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Question:
$\text{Consider the following reactions:}$ $\text{(i) COCl}_2 \text{ (g)} \rightleftharpoons \text{CO (g) + Cl}_2 \text{ (g)}$ $\text{(ii) CO}_2 \text{ (g) + C (s)} \rightleftharpoons \text{2CO (g)}$ $\text{(iii) 2H}_2 \text{ (g) + CO (g)} \rightleftharpoons \text{CH}_3\text{OH (g)}$ $\text{Which reaction(s) will shift in a backward direction when the pressure is increased?}$
Options:
  • 1. $\text{Only (iii)}$
  • 2. $\text{Only (ii)}$
  • 3. $\text{Both (i) and (ii)}$
  • 4. $\text{None of the above.}$
Solution:
$\text{Hint: Use Le Chatelier's Principle}$ $\text{According to Le Chatelier's principle, when pressure is increased, a system at equilibrium will shift in the direction that reduces the pressure. This means the reaction will favor the side with fewer moles of gas.}$ $\text{Let's analyze each reaction:}$ $\text{(i) COCl}_2 \text{ (g)} \rightleftharpoons \text{CO (g) + Cl}_2 \text{ (g)}$ $\text{Left side: 1 mole of gas (COCl}_2\text{)}$ $\text{Right side: 2 moles of gas (CO and Cl}_2\text{)}$ $\text{When pressure increases, the reaction will shift to the side with fewer moles of gas to reduce pressure. Therefore, reaction (i) will shift in the backward direction (toward COCl}_2\text{).}$ $\text{(ii) CO}_2 \text{ (g) + C (s)} \rightleftharpoons \text{2CO (g)}$ $\text{Left side: 1 mole of gas (CO}_2\text{) and 1 mole of solid (C). Note that solids don't contribute to pressure.}$ $\text{Right side: 2 moles of gas (2CO)}$ $\text{When pressure increases, this reaction will also shift in the backward direction (toward CO}_2 \text{ and C) since there are fewer moles of gas on the left side.}$ $\text{(iii) 2H}_2 \text{ (g) + CO (g)} \rightleftharpoons \text{CH}_3\text{OH (g)}$ $\text{Left side: 3 moles of gas (2H}_2 \text{ and CO)}$ $\text{Right side: 1 mole of gas (CH}_3\text{OH)}$ $\text{When pressure increases, this reaction will shift in the forward direction (toward CH}_3\text{OH) since there are fewer moles of gas on the right side. It will NOT shift in the backward direction.}$ $\text{Therefore, reactions (i) and (ii) will shift in the backward direction when pressure is increased.}$ $\text{The correct answer is option 3: Both (i) and (ii).}$

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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
}