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Current Question (ID: 18281)

Question:

$\text{The correct expression for the 3/4th life of a first-order reaction is:}$

Options:

1. $\frac{k}{2.303} \log \frac{4}{3}$
2. $\frac{k}{2.303} \log 4$
3. $\frac{2.303}{k} \log 4$
4. $\frac{2.303}{k} \log 3$

Solution:

$\text{(3) HINT: Use first order kinetics formula.}$ $\text{Explanation:}$ $\text{We know that, for First order reaction}$ $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$ $\text{where, } [A]_0 \text{ is the initial concentration of reactant and } [A]_t \text{ is the concentration of reactant at time 't'}$ $\text{For 3/4 th half life}$ $[A]_t = [A]_0 - \frac{3}{4} [A]_0 = \frac{1}{4} [A]_0$ $\text{Thus, } t_{3/4} = \frac{2.303}{k} \log \frac{[A]_0}{[A]_0/4}$ $\Rightarrow t_{3/4} = \frac{2.303}{k} \log 4$

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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
}