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Current Question (ID: 18297)

Question:

$\text{The rate constant, the activation energy, and the Arrhenius parameter of a}$ $\text{chemical reaction at } 25^\circ\text{C are } 3.0 \times 10^{-4} \text{ s}^{-1}, 104.4 \text{ kJ mol}^{-1} \text{ and } 6.0 \times 10^{14}\text{s}^{-1}$ $\text{ respectively.}$ $\text{The value of the rate constant as } T \rightarrow \infty \text{ will be:}$

Options:

1. $2.0 \times 10^{18} \text{ s}^{-1}$
2. $6.0 \times 10^{14} \text{ s}^{-1}$
3. $\infty$
4. $3.6 \times 10^{30} \text{ s}^{-1}$

Solution:

$\text{HINT: Use of Arrhenius equation}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{According to Arrhenius Equation}$ $K = A e^{-E_a/RT}$ $\text{Step 2:}$ $\text{When } T \rightarrow \infty$ $e^{-E_a/RT} = 1$ $K \rightarrow A$ $A = 6 \times 10^{14}\text{s}^{-1}$ $K = 6 \times 10^{14}\text{s}^{-1}$ $\text{Hence, option 2 is the correct answer.}$

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
}