Import Question JSON

Current Question (ID: 18266)

Question:

$\text{For the reaction,}$ $\text{N}_2\text{O}_5(g) \rightarrow 2\text{NO}_2(g) + \frac{1}{2}\text{O}_2(g)}$ $\text{the value of the rate of disappearance of } \text{N}_2\text{O}_5 \text{ is given as } 6.25 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1}. \text{ The rate of formation of } \text{NO}_2 \text{ and } \text{O}_2 \text{ is given respectively as:}$

Options:

1. $6.25 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1} \text{ and } 6.25 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1}$
2. $1.25 \times 10^{-2} \text{ mol L}^{-1}\text{s}^{-1} \text{ and } 3.125 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1}$
3. $6.25 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1} \text{ and } 3.125 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1}$
4. $1.25 \times 10^{-2} \text{ mol L}^{-1}\text{s}^{-1} \text{ and } 6.25 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1}$

Solution:

$\text{HINT: Rate of disappearance of reactant = rate of appearance of product}$ $\text{or } -\frac{1}{\text{stoichiometric coefficient}} \frac{d[\text{reactant}]}{dt} = +\frac{1}{\text{stoichiometric coefficient}} \frac{d[\text{product}]}{dt}$ $\text{Explanation:}$ $\text{For the reaction,}$ $\text{N}_2\text{O}_5(g) \rightarrow 2\text{NO}_2(g) + \frac{1}{2}\text{O}_2(g)$ $-\frac{d[\text{N}_2\text{O}_5]}{dt} = +\frac{1}{2} \frac{d[\text{NO}_2]}{dt} = +2 \frac{d[\text{O}_2]}{dt}$ $\therefore \frac{d[\text{NO}_2]}{dt} = 2 \times 6.25 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1} = 12.5 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1} = 1.25 \times 10^{-2} \text{ mol L}^{-1}\text{s}^{-1}$ $\frac{d[\text{O}_2]}{dt} = -\frac{d[\text{N}_2\text{O}_5]}{dt} \times \frac{1}{2} = \frac{6.25 \times 10^{-3}}{2} \text{ mol L}^{-1}\text{s}^{-1} = 3.125 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1}$

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{
  "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
}