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Current Question (ID: 18358)

Question:

$\text{The decomposition of sucrose follows the first-order rate law. For this decomposition, } t_{1/2} = 3.00 \text{ hours. The fraction of a sample of sucrose that remains after 8 hours would be:}$

Options:

1. $0.13$
2. $0.42$
3. $0.16$
4. $0.25$

Solution:

$\text{HINT: } k = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$ $\text{For a first-order reaction, rate constant } (k) = \frac{0.693}{t_{1/2}}$ $\text{So, } k = \frac{0.693}{3} \text{ hours} = 0.231 \text{ h}^{-1}$ $\text{Now, } k = \frac{2.303}{t} \log \frac{[R]_0}{[R]_t}$ $\log \frac{[R]_0}{[R]_t} = 0.231 \text{ h}^{-1} \times 8 \text{ h} \cdot 2.303$ $\log \frac{[R]_0}{[R]_t} = 0.802$ $\frac{[R]_0}{[R]_t} = \text{antilog} (0.802)$ $\frac{[R]_0}{[R]_t} = 6.34$ $\frac{[R]_t}{[R]_0} = 0.1576 \approx 0.16$ $\text{Hence, the fraction of the sample of sucrose that remains after 8 hours is 0.16.}$

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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
}