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Current Question (ID: 8377)

Question:

$\text{HI was heated in a sealed tube at 440}^{\circ} \text{C till the equilibrium was reached. At this point, HI was found to be 22 \% decomposed. The equilibrium constant for this dissociation is:}$

Options:

1. $0.28$
2. $0.08$
3. $0.02$
4. $1.99$

Solution:

$\text{HINT: K}_C = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{Consider the following reaction and find the concentration of reactant and products at equilibrium.}$ $2\text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2;$ $\text{t = 0} \quad 1 \quad 0 \quad 0$ $\text{t = t'} \quad 1-\alpha \quad \alpha/2 \quad \alpha/2$ $\text{The value of } \alpha = 22\% = 0.22$ $\text{moles of [HI] = 1-0.22}$ $\text{moles of [HI] = 0.78}$ $\text{The moles of [H}_2\text{] and [I}_2\text{] = } \alpha/2 = 0.11$ $\text{Step 2:}$ $\text{Calculate the value of K}_C \text{ as follows:}$ $\text{K}_C = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}$ $\text{K}_C = \frac{\left[ \frac{0.11}{v} \right] \left[ \frac{0.11}{v} \right]}{\left[ \frac{0.78}{v} \right]^2}$ $\text{K}_C = \frac{0.11 \times 0.11}{(0.78)^2} = 0.019 \approx 0.02$

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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
}