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Current Question (ID: 21495)

Question:

$\text{NO}_2 \text{ required for a reaction is produced by the decomposition of } \text{N}_2\text{O}_5 \text{ in CCl}_4 \text{ as per the equation,}$ $2\text{N}_2\text{O}_5(g) \rightarrow 4\text{NO}_2(g) + \text{O}_2(g)$ $\text{The initial concentration of } \text{N}_2\text{O}_5 \text{ is } 3.00 \text{ mol L}^{-1} \text{ and it is } 2.75 \text{ mol L}^{-1}$ $\text{after 30 minutes. The rate of formation of } \text{NO}_2 \text{ is:}$

Options:

1. $2.083 \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1}$
2. $8.333 \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1}$
3. $4.167 \times 10^{-3} \text{ mol L}^{-1} \text{ min}^{-1}$
4. $1.667 \times 10^{-2} \text{ mol L}^{-1} \text{ min}^{-1}$

Solution:

$\text{Hint: Rate of reaction } = -\frac{1}{2} \frac{d[\text{N}_2\text{O}_5]}{dt} = \frac{1}{4} \frac{d[\text{NO}_2]}{dt} = \frac{d[\text{NO}_2]}{dt}$ $\text{The rate of reaction is as follows:}$ $-\frac{1}{2} \frac{d[\text{N}_2\text{O}_5]}{dt} = \frac{1}{4} \frac{d[\text{NO}_2]}{dt} = \frac{d[\text{NO}_2]}{dt}$ $\text{The rate of formation of } \text{NO}_2 \text{ with respect to } \text{N}_2\text{O}_5 \text{ is as follows:}$ $-\frac{4}{2} \frac{d[\text{N}_2\text{O}_5]}{dt} = \frac{d[\text{NO}_2]}{dt}$ $\text{The initial concentration of } \text{N}_2\text{O}_5 \text{ is } 3.00 \text{ mol L}^{-1} \text{ and it is } 2.75 \text{ mol L}^{-1}$ $\text{after 30 minutes.}$ $\text{Calculate rate of formation of } \text{NO}_2 \text{ as follows:}$ $-2 \times \frac{(2.75 - 3.00)}{30} = \frac{d[\text{NO}_2]}{dt}$ $= 1.66 \times 10^{-2} \text{ mol L}^{-1} \text{ min}^{-1}$

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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
}