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Question:
$\text{(a) PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$ $\text{(b) CaO}_{\text{(s)}} + \text{CO}_2\text{(g)} \rightleftharpoons \text{CaCO}_3\text{(s)}$ $\text{(c) 3Fe}_{\text{(s)}} + 4\text{H}_2\text{O}_{\text{(g)}} \rightleftharpoons \text{Fe}_3\text{O}_4\text{(s)} + 4\text{H}_2\text{(g)}$ $\text{The effect of an increase in the volume on the number of moles of products in the above-mentioned reactions would be, respectively:}$
Options:
  • 1. $\text{a) Increase, b) decrease, c) same}$
  • 2. $\text{a) Decrease, b) same, c) increase}$
  • 3. $\text{a) Increase, b) increase, c) same}$
  • 4. $\text{a) Increase, b) decrease, c) increase}$
Solution:
$\text{Hint: Use Le Chatelier's principle}$ $\text{Explanation:}$ $\text{(a) The number of moles of reaction products will increase. According to Le Chatelier's principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.}$ $\text{Analysis of each reaction:}$ $\text{(a) PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$ $\text{Gaseous moles: 1 → 2 (increase)}$ $\text{Volume increase (pressure decrease) → equilibrium shifts right → products INCREASE}$ $\text{(b) CaO}_{\text{(s)}} + \text{CO}_2\text{(g)} \rightleftharpoons \text{CaCO}_3\text{(s)}$ $\text{Gaseous moles: 1 → 0 (decrease)}$ $\text{Volume increase (pressure decrease) → equilibrium shifts left → products DECREASE}$ $\text{(c) 3Fe}_{\text{(s)}} + 4\text{H}_2\text{O}_{\text{(g)}} \rightleftharpoons \text{Fe}_3\text{O}_4\text{(s)} + 4\text{H}_2\text{(g)}$ $\text{Gaseous moles: 4 → 4 (same)}$ $\text{Volume increase has no effect → products remain SAME}$

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{
  "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
}