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Current Question (ID: 18322)

Question:

$\text{A first-order reaction's 10 percent completion time at 298 K is the same as its 25 percent completion time at 308 K. The value of } E_a \text{ will be:}$

Options:

1. $76.64 \text{ J mol}^{-1}$
2. $66.64 \text{ kJ mol}^{-1}$
3. $76.64 \text{ kJ mol}^{-1}$
4. $70.34 \text{ kJ mol}^{-1}$

Solution:

$\text{HINT: Use Arrhenius equation for different temperatures.}$ $\text{Step 1: Calculate the value of the rate constant } (K_1) \text{ when reaction completed 10\% as follows:}$ $k_1 = \frac{2.303}{t_1} \log \frac{[A_0]}{[A]}$ $k_1 = \frac{2.303}{t_1} \log \frac{[100]}{[90]}$ $\text{Step 2:}$ $\text{Calculate the value of the rate constant } (K_2) \text{ when reaction completed 25\% as follows:}$ $k_2 = \frac{2.303}{t_2} \log \frac{[A_0]}{[A]}$ $k_2 = \frac{2.303}{t_2} \log \frac{[100]}{[75]}$ $k_2 = \frac{0.287}{t_2}$ $\text{Step 3:}$ $t_1 = t_2$ $t_1 = \frac{0.105}{k_1}$ $t_2 = \frac{0.287}{k_2}$ $\frac{0.105}{k_1} = \frac{0.287}{k_2}$ $\frac{k_2}{k_1} = \frac{0.287}{0.105} = 2.74$ $\text{Step 4:}$ $\text{Calculate the value of activation energy } (E_a) \text{ as follows:}$ $\log \frac{k_2}{k_1} = \frac{E_a}{2.303} R \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$ $\log 2.74 = \frac{E_a}{2.303} \times 8.314 \left( \frac{1}{298} - \frac{1}{308} \right)$ $0.438 \times 2.303 \times 8.314 = E_a \times \frac{10}{298 \times 308}$ $E_a = 76640.1 \text{ J mol}^{-1}$ $E_a = 76.64 \text{ kJ mol}^{-1}$

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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
}