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Current Question (ID: 21488)

Question:

$\text{The rate constant of a reaction increases by five times on increase in temperature from } 27^\circ\text{C to } 52^\circ\text{C. The value of activation energy in kJ mol}^{-1} \text{ is-}$ $\text{(Rounded-off to the nearest integer) [R = 8.314 J K}^{-1} \text{ mol}^{-1}]$

Options:

1. 50
2. 56
3. 52
4. 60

Solution:

$\text{Hint: Use the formula, } \ln \frac{K_2}{K_1} = \frac{E_a}{R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$ $\text{Step 1:}$ $\text{The given values are as follows:}$ $T_1 = 300 \text{ K, } K_2 = 5 \text{ K}$ $T_2 = 325 \text{ K}$ $\text{The formula is as follows:}$ $\ln \frac{K_2}{K_1} = \frac{E_a}{R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$ $\text{Step 2:}$ $\text{Calculate the value of activation energy is as follows:}$ $\ln \frac{5K}{K} = \frac{E_a}{8.314} \left[ \frac{1}{300} - \frac{1}{325} \right]$ $\frac{1.61 \times 8.314 \times 300 \times 325}{25} = E_a$ $E_a = 52203.6 \text{ J mol}^{-1}$ $E_a = 52.2 \text{ kJ mol}^{-1}$ $\text{The nearest integer answer will be } 52 \text{ kJ mol}^{-1}$

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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
}