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Current Question (ID: 18362)

Question:

$1 \text{ mole of a gas changes linearly from its initial state } (2 \text{ atm}, 10 \text{ lt}) \text{ to its final state } (8 \text{ atm}, 4 \text{ lt}).$ $\text{The maximum rate constant is equal to } 20 \text{ sec}^{-1} \text{ and the value of activation energy is } 40 \text{ kJ}, \text{ assuming that the activation energy does not change in this temperature range.}$ $\text{The value of the rate constant, at the maximum temperature that the gas can attain, is:}$

Options:

1. $0.56 \times 10^{-3} \text{ sec}^{-1}$
2. $3.16 \times 10^{-3} \text{ sec}^{-1}$
3. $1.56 \times 10^{-3} \text{ sec}^{-1}$
4. $5.12 \times 10^{-3} \text{ sec}^{-1}$

Solution:

$\text{Use Arrhenius equation.}$ $\text{Let the equation of straight line is } P = mV + c$ $\text{Now putting the values, we get } P + V = 12$ $\text{From the ideal gas equation,}$ $T = \frac{PV}{nR} = \frac{PV}{R} ; T = \frac{(12-V)V}{R}$ $\text{For } T \text{ to be maximum } \frac{dT}{dV} = 0$ $V = 6 \text{ lt}, P = 6 \text{ atm.}$ $\text{Value of } T_{\text{max}} = \frac{36}{0.082} = 439 \text{ K}$ $\text{Now Putting the values of } A = 20 \text{ sec}^{-1}$ $E_a = 40 \times 10^3 \text{ J mole}^{-1}$ $\text{we get, } k = A e^{-E_a/RT}$ $k = 20 e^{-\frac{40 \times 10^3}{8.314 \times 439}} ; k = 1.56 \times 10^{-3} \text{ sec}^{-1}$

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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
}