Import Question JSON

Current Question (ID: 21485)

Question:

$\text{For a 1}^{\text{st}} \text{ order reaction following graph is obtained between } \ln k \text{ and } \frac{1000}{T}. \text{ Then activation energy of reaction in kcal is:}$ $\begin{array}{c} \text{Slope} = -18.5 \\ \ln k \\ \frac{1000}{T(\text{K})} \end{array}$

Options:

1. $37 \text{ kcal}$
2. $40 \text{ kcal}$
3. $42 \text{ kcal}$
4. $34 \text{ kcal}$

Solution:

$\text{Hint: } \ln k = \ln A - \frac{E_a}{RT}$ $\text{Step 1:}$ $\text{The formula of rate constant according to Arrhenius is as follows:}$ $k = A e^{-\frac{E_a}{RT}}$ $\ln k = \ln A - \frac{E_a}{RT} \quad \ldots 1$ $\text{When a graph is plotted between } \ln k \text{ vs } \frac{1}{T} \text{ in that case, slope is } \frac{-E_a}{R}.$ $\text{But in this question graph between } \ln K \text{ vs } \frac{1000}{T} \text{ is plotted.}$ $\text{Equation 1 is modified and a new equation is obtained.}$ $\ln k = \ln A + \left[ \frac{-E_a}{1000R} \right] \frac{1000}{T}$ $\text{Here the slope is } \frac{-E_a}{1000R}.$ $\text{Slope} = \frac{-E_a}{1000R} = -18.5$ $\text{Step 2:}$ $\text{Calculate the value of } E_a \text{ as follows:}$ $E_a = 18.5 \times 1000 \times 2$ $= 37 \times 10^3 \text{Cal}$ $= 37 \text{kCal}$ $\text{Thus, option 1 is the correct answer.}$

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
}