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Current Question (ID: 18356)

Question:

$\text{The elementary reaction A + B} \rightarrow \text{products has } k = 2 \times 10^{-5} \text{ M}^{-1} \text{ s}^{-1} \text{ at a temperature of } 27^\circ \text{C.}$ $\text{Several experimental runs are carried out using stoichiometric proportion.}$ $\text{The reaction has a temperature coefficient value of } 2.0.$ $\text{At what temperature should the reaction be carried out if, in spite of halving the concentrations,}$ $\text{the rate of reaction is desired to be } 50\% \text{ higher than in the previous run?}$ $\left( \text{Given } \frac{\ln 6}{\ln 2} = 2.585 \right)$

Options:

1. $47^\circ \text{C}$
2. $53^\circ \text{C}$
3. $57^\circ \text{C}$
4. $37^\circ \text{C}$

Solution:

$\text{(2) HINT: Rate = k[A][B]}$ $\text{Explanation:}$ $r_2 = k_2[A]_2^{1}[B]_2^{1} \quad \text{for a certain run}$ $r_1 = k_1[A]_1^{1}[B]_1^{1} \quad \text{for a previous run}$ $\text{dividing we get,}$ $\frac{r_2}{r_1} = \frac{k_2}{k_1} \frac{[A]_2}{[A]_1} \frac{[B]_2}{[B]_1}$ $\text{Substituting the given information}$ $1.5 = 2 \left( \frac{t_2 - 27}{10} \right) \times \frac{1}{2} \times \frac{1}{2}$ $\Rightarrow 6 = 2 \Rightarrow \frac{t_2 - 27}{10} \ln 2 = \ln 6$ $\Rightarrow \frac{t_2 - 27}{10} = \frac{\ln 6}{\ln 2} \Rightarrow \frac{t_2 - 27}{10} = 2.585$ $\Rightarrow t_2 = 52.85^\circ \text{C} \approx 53^\circ \text{C}$

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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
}