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Current Question (ID: 18365)

Question:

$\text{Half-life is independent of the concentration of a reactant. After 10 minutes, the volume of } \text{N}_2 \text{ gas is 10 L and after complete reaction, it is 50 L. Hence, the rate constant is:}$

Options:

1. $\frac{2.303}{10} \log 5 \text{ min}^{-1}$
2. $\frac{2.303}{10} \log 1.25 \text{ min}^{-1}$
3. $\frac{2.303}{10} \log 2 \text{ min}^{-1}$
4. $\frac{2.303}{10} \log 4 \text{ min}^{-1}$

Solution:

$\text{The half-life of a reaction is independent of the concentration of the reactant. Thus, the reaction is first order.}$ $\text{Initially, the only reactant is present hence the concentration of diazonium salt is } 'a'.$ $\text{At } t=0, \ a \ 0 \ 0$ $\text{At } t=10, \ (a-x) \ x \ x$ $\text{At } t=\infty, \ 0 \ a \ a$ $\text{When the time is 10 min the concentration of chlorobenzene is 10 L.}$ $\text{When the reaction gets completed then at } t = \infty \ 50 \text{ L of chlorobenzene is obtained.}$ $\text{Hence, from the reaction, it is clear that 1 mole of diazonium salt gives 1 mole of chlorobenzene.}$ $\text{Thus, 50 L of chlorobenzene is obtained from 50 L of a diazonium salt.}$ $\text{The initial concentration of diazonium salt is 50 L. The value of } 'a' \text{ is 50 L.}$ $\text{At } t = 10 \text{ min, } x = 10 \text{ L}$ $\text{At } t = \infty, \ a = 50 \text{ L}$ $\text{Calculate the value of } k \text{ as follows:}$ $k = \frac{2.303}{t} \log \frac{a}{a-x}$ $k = \frac{2.303}{10} \log \left( \frac{50}{50-10} \right)$ $k = \frac{2.303}{10} \log 1.25$

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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
}