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Current Question (ID: 18289)

Question:

$\text{A gaseous reaction } \text{A}_2(g) \rightarrow \text{B}(g) + \frac{1}{2}\text{C}(g) \text{ shows increase in pressure from } 100 \text{ mm to } 120 \text{ mm in 5 minutes. The rate of disappearance of } \text{A}_2 \text{ will be:}$

Options:

1. $4 \text{ mm min}^{-1}$
2. $8 \text{ mm min}^{-1}$
3. $16 \text{ mm min}^{-1}$
4. $2 \text{ mm min}^{-1}$

Solution:

$\text{Rate of disappearance of } \text{A}_2 = -\frac{d[\text{A}_2]}{dt}$ $\text{For the reaction condition as pressure is directly proportional to the number of moles (assuming constant temperature and volume), we can use pressure in place of moles.}$ $\text{A}_2(g) \rightarrow \text{B}(g) + \frac{1}{2}\text{C}(g)$ $\text{At } t = 0: 100 \quad 0 \quad 0$ $\text{At } t = 5 \text{ min}: (100 - x) \quad x \quad x/2$ $\text{At } t = 5 \text{ min, total pressure } = (100 + X/2) \text{ mm and the given pressure in question is } 120 \text{ mm.}$ $\text{On equating both we will get the value of } X.$ $\Rightarrow 100 + \frac{x}{2} = 120$ $\Rightarrow x = 40 \text{ mm}$ $\therefore \text{Decrease in pressure of } \text{A}_2 = 40 \text{ mm}$ $\text{Rate of disappearance of } \text{A} = -\frac{d[\text{A}_2]}{dt}$ $\Rightarrow -\frac{d[\text{A}_2]}{dt} = \frac{40}{5} = 8 \text{ mm min}^{-1}$

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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
}