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Current Question (ID: 18302)

Question:

$\text{In a first-order reaction A} \rightarrow \text{products, the concentration of the reactant decreases to } 6.25\% \text{ of its initial value in } 80 \text{ minutes. The value of the rate constant, if the initial concentration is } 0.2 \text{ mole/litre, will be:}$

Options:

1. $2.17 \times 10^{-2} \text{ min}^{-1}$
2. $3.46 \times 10^{-2} \text{ min}^{-1}$
3. $3.46 \times 10^{-3} \text{ min}^{-1}$
4. $2.16 \times 10^{-3} \text{ min}^{-1}$

Solution:

$\text{Given } [A_0] = 0.2 \text{ M } \& \ [A] = 0.2 \times \frac{6.25}{100} = 0.0125$ $k = \frac{2.303}{t} \log \frac{[A_0]}{[A]} = \frac{2.303}{80} \log \frac{0.2}{0.0125}$ $= 0.0346 \text{ min}^{-1}$

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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
}