Import Question JSON

Current Question (ID: 21497)

Question:

$\text{Decomposition of } \text{H}_2\text{O}_2 \text{ follows a first order reaction. In fifty minutes the concentration of } \text{H}_2\text{O}_2 \text{ decreases from } 0.5 \text{ to } 0.125 \text{ M in one such decomposition. When the concentration of } \text{H}_2\text{O}_2 \text{ reaches } 0.05 \text{ M, the rate of formation of } \text{O}_2 \text{ will be:}$ $\text{ln } 4 = 1.39$

Options:

1. $7 \times 10^{-2} \text{ M min}^{-1}$
2. $7 \times 10^{-4} \text{ M min}^{-1}$
3. $2.66 \times 10^{-2} \text{ M min}^{-1}$
4. $2.66 \times 10^{-4} \text{ M min}^{-1}$

Solution:

$\text{Hint: Use the formula of the first law of thermodynamic}$ $\text{Step 1:}$ $\text{The formula for a first-order reaction is as follows:}$ $k = \frac{1}{t} \ln \frac{[A_0]}{[A]}$ $\text{Calculate the value of the rate constant as follows:}$ $k = \frac{1}{50} \ln \frac{0.5}{0.125}$ $k = \frac{\ln 4}{50} \text{ min}^{-1}$ $k = 0.028 \text{ min}^{-1}$ $\text{Step 2:}$ $\text{The decomposition of } \text{H}_2\text{O}_2 \text{ is as follows:}$ $\text{H}_2\text{O}_2 \rightarrow \text{H}_2\text{O} + \frac{1}{2} \text{O}_2$ $\text{Calculate the rate of reaction as follows:}$ $\text{Rate of reaction} = k[\text{H}_2\text{O}_2]$ $= 0.028 \times 0.05$ $\text{Rate of reaction} = 0.0014 \text{ M min}^{-1}$ $\text{Step 3:}$ $\text{Calculate the rate of appearance of } \text{O}_2 \text{ as follows:}$ $\text{Rate of reaction} = \frac{1}{\frac{1}{2}} \frac{d[\text{O}_2]}{dt}$ $\text{Rate of reaction} = 2 \frac{d[\text{O}_2]}{dt}$ $0.0014 = 2 \frac{d[\text{O}_2]}{dt}$ $\frac{d[\text{O}_2]}{dt} = 7.0 \times 10^{-4} \text{ M min}^{-1}$

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
}