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Current Question (ID: 18308)

Question:
$\text{The following reaction was carried out at 300 K.}$ $2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g)$ $\text{The rate of formation of } \text{SO}_3 \text{ is related to the rate of disappearance of } \text{O}_2 \text{ by the following expression:}$
Options:
  • 1. $-\frac{\Delta [\text{O}_2]}{\Delta t} = +\frac{1}{2} \frac{\Delta [\text{SO}_3]}{\Delta t}$
  • 2. $-\frac{\Delta [\text{O}_2]}{\Delta t} = \frac{\Delta [\text{SO}_3]}{\Delta t}$
  • 3. $-\frac{\Delta [\text{O}_2]}{\Delta t} = -\frac{1}{2} \frac{\Delta [\text{SO}_3]}{\Delta t}$
  • 4. $\text{None of the above.}$
Solution:
$\text{Given reaction:}$ $2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g)$ $\text{Rate of reaction} = -\frac{\Delta [\text{O}_2]}{\Delta t} = +\frac{1}{2} \frac{\Delta [\text{SO}_3]}{\Delta t}$ $\text{Therefore, the rate of disappearance is related to the rate of formation of } \text{SO}_3 \text{ as } -\frac{\Delta [\text{O}_2]}{\Delta t} = +\frac{1}{2} \frac{\Delta [\text{SO}_3]}{\Delta t}$

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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
}