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Current Question (ID: 21500)

Question:

$\text{The time for the half-life period of a certain reaction } A \rightarrow \text{ Products is 1 hour.}$ $\text{When the initial concentration of the reactant 'A' is } 2.0 \text{ mol L}^{-1},$ $\text{the time taken for its concentration to come from } 0.50 \text{ to } 0.25 \text{ mol L}^{-1},$ $\text{if it is a zero-order reaction, is:}$

Options:

1. $1 \text{ h}$
2. $4 \text{ h}$
3. $0.5 \text{ h}$
4. $0.25 \text{ h}$

Solution:

$\text{Hint: Half life formula for zero order reaction is } T_{\frac{1}{2}} = \frac{[A_0]}{2k}$ $\text{The half life formula for zero order reaction is as follows:}$ $T_{\frac{1}{2}} = \frac{[A_0]}{2k}$ $\text{Calculate the value of } k \text{ as follows:}$ $[A_0] = 2.0 \text{ mol L}^{-1}$ $1 \text{ hr} = \frac{2.0}{2k}$ $k = 1 \text{ mol L}^{-1}\text{hr}^{-1}$ $\text{Step 2:}$ $\text{Calculate the time by using zero order rate law equation as follows:}$ $[A] - [A_0] = -kt$ $0.25 - 0.50 = -1 \times t$ $t = 0.25 \text{ hr}$

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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
}