Import Question JSON

Current Question (ID: 18367)

Question:

$\text{A first-order reaction is 15 \% completed in 20 minutes. The amount of time required to complete 60 \% of the reaction is:}$

Options:

1. $112.8 \text{ min}$
2. $120.7 \text{ min}$
3. $100.4 \text{ min}$
4. $140.7 \text{ min}$

Solution:

$\text{For the first order reaction}$ $K = \frac{2.303}{t} \log \frac{a}{a-x}$ $\text{First case: } a = 100, x = 15, t = 20 \text{ minutes}$ $(a-x) = 100 - 15 = 85$ $t_{15\%} = \frac{2.303}{K} \log \frac{100}{85}$ $\therefore 20 = \frac{2.303}{K} \log \frac{100}{85} \quad \text{(1)}$ $\text{Second case: } a = 100, x = 60, t_{60\%} = ?$ $(a-x) = 100 - 60 = 40$ $t_{60\%} = \frac{2.303}{K} \log \frac{100}{40} \quad \text{(2)}$ $\text{Dividing equation (2) by (1), we get}$ $\frac{t_{60\%}}{20} = \frac{\log \frac{100}{40}}{\log \frac{100}{85}} = \frac{\log 100 - \log 40}{\log 100 - \log 85}$ $= \frac{2.000 - 1.6021}{2.000 - 1.9294}$ $= \frac{0.3979}{0.0706} = 5.64$ $\therefore t_{60\%} = 20 \times 5.64 = 112.8 \text{ minutes}$

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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
}