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Current Question (ID: 18343)

Question:

$\text{The correct graphical representation of first-order reaction is:}$

Options:

1. $(a) \text{ and } (b)$
2. $(b) \text{ and } (c)$
3. $(c) \text{ and } (d)$
4. $(a) \text{ and } (d)$

Solution:

$\text{Step 1:}$ $\text{For the first-order reaction}$ $k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$ $\log \frac{[R]_0}{[R]} = \frac{k}{2.303} t \quad \ldots (1)$ $\text{The first equation is the same as } y=mx+c. \text{ Thus, straight-line graph is obtained. The slope is } \frac{k}{2.303}.$ $\text{The correct plot between } \log \frac{[R]_0}{[R]} \text{ vs } t \text{ can be represented by (d)}$ $\text{Step 2:}$ $\text{The time taken for any fraction of the reaction to complete is independent of the initial concentration. Let, us consider it for half of the reaction to complete.}$ $t = \frac{2.303}{k} \log \frac{a}{a-x}$ $\text{For half life}$ $t = t_{1/2}; \; x = \frac{a}{2}$ $t_{1/2} = \frac{2.303}{k} \log \frac{a}{a-\frac{a}{2}}$ $t_{1/2} = \frac{2.303}{k} \log 2$ $t_{1/2} = \frac{0.693}{k}$ $t_{1/2} \text{ is independent of initial concentration. Hence, the correct plot of } t_{1/2}, \text{ and } [R]_0 \text{ can be represented by (a)}$

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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
}