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Current Question (ID: 21499)

Question:

$\text{A reactant (A) forms two products:}$ $\text{A} \xrightarrow{k_1} \text{B, Activation Energy } \text{Ea}_1$ $\text{A} \xrightarrow{k_2} \text{C, Activation Energy } \text{Ea}_2$ $\text{If Ea}_2 = 2 \text{Ea}_1, \text{ then } k_1 \text{ and } k_2 \text{ are related as:}$

Options:

1. $k_2 = k_1 e^{\text{Ea}_1 / RT}$
2. $k_2 = k_1 e^{\text{Ea}_2 / RT}$
3. $k_1 = k_2 e^{\text{Ea}_1 / RT}$
4. $k_1 = 2k_2 e^{\text{Ea}_2 / KT}$

Solution:

$\text{Hint: Use the Arrhenius equation}$ $\text{Step 1:}$ $\text{Calculate the relation between } k_1 \text{ and } k_2 \text{ as follows:}$ $k_1 = A e^{-\text{Ea}_1 / RT}$ $k_2 = A e^{-\text{Ea}_2 / RT}$ $\text{Step 2:}$ $\text{Divide above two equations as follows:}$ \frac{k_1}{k_2} = \frac{A e^{-\text{Ea}_1 / RT}}{A e^{-\text{Ea}_2 / RT}}$ $\frac{k_1}{k_2} = e^{\text{Ea}_1 / RT + \text{Ea}_2 / RT}$ $\text{Ea}_2 = 2 \text{Ea}_1$ $\frac{k_1}{k_2} = e^{\text{Ea}_1 / RT + 2 \text{Ea}_1 / RT}$ $k_1 = k_2 e^{\text{Ea}_1 / RT}$

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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
}