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Current Question (ID: 18317)

Question:

$\text{The activation energy of a chemical reaction can be calculated by:}$

Options:

1. $\text{Determining the rate constant at standard temperature.}$
2. $\text{Determining the rate constant at two temperatures.}$
3. $\text{Determining probability of collision.}$
4. $\text{Using the catalyst.}$

Solution:

$\text{HINT: The formula used is } \ln \left( \frac{k_1}{k_2} \right) = \frac{E_a}{R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$ $\text{Explanation:}$ $\text{The activation energy of a chemical reaction is related to the rate constant}$ $\text{of a reaction at two different temperatures i.e., } k_1 \text{ and } k_2 \text{ respectively}$ $\ln \left( \frac{k_1}{k_2} \right) = \frac{E_a}{R} \left[ \frac{1}{T_1} - \frac{1}{T_2} \right]$ $\text{where,}$ $E_a = \text{activation energy}$ $T_2 = \text{higher temperature}$ $T_1 = \text{lower temperature}$ $k_1 = \text{rate constant at temperature } T_1$ $k_2 = \text{rate constant at temperature } T_2$ $\text{This equation is known as the Arrhenius equation.}$

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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
}