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Current Question (ID: 18361)

Question:

$\text{The gas phase decomposition } 2\text{N}_2\text{O}_5 \rightarrow 4\text{NO}_2 + \text{O}_2 \text{ follows the first order rate law, } K = 7.5 \times 10^{-3} \text{ sec}^{-1}. \text{ The initial pressure of } \text{N}_2\text{O}_5 \text{ is } 0.1 \text{ atm. The time of decomposition of } \text{N}_2\text{O}_5 \text{ so that the total pressure becomes } 0.15 \text{ atm will be -}$

Options:

1. $54 \text{ sec}$
2. $5.4 \text{ sec}$
3. $3.45 \text{ sec}$
4. $34.55 \text{ sec}$

Solution:

$\text{HINT: Follow first order kinetics in terms of pressure.}$ $\text{Explanation:}$ $2\text{N}_2\text{O}_5 \rightarrow 4\text{NO}_2 + \text{O}_2$ $\text{Initial pressure } 0.1 \text{ atm} \quad 0 \quad 0$ $\text{pressure at time, } t \quad (0.1 - x) \text{ atm} \quad 2x \text{ atm} \quad x/2 \text{ atm}$ $\text{Total pressure } = (0.1 - x) + 2x + x/2 = 0.1 + 3x/2 = 0.15$ $\therefore x = 0.1/3$ $\text{Now, from 1st order kinetics,}$ $t = \frac{2.303}{K} \log \frac{P_0}{P} = \frac{2.303}{7.5 \times 10^{-3}} \log \frac{0.1}{0.1-x} = 54 \text{ sec.}$

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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
}