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Current Question (ID: 18324)

Question:

$\text{The first order rate constant for a certain reaction increases from}$ $1.667 \times 10^{-6} \text{ s}^{-1} \text{ at } 727^\circ\text{C to } 1.667 \times 10^{-4} \text{ s}^{-1} \text{ at } 1571^\circ\text{C. The rate constant at } 1150^\circ\text{C is:}$ $\text{(assume activation energy is constant over the given temperature range)}$

Options:

1. $3.911 \times 10^{-5} \text{ s}^{-1}$
2. $1.139 \times 10^{-5} \text{ s}^{-1}$
3. $3.318 \times 10^{-5} \text{ s}^{-1}$
4. $1.193 \times 10^{-5} \text{ s}^{-1}$

Solution:

$\text{HINT: Use Arrhenius Equation:}$ $2.303 \log \left[ \frac{1.667 \times 10^{-4}}{1.667 \times 10^{-6}} \right] = -\frac{E_a}{R} \left[ \frac{1}{1844} - \frac{1}{1000} \right]$ $2.303 \times 2 = \frac{E_a}{R} \times \frac{844}{1844 \times 1000} \quad \ldots(1)$ $\therefore \frac{E_a}{R} = \frac{4.606 \times 1844 \times 1000}{844}$ $2.303 \log \left[ \frac{K}{1.667 \times 10^{-6}} \right] = \frac{E_a}{R} \times \frac{1423 - 1000}{1423 \times 1000}$ $\therefore \frac{E_a}{R} \times \frac{423}{1423 \times 1000} \quad \ldots(2)$ $\text{Dividing equation (2) by equation (1)}$ $\log \left[ \frac{K}{1.667 \times 10^{-6}} \right] = \frac{423}{1423 \times 1000} \times \frac{1844 \times 1000}{844}$ $\therefore \log \left[ \frac{K}{1.667 \times 10^{-6}} \right] = 2 \times \frac{423 \times 1844}{1423 \times 844} = 1.299$ $\therefore K = 19.9 \times 1.667 \times 10^{-6} = 3.318 \times 10^{-5} \text{ s}^{-1}$

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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
}