C = \frac{[\text{F}_2]^5 [\text{I}_2]}{[\text{IF}_2]^2}$
4. $K_C = \frac{[\text{F}_2]^5}{[\text{IF}_5]^2}$

Solution:

$\text{Hint: The solid reactants or products did not include in the equilibrium constant expression.}$ $\text{Explanation:}$ $\text{The equilibrium constant } (K_C) \text{ is a mathematical relationship that shows how the concentrations of the products vary with the concentration of the reactants.}$ $\text{In equilibrium constant expression, solid reactants and products did not include because solid reactants or products concentration remains constant. The equilibrium constant expression for the given reaction is as follows:}$ $K_C = \frac{[\text{IF}_5]^2}{[\text{F}_2]^5}$ $\text{Note that } \text{I}_2\text{(s)} \text{ is not included in the expression because it is a solid, and solid concentrations are considered constant and are incorporated into the equilibrium constant value itself.}$

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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
}