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Current Question (ID: 18363)

Question:

$\text{From the plot of } \log k \text{ vs } 1/T, \text{ following parameter can be calculated:}$ $1. \text{ Activation energy.}$ $2. \text{ Rate constant of reaction.}$ $3. \text{ Order of reaction.}$ $4. \text{ Activation energy as well as the frequency factor.}$

Options:

1. $\text{Activation energy.}$
2. $\text{Rate constant of reaction.}$
3. $\text{Order of reaction.}$
4. $\text{Activation energy as well as the frequency factor.}$

Solution:

$\text{HINT: Use Arrhenius equation.}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{The Arrhenius equation is as follows:}$ $\log k = \log A - \frac{E_a}{2.303RT}$ $\text{Comparing with } y = mx + c$ $\text{Step 2:}$ $\log A = \text{intercept}$ $-\frac{E_a}{2.303RT} = \text{slope}$ $E_a = \text{slope} \times (-2.303R)$ $\text{So, } E_a \text{ (activation energy) and } A \text{ (Frequency factor) can be calculated from the above plot.}$

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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
}