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Current Question (ID: 8410)

Question:

$\text{For the given reaction:}$ $\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}, \Delta_r H° = 124.0 \text{ kJ mol}^{-1} \text{ and } K_c = 8.3 \times 10^{-3} \text{ mol L}^{-1}$ $\text{at 473 K}$ $\text{The effect on } K_c \text{ if (i) pressure is increased and (ii) the temperature is increased will be, respectively-}$

Options:

1. $\text{(i) Will increase; (ii) will decrease}$
2. $\text{(i) Will decrease; (ii) will remain the same}$
3. $\text{(i) Will remain the same; (ii) will increase}$
4. $\text{(i) will remain the same; (ii) Will decrease}$

Solution:

$\text{Hint: } K_c \text{ depends only on temperature}$ $\text{Explanation:}$ $\text{(i) } K_c \text{ is constant at a constant temperature. Thus, in this case, } K_c \text{ would not change when pressure is increased.}$ $\text{(ii) In an endothermic reaction, the value of } K_c \text{ increases with an increase in temperature. Since the given reaction is an endothermic reaction } (\Delta_r H° = +124.0 \text{ kJ mol}^{-1}), \text{ the value of } K_c \text{ will increase if the temperature is increased.}$ $\text{Therefore, pressure has no effect on } K_c, \text{ while increasing temperature increases } K_c \text{ for this endothermic reaction.}$

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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
}