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Current Question (ID: 21507)

Question:

$\text{Gaseous cyclobutene isomerizes to butadiene in a first-order process which has a 'k' value of}$ $3.3 \times 10^{-4} \text{s}^{-1} \text{ at } 153 \degree \text{C. The time in minutes it takes for the isomerization to proceed } 40 \% \text{ to}$ $\text{completion at this temperature is } \_\_\_. \text{ (Rounded off to the nearest integer)}$

Options:

1. $26 \text{ min}$
2. $30 \text{ min}$
3. $22 \text{ min}$
4. $24 \text{ min}$

Solution:

$\text{Hint: The formula of rate constant for a first-order reaction is}$ $k = \frac{1}{t} \ln \left[ \frac{[A]_0}{[A]} \right]$ $\text{Step 1:}$ $\text{The reaction is as follows:}$ $\rightarrow \text{H}_2\text{C} = \text{HC} - \text{CH} = \text{CH}_2$ $\text{The formula of rate constant for a first-order reaction is}$ $K = \frac{1}{t} \times \ln \left[ \frac{[A]_0}{[A]}_t \right]$ $\text{Step 2:}$ $\text{The reaction proceeds } 40\%. \text{ It indicates that if the reactant amount is } 100\% \text{ is } t=0 \text{ sec, hence the amount of reactant is } 60\% \text{ at } t=t' \text{ time.}$ $\text{Calculate the time at which } 40\% \text{ reaction get completed.}$ $3.3 \times 10^{-4} = \frac{1}{t} \times \ln \frac{100}{60}$ $t = \frac{0.513}{3.3 \times 10^{-4}} \text{ sec}$ $t = 1554 \text{ sec}$ $\text{inmin}=26\text{min}$

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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
}