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Current Question (ID: 18333)

Question:

$\text{The rate constant for a first-order reaction is } 4.606 \times 10^{-3} \text{ s}^{-1}. \text{ The time required to reduce } 2.0 \text{ g of the reactant to } 0.2 \text{ g will be:}$

Options:

1. $200 \text{ s}$
2. $500 \text{ s}$
3. $1000 \text{ s}$
4. $100 \text{ s}$

Solution:

$\text{The formula of first order reaction is as follows:}$ $k = \frac{2.303}{t} \log \frac{a}{(a-x)}$ $\text{The given values are as follows:}$ $a = 2.0 \text{ g}$ $a-x = 0.2 \text{ g}$ $\text{Calculate the value of } t \text{ as follows:}$ $4.606 \times 10^{-3} = \frac{2.303}{t} \log \frac{2}{0.2}$ $t = \frac{2.303}{4.606 \times 10^{-3}} \log \frac{2}{0.2}$ $= 500 \text{ s}$

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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
}