Import Question JSON

Current Question (ID: 18386)

Question:

$\text{Match the statements in Column I and Column II:}$ $\text{Column I (Thermodynamic/Kinetic Principle)}$ $\text{A. Catalyst alters the rate of reaction}$ $\text{B. } e^{-E_a/RT}$ $\text{C. Energetically favorable reactions are sometimes slow}$ $\text{D. The area under the Maxwell-Boltzmann curve is constant}$ $\text{Column II (Impact on Reaction Dynamics)}$ $\text{1. Proper orientation is not there always}$ $\text{2. By lowering the activation energy}$ $\text{3. Total probability is one}$ $\text{4. Refers to the fraction of molecules with energy equal to or greater than the activation energy}$

Options:

1. $2 \ 4 \ 1 \ 3$
2. $3 \ 1 \ 4 \ 2$
3. $1 \ 4 \ 3 \ 2$
4. $3 \ 4 \ 1 \ 2$

Solution:

$\text{1. Catalyst alters the rate of reaction by lowering the activation energy.}$ $\text{2. } e^{-E_a/RT} \text{ refers to the fraction of molecules with kinetic energy equal to or greater than the activation energy.}$ $\text{3. Energetically favorable reactions are sometimes slow due to the improper orientation of molecules that cause some ineffective collision of molecules.}$ $\text{4. Area under the Maxwell-Boltzmann curve is constant because the total probability of a molecule taking part in a chemical reaction is equal to one.}$

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
}