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Current Question (ID: 18268)

Question:

$\text{For the reaction, } \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3, \text{ if, } \frac{d[\text{NH}_3]}{dt} = 2 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1},$ $\text{the value of } -\frac{d[\text{H}_2]}{dt} \text{ would be:}$

Options:

1. $3 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$
2. $4 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$
3. $6 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$
4. $1 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$

Solution:

$\text{The rate of such a reaction where stoichiometric coefficients of}$ $\text{reactants or products are not equal to one, rate of disappearance of any of}$ $\text{the reactants or the rate of appearance of products is divided by their}$ $\text{respective stoichiometric coefficients.}$ $\text{The rate of reaction according to reactant and product is as follows:}$ $-\frac{d}{dt}(\text{N}_2) = -\frac{1}{3} \frac{d}{dt}(\text{H}_2) = \frac{1}{2} \frac{d}{dt}(\text{NH}_3)$ $-\frac{d}{dt}(\text{H}_2) = \frac{3}{2} \frac{d}{dt}(\text{NH}_3)$ $= \frac{3}{2} \times 2 \times 10^{-4}$ $= 3 \times 10^{-4} \text{ m L}^{-1} \text{ mol}^{-1}$

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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
}