Import Question JSON

Current Question (ID: 18285)

Question:

$\text{For the reaction } 2\text{N}_2\text{O}_5(g) \rightarrow 4\text{NO}_2(g) + \text{O}_2(g), \text{ the concentration of } \text{NO}_2$ $\text{increases by } 2.4 \times 10^{-2} \text{ mol L}^{-1}$ $\text{in 6 seconds. The rate of appearance of } \text{NO}_2 \text{ and the rate of disappearance of } \text{N}_2\text{O}_5, \text{ respectively, are:}$

Options:

1. $2 \times 10^{-3} \text{ mol L}^{-1} \text{ sec}^{-1}, 4 \times 10^{-3} \text{ mol L}^{-1} \text{ sec}^{-1}$
2. $2 \times 10^{-3} \text{ mol L}^{-1} \text{ sec}^{-1}, 1 \times 10^{-3} \text{ mol L}^{-1} \text{ sec}^{-1}$
3. $2 \times 10^{-3} \text{ mol L}^{-1} \text{ sec}^{-1}, 2 \times 10^{-3} \text{ mol L}^{-1} \text{ sec}^{-1}$
4. $4 \times 10^{-3} \text{ mol L}^{-1} \text{ sec}^{-1}, 2 \times 10^{-3} \text{ mol L}^{-1} \text{ sec}^{-1}$

Solution:

$\text{HINT: Rate of reaction} = -\frac{1}{2} \frac{\Delta [\text{N}_2\text{O}_5]}{\Delta t} = \frac{1}{4} \frac{\Delta [\text{NO}_2]}{\Delta t}$ $\text{Explanation:}$ $\text{Since } \text{NO}_2 \text{ is the product, therefore, its concentration when } t = 0 \text{ is zero.}$ $\therefore \text{Rate of appearance of } \text{NO}_2 \text{ i.e.}$ $\frac{\Delta [\text{NO}_2]}{\Delta t} = \frac{2.4 \times 10^{-2}}{6} = 4 \times 10^{-3} \text{ mol L}^{-1} \text{ s}^{-1}$ $\text{Thus, rate of reaction} = \frac{1}{4} \frac{\Delta [\text{NO}_2]}{\Delta t} = \frac{4 \times 10^{-3}}{4} \text{ mol lit}^{-1} \text{ sec}^{-1} = 1 \times 10^{-3} \text{ mol lit}^{-1} \text{ sec}^{-1}$ $\therefore \text{Rate of disappearance of } \text{N}_2\text{O}_5 \text{ i.e.}$ $\frac{\Delta [\text{N}_2\text{O}_5]}{\Delta t} = 2 \times \text{Rate of reaction} = 2 \times 10^{-3} \text{ mol lit}^{-1} \text{ sec}^{-1}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}