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Current Question (ID: 8387)

Question:
$\text{For the reaction,}$ $\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$ $\text{The forward reaction at constant temperature is favoured by:}$
Options:
  • 1. $\text{Introducing an inert gas at a constant volume}$
  • 2. $\text{Introducing chlorine gas at a constant volume}$
  • 3. $\text{Introducing an inert gas at constant pressure}$
  • 4. $\text{None of the above}$
Solution:
$\text{Hint: Follow Le Chatelier's principle}$ $\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$ $\text{Equilibrium moles: (a-x) \quad x \quad x}$ $\text{The addition of inert gas at constant V has no effect on reactions having } \Delta n = 0 \text{ or } \Delta n \neq 0. \text{ But the addition of inert gas at constant P has an effect on reactions having } \Delta n \neq 0, \text{ and no effect if } \Delta n = 0. \text{ The given reaction has } \Delta n \neq 0 \text{ and thus, only choice (c) is correct.}$ $\text{Analysis of each option:}$ $\text{Option 1: Inert gas at constant volume}$ $\text{• No effect on equilibrium because partial pressures remain unchanged}$ $\text{• Total pressure increases but individual concentrations stay the same}$ $\text{Option 2: Adding Cl}_2 \text{ at constant volume}$ $\text{• Increases [Cl}_2\text{] → shifts equilibrium backward (Le Chatelier's principle)}$ $\text{• Does not favor forward reaction}$ $\text{Option 3: Inert gas at constant pressure}$ $\text{• Volume must increase to maintain constant pressure}$ $\text{• Effectively decreases pressure of reacting gases}$ $\text{• For } \Delta n > 0 \text{ (1 → 2 moles), lower pressure favors forward reaction}$ $\text{• This favors the forward reaction}$ $\text{Therefore, introducing an inert gas at constant pressure favors the forward reaction.}$

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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
}