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Current Question (ID: 21521)

Question:

$\text{For a reaction, } \text{N}_2\text{O}_5 (g) \rightarrow 2\text{NO}_2 (g) + \frac{1}{2}\text{O}_2 (g) \text{ in a constant volume container, no products were present initially. The final pressure of the system when 50\% of the reaction gets completed is:}$

Options:

1. $\frac{7}{2} \text{ times of initial pressure}$
2. $5 \text{ times of initial pressure}$
3. $\frac{5}{2} \text{ times of initial pressure}$
4. $\frac{7}{4} \text{ times of initial pressure}$

Solution:

$\text{N}_2\text{O}_5 (g) \rightarrow 2\text{NO}_2 (g) + \frac{1}{2}\text{O}_2 (g)$ $t = 0 \quad P_0$ $t = t \quad P_0 - x \quad 2x \quad -\frac{x}{2}$ $x = \frac{P_0}{2}$ $P_{\text{total}} = P_0 - \frac{P_0}{2} + P_0 + \frac{P_0}{4} = \frac{7}{4} P_0$

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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
}