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Current Question (ID: 8548)

Question:

$\text{Consider the nitration of benzene using mixed conc. H}_2\text{SO}_4 \text{ and HNO}_3\text{. If a large amount of KHSO}_4 \text{ is added to the mixture, the rate of nitration will be:}$

Options:

1. $\text{Slower}$
2. $\text{Unchanged}$
3. $\text{Doubled}$
4. $\text{Faster}$

Solution:

$\text{Hint: Common ion effect}$ $\text{Explanation:}$ $\text{When nitration of benzene occurs in the presence of conc. H}_2\text{SO}_4 \text{ and HNO}_3\text{, nitrobenzene is formed.}$ $\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightleftharpoons \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}$ $\text{Electrophile} \quad \text{Nucleophile}$ $\text{If a large amount of KHSO}_4 \text{ is added to this mixture, more HSO}_4^- \text{ ion is furnished which decrease the concentration of NO}_2^+ \text{ i.e., electrophile. As the concentration of electrophile decreases, the rate of electrophilic aromatic substitution slows down.}$

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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
}