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Current Question (ID: 18279)

Question:

$\text{The average rate of reaction between the time interval of 30 to 60 seconds in}$ $\text{the below mentioned pseudo 1st order reaction is:}$ $\begin{array}{|c|c|c|c|c|} \hline \text{t/s} & 0 & 30 & 60 & 90 \\ \hline [\text{Ester}]/\text{mol L}^{-1} & 0.55 & 0.31 & 0.17 & 0.085 \\ \hline \end{array}$

Options:

1. $6.67 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}$
2. $2.67 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$
3. $4.67 \times 10^{-3} \text{ mol L}^{-1} \text{ s}^{-1}$
4. $4.27 \times 10^{3} \text{ mol L}^{-1} \text{ s}^{-1}$

Solution:

$\text{The average rate of reaction between the time interval, 30 to 60 seconds is,}$ $= \frac{0.31 - 0.17}{60 - 30}$ $= \frac{0.14}{30}$ $= 4.67 \times 10^{-3} \text{ mol L}^{-1} \text{ s}^{-1}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}