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Current Question (ID: 18321)

Question:

$\text{For the reaction, } \text{C}_2\text{H}_5\text{I} + \text{OH}^- \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{I}^- \text{ the rate constant was found to have a value of } 5.03 \times 10^{-2} \text{ mol}^{-1} \text{ dm}^3 \text{ s}^{-1} \text{ at } 289 \text{ K and } 6.71 \text{ mol}^{-1} \text{ dm}^3 \text{ s}^{-1} \text{ at } 333 \text{ K.}$ $\text{The rate constant at } 305 \text{ K will be:}$

Options:

1. $1.35 \text{ mol}^{-1} \text{ dm}^3 \text{ s}^{-1}$
2. $0.35 \text{ mol}^{-1} \text{ dm}^3 \text{ s}^{-1}$
3. $3.15 \text{ mol}^{-1} \text{ dm}^3 \text{ s}^{-1}$
4. $7.14 \text{ mol}^{-1} \text{ dm}^3 \text{ s}^{-1}$

Solution:

$\text{HINT: Use Arrhenius equation at different temperatures.}$ $\text{Explanation:}$ $k_1 = 5.03 \times 10^{-2} \text{ mol}^{-1} \text{ dm}^3 \text{ s}^{-1} \text{ at } T_1 = 289 \text{ K}$ $k_2 = 6.71 \text{ mol}^{-1} \text{ dm}^3 \text{ s}^{-1} \text{ at } T_2 = 333 \text{ K}$ $\log \left( \frac{6.71}{5.03 \times 10^{-2}} \right) = \frac{E_a}{2.303 \times 8.314} \left( \frac{333 - 289}{333 \times 289} \right)$ $\text{On solving we get, } E_a = 88.914 \text{ kJ The rate constant at 305 K may be determined from the relation,}$ $\log \left( \frac{k_3}{5.03 \times 10^{-2}} \right) = \frac{88.914}{2.303 \times 8.314} \left( \frac{1}{289} - \frac{1}{305} \right)$ $\text{On solving we get, } k_3 = 0.35 \text{ mol}^{-1} \text{ dm}^3 \text{ s}^{-1}$

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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
}