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Current Question (ID: 21492)

Question:

$\text{The rate constant (k) of a reaction is measured at different temperatures (T), and the data are plotted in the given figure. The activation energy of the reaction in kJ mol}^{-1} \text{ is:}$ $\text{(R is gas constant)}$

Options:

1. $2R$
2. $R$
3. $1/R$
4. $2/R$

Solution:

$\text{Hint: Use the formula, that is, } \ln k = -\frac{E_a}{RT} + \ln A$ $\text{Step 1:}$ $\text{The Arrhenius equation is as follows:}$ $\ln k = -\frac{E_a}{RT} + \ln A$ $\text{Here, the slope of the equation is } -\frac{E_a}{R}$ $\text{Step 2:}$ $\text{The slope given by } \left(\frac{Y_2 - Y_1}{X_2 - X_1}\right)$ $\text{Here, } Y_1 = 10, Y_2 = 0, X_2 = 5, \text{ and } X_1 = 0$ $\left(\frac{10 - 0}{5 - 0}\right) = -\frac{E_a}{R}$ $-\left(\frac{10}{5}\right) = -\frac{E_a}{R}$ $E_a = 2R$

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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
}