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Current Question (ID: 18306)

Question:
$\text{The rate constant for a chemical reaction that takes place at 500 K is expressed as } K = A \cdot e^{-1000}. \text{ The activation energy of the reaction will be:}$
Options:
  • 1. $100 \text{ cal/mol}$
  • 2. $1000 \text{ kcal/mol}$
  • 3. $10^4 \text{ kcal/mol}$
  • 4. $10^6 \text{ kcal/mol}$
Solution:
$\text{According to the Arrhenius equation, } K = A \times e^{-E_a/RT}$ $\text{where } K = \text{rate constant of the reaction, } A = \text{pre-exponential factor, } E_a = \text{activation energy, } T = \text{temperature.}$ $\text{Given, } K = A \times e^{-1000}$ $\text{By comparing with the Arrhenius equation, we got}$ $-\frac{E_a}{RT} = -1000$ $E_a = 1000 \times R \times T$ $= 1000 \times 500 \times 2$ $= 1000000$ $= 1000 \text{ kcal/mol}$

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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
}