Import Question JSON

Current Question (ID: 18375)

Question:

$\text{Match the graph given in Column I with the order of reaction given in Column II.}$ $\text{More than one item in Column I may be linked to the same item in Column II:}$ $\begin{array}{|c|c|} \hline \text{Column I} & \text{Column II} \\ \hline (i) & (a) \text{1st order} \\ (ii) & (b) \text{Zero order} \\ (iii) & \\ (iv) & \\ \hline \end{array}$

Options:

1. $(i) \rightarrow (a), (ii) \rightarrow (b), (iii) \rightarrow (a), (iv) \rightarrow (b)$
2. $(i) \rightarrow (a), (ii) \rightarrow (b), (iii) \rightarrow (b), (iv) \rightarrow (a)$
3. $(i) \rightarrow (a), (ii) \rightarrow (b), (iii) \rightarrow (b), (iv) \rightarrow (b)$
4. $(i) \rightarrow (b), (ii) \rightarrow (b), (iii) \rightarrow (a), (iv) \rightarrow (a)$

Solution:

$\text{HINT: } k = \frac{[R]_0 - [R]}{t} \text{ for zero order reaction.}$ $\text{Explanation:}$ $\text{For zero-order reaction rate equation may be written as}$ $A = -kt + [R]_0 \quad \text{(i)}$ $\text{Which denotes a straight line equation similar to } y = mx + C$ $\text{On transforming (i)}$ $\frac{[R] - [R]_0}{t} = -k$ $k = \frac{[R]_0 - [R]}{t}$ $\text{For a first-order reaction } \frac{dx}{dt} \propto \text{[concentration]}$ $\text{Therefore the graph between rate and concentration may be drawn as}$ $\text{Rate} = k \cdot [t]^r$ $\text{Rate} \propto [t]^0$ $k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$ $-\frac{kt}{2.303} = \log \frac{R_0}{R}$ $-\frac{kt}{2.303} = \log [R]_0 - \log [R]$ $\log [R] = \left( -\frac{k}{2.303} \right) t + \log [A]_0$ $\text{slope} \quad \text{intercept}$

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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
}