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Current Question (ID: 18369)

Question:

$\text{The following data were obtained during the first-order thermal decomposition of } \text{SO}_2\text{Cl}_2 \text{ at a constant volume.}$ $\text{SO}_2\text{Cl}_2(g) \rightarrow \text{SO}_2(g) + \text{Cl}_2(g)$ $\begin{array}{|c|c|c|} \hline \text{Experiment} & \text{Time/s} & \text{Total pressure/atm} \\ \hline 1 & 0 & 0.5 \\ \hline 2 & 100 & 0.6 \\ \hline \end{array}$ $\text{The rate of the reaction when total pressure is 0.65 atm will be:}$

Options:

1. $7.8 \times 10^{-4} \text{ s}^{-1} \text{ atm.}$
2. $0.8 \times 10^{-4} \text{ s}^{-1} \text{ atm.}$
3. $2.4 \times 10^{-2} \text{ s}^{-1} \text{ atm.}$
4. $6.1 \times 10^{-8} \text{ s}^{-1} \text{ atm.}$

Solution:

$\text{HINT: Use of Log expression for First-order reaction}$ $\text{Explanation:}$ $k = \frac{2.303}{t} \log \frac{p_f - p_i}{p_f - p_i} = \frac{2.303}{t} \log \frac{(1 - 0.5) \text{ atm}}{(1 - 0.6) \text{ atm}}$ $\text{For } t = 100 \text{ s}$ $k = \frac{2.303}{t} \log \frac{0.5 \text{ atm}}{0.4 \text{ atm}} = 2.23 \times (10)^{-3} \text{ s}^{-1}$ $\text{For total pressure of 0.65 atm,}$ $\text{Rate the reaction} = kp(\text{SO}_2\text{Cl}_2)$ $\text{From the reaction stoichiometry,}$ $p(\text{SO}_2\text{Cl}_2) = 2p_i - p_t = 1 \text{ atm} - 0.65 \text{ atm} = 0.35 \text{ atm}$ $\text{So, the rate of reaction} = 2.23 \times (10)^{-3} \text{ s}^{-1} \times 0.35 \text{ atm}$ $= 7.8 \times (10)^{-4} \text{ s}^{-1} \text{ atm.}$

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{
  "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
}