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Current Question (ID: 18295)

Question:

$\text{For a reaction } A \rightarrow B, \text{ the Arrhenius equation is given as } \log_e k = 4 - \frac{1000}{T} \text{ the activation energy in J/mol for the given reaction will be:}$ $1. \ 8314$ $2. \ 2000$ $3. \ 2814$ $4. \ 3412$

Options:

1. $8314$
2. $2000$
3. $2814$
4. $3412$

Solution:

$\text{HINT: Use Arrhenius equation}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{According to Arrhenius equation;} \log_e k = \log_e A - \frac{E_a}{RT} \quad \text{(I)}$ $\text{Given Equation is:}$ $\log_e k = 4 - \frac{1000}{T} \quad \text{(II)}$ $\text{Step 2:}$ $\text{By comparing (I) and (II), we got}$ $\frac{-E_a}{RT} = \frac{-1000}{T}$ $E_a = 1000 \times 8.314$ $E_a = 8314 \text{ J}$

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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
}