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Current Question (ID: 21491)

Question:
$\text{If 75 \% of a first-order reaction was completed in 90 minutes, 60 \% of the same reaction would be completed in approximately (in minutes):}$ $\text{(Take : log 2 = 0.30 ; log 2.5 = 0.40)}$
Options:
  • 1. $50 \text{ min}$
  • 2. $60 \text{ min}$
  • 3. $70 \text{ min}$
  • 4. $65 \text{ min}$
Solution:
$\text{Hint: The formula of first order reaction formula is}$ $k = \frac{2.303}{t} \times \log \left[ \frac{[A_o]}{[A_t]} \right]$ $\text{Step 1:}$ $\text{First, write the equation as follows:}$ $k_1 = \frac{2.303}{90} \times \log \left[ \frac{100}{25} \right]$ $k_2 = \frac{2.303}{t_2} \times \log \left[ \frac{100}{40} \right]$ $\text{Step 2:}$ $\text{Equate equation 1 and 2 as follows:}$ $\frac{2.303}{90} \times \log 4 = \frac{2.303}{t_2} \times \log 2.5$ $t_2 = \frac{90 \times \log 2.5}{2 \log 2}$ $t_2 = \frac{90 \times 0.40}{0.60}$ $t_2 = 60 \text{ min}$ $\text{Hence, option second is the correct answer.}$

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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
}