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Current Question (ID: 18313)

Question:

$\text{The decomposition of A into product has value of k as } 4.5 \times 10^3 \text{ s}^{-1} \text{ at } 10^\circ \text{C.}$ $\text{Energy of activation of the reaction is } 60 \text{ kJ mol}^{-1}.$ $\text{The temperature at which value k would become } 1.5 \times 10^4 \text{ s}^{-1} \text{ is:}$

Options:

1. $12^\circ \text{C}$
2. $24^\circ \text{C}$
3. $48^\circ \text{C}$
4. $36^\circ \text{C}$

Solution:

$\text{Using the Arrhenius equation at different temperatures:}$ $k_{10} = k_{283} = 4.5 \times (10)^3 \text{ s}^{-1}, k_T = 1.5 \times (10)^4 \text{ s}^{-1}, T = ?$ $E_a = 60 \text{kJ mol}^{-1} = 60,000 \text{J mol}^{-1}$ $\text{Using the Arrhenius equation, one can write}$ $\log \frac{k_T}{k_{283}} = \frac{E_a}{2.303R} \left[ \frac{1}{283K} - \frac{1}{T} \right]$ $\log \frac{1.5 \times (10)^4 \text{ s}^{-1}}{4.5 \times (10)^3 \text{ s}^{-1}} = \frac{60000 \text{J mol}^{-1}}{2.303 \times 8.3 \text{J K}^{-1} \text{mol}^{-1}} \left[ \frac{T - 283K}{283K \times T} \right]$ $\text{On solving this gives } T = 297K = 24^\circ \text{C}$

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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
}