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Current Question (ID: 21513)

Question:

$\text{Consider the following first-order reactions:}$ $A \rightarrow C; \ t_{1/2} = 15 \ \text{min}, \ B \rightarrow D; \ t_{1/2} = 5 \ \text{min}$ $\text{The initial concentrations of A and B are 1 molar and 8 molar respectively.}$ $\text{The time when the concentration of A and B becomes equal is 'X' minutes.}$ $\text{The value of 2X is:}$

Options:

1. $40 \ \text{min}$
2. $50 \ \text{min}$
3. $45 \ \text{min}$
4. $54 \ \text{min}$

Solution:

$\text{Step 1: For a first order reaction:} \ k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ $\text{where, } k = \text{rate constant, } [A]_t \text{ is the concentration of reactant at time 't' and } [A]_0 \text{ is the initial concentration of reactant.}$ $\text{Step 2: } [A]_0 = 1 \text{ and } [B]_0 = 8$ $\text{Also, } (t_{1/2})_1 = 15 \ \text{min and } (t_{1/2})_2 = 5 \ \text{min.}$ $\text{We also know that for a first order reaction,}$ $\frac{1}{K_1} \ln \frac{1}{[A]} = \frac{1}{K_2} \ln \frac{8}{[A]}$ $A = \frac{1}{\sqrt{8}}$ $t_1/2 = \frac{15}{0.693} \ln \sqrt{8}$ $= 22.5 \ \text{min}$ $2x = 45$

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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
}