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Current Question (ID: 18291)

Question:
$\text{The correct statement about the rate constant of a reaction is:}$
Options:
  • 1. $\text{Rate constant is nearly doubled with a rise in temperature by } 10 \, ^\circ \text{C}$
  • 2. $\text{Rate constant becomes half with a rise in temperature by } 10 \, ^\circ \text{C}$
  • 3. $\text{Rate constant remains unchanged with a rise in temperature by } 10 \, ^\circ \text{C}$
  • 4. $\text{None of the above}$
Solution:
$\text{Hint: Arrhenius equation.}$ $\text{Explanation:}$ $\text{The rate constant is nearly doubled with a rise in temperature by } 10^\circ \text{ for a chemical reaction.}$ $\text{The temperature effect on the rate constant can be represented quantitatively by the Arrhenius equation,}$ $k = A e^{-E_a / RT}$ $\text{where } k \text{ is the rate constant,}$ $A \text{ is the Arrhenius factor or the frequency factor,}$ $R \text{ is the gas constant,}$ $T \text{ is the temperature, and}$ $E_a \text{ is the energy of activation for the reaction}$

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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
}