Import Question JSON

Current Question (ID: 21519)

Question:

$\text{Given below are two graphs:}$ $\text{Graph I:}$ \quad t_{1/2} \quad \text{vs} \quad [R]_0 \quad \text{is valid for first order reaction.}$ $\text{Graph II:}$ \quad \log \frac{[R]}{[R]_0} \quad \text{vs} \quad \text{time with slope} \quad = \frac{-k}{2.303} \quad \text{is valid for first order reaction.}$ $\text{In the light of the above two graphs, choose the correct answer from the options given below:}$

Options:

1. $\text{Both graph I and graph II are not correct.}$
2. $\text{Graph I is not correct but graph II is correct.}$
3. $\text{Both graph I and graph II are correct.}$
4. $\text{Graph I is correct, but graph II is not correct.}$

Solution:

$\text{Hint: For first-order reaction } t_{1/2} = \frac{\ln 2}{k}$ $\text{Explanation:}$ $\text{(i) The half-life } (t_{1/2}) \text{ of a first-order reaction is constant and independent of the initial concentration of the reactant.}$ $\text{For first-order reaction } t_{1/2} = \frac{\ln 2}{k}$ $\text{(ii) For first-order reaction:}$ $\log \frac{[R]_0}{[R]} = \frac{1}{2.303} kt$ $\log \frac{[R]_0}{[R]} = \left( \frac{k}{2.303} \right) \times t$ $\text{The plot of } \log \left( \frac{[R]_0}{[R]} \right) \text{ vs. time is a straight line with slope } \frac{k}{2.303}.$ $\text{Statement I is correct but statement II is incorrect because the graph of } \log \left( \frac{[R]}{[R]_0} \right) \text{ vs time will not have a slope of } \frac{-k}{2.303}.$ $\text{Hence, option 4 is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}