Import Question JSON

Current Question (ID: 21516)

Question:

$\text{Given two first-order reactions, with the ratio of their half-lives:}$ $\frac{t_{1/2}}{t'_{1/2}} = \frac{2}{5}.\text{Calculate the ratio of their fractional half-lives:}$ $\frac{t_{1/2/3}}{t'_{4/5}} = ?$

Options:

1. 0.273
2. 0.468
3. 0.318
4. 2.55

Solution:

$\text{Hint: } t_{1/2} = \frac{0.693}{k}$ $\frac{K_2}{K_1} = \frac{2}{5}$ $t_{1/2/3} = \frac{2.303}{K_1} \log 3$ $t'_{4/5} = \frac{2.303}{K_2} \log 5$ $\frac{t_{1/2/3}}{t'_{4/5}} = \frac{K_2}{K_1} \frac{\log 3}{\log 5}$ $= \frac{2}{5} \times \frac{0.477}{0.699}$ $= 0.273$

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
}