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Current Question (ID: 18368)

Question:

$\text{For a reaction A} \rightarrow \text{Product, with } k = 2.0 \times 10^{-2} \text{ s}^{-1}, \text{ if the initial concentration}$ $\text{of A is } 1.0 \text{ mol L}^{-1}, \text{ the concentration of A after 100 seconds would be:}$

Options:

1. $0.23 \text{ mol L}^{-1}$
2. $0.18 \text{ mol L}^{-1}$
3. $0.11 \text{ mol L}^{-1}$
4. $0.13 \text{ mol L}^{-1}$

Solution:

$\text{Hint: First order reaction.}$ $\text{Step 1:}$ $\text{Given data: } k = 2.0 \times 10^{-2} \text{ s}^{-1}; \text{ Time}(t) = 100 \text{ s}; \text{ Initial reactant}$ $\text{concentration (A) = 1.0 mol L}^{-1}$ $\text{As the unit of rate constant is s}^{-1}, \text{ so the reaction follows first order.}$ $\text{Step 2:}$ $\text{The formula of } k \text{ for first order reaction is}$ $k = \frac{2.303}{t} \log \frac{[A_0]}{[A]}$ $2.0 \times 10^{-2} = \frac{2.303}{100} \log \frac{[1]}{[A]}$ $\frac{2.0 \times 10^{-2} \times 100}{2.303} = \log \frac{1}{[A]}$ $0.868 = \log \frac{1}{[A]}$ $10^{0.868} = \frac{1}{[A]}$ $[A] = \frac{1}{7.38}$ $[A] = 0.135 \text{ mol L}^{-1}$ $\text{Hence, the remaining concentration of A is } 0.135 \text{ mol L}^{-1}$

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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
}