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Current Question (ID: 18315)

Question:

$\text{At 400 K, the energy of activation of a reaction is decreased by 0.8 kcal in the presence of a catalyst. As a result, the rate will be:}$

Options:

1. $\text{Increased by 2.71 times.}$
2. $\text{Increased by 1.18 times.}$
3. $\text{Decreased by 2.72 times.}$
4. $\text{Increased by 6.26 times.}$

Solution:

$\text{HINT: Use Arrhenius equation } K = Ae^{-E_a/RT}$ $\text{Step 1:}$ $\text{The Arrhenius Equation for catalytic reaction and non-catalytic reaction is as follows:}$ $K_1(\text{no catalyst}) = Ae^{-E_a/RT}$ $K_2(\text{catalyst}) = Ae^{-(E_a-0.8)/RT}$ $\text{Step 2:}$ $\text{Calculate the rate in the presence of a catalyst as follows:}$ $\log \frac{K_2}{K_1} = \frac{0.8 \text{ kcal}}{2.303 \ RT}$ $\log \frac{K_2}{K_1} = \frac{0.8}{2.303} \times \frac{2}{1000} \times 400$ $\log \frac{K_2}{K_1} = 0.434$ $\text{antilog } 0.434 = 2.71$ $K_2 = 2.71 \ K_1$ $\text{Hence, the rate of reaction will increase 2.71 times.}$

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
}