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Current Question (ID: 18271)

Question:

$\text{The decomposition of NH}_3 \text{ on a platinum surface is a zero-order reaction. The rates of production of N}_2 \text{ and H}_2 \text{ will be respectively:}$ $\text{(given ; } k = 2.5 \times 10^{-4} \text{ mol}^{-1} \text{ L s}^{-1} \text{)}$

Options:

1. $2.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1} \text{ and } 5.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$
2. $2.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1} \text{ and } 7.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$
3. $1.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1} \text{ and } 4.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$
4. $0.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1} \text{ and } 3.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$

Solution:

$\text{Hint: For zero order reaction, rate of reaction = rate constant}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{The decomposition of NH}_3 \text{ on the platinum surface is represented by the following equation.}$ $2 \text{NH}_3(g) \rightarrow \text{N}_2(g) + 3\text{H}_2(g)$ $\text{Therefore,}$ $\text{Rate} = -\frac{1}{2} \frac{d[\text{NH}_3]}{dt} = \frac{d[\text{N}_2]}{dt} = \frac{1}{3} \frac{d[\text{H}_2]}{dt}$ $\text{Step 2:}$ $\text{However, it is given that the reaction is of zero order.}$ $\text{Therefore,}$ $-\frac{1}{2} \frac{d[\text{NH}_3]}{dt} = \frac{d[\text{N}_2]}{dt} = \frac{1}{3} \frac{d[\text{H}_2]}{dt} = K$ $= 2.5 \times 10^{-4} \text{ mol L}^{-1}\text{s}^{-1}$ $\text{Step 3:}$ $\text{Calculate the rate of production of N}_2 \text{ and H}_2$ $\text{Therefore, the rate of production of N}_2 \text{ is}$ $\frac{d[\text{NH}_3]}{dt} = 2.5 \times 10^{-4} \text{ mol L}^{-1}\text{s}^{-1}$ $\text{And, the rate of production of H}_2 \text{ is}$ $\frac{d[\text{H}_2]}{dt} = 3 \times 2.5 \times 10^{-4} \text{ mol L}^{-1}\text{s}^{-1}$ $= 7.5 \times 10^{-4} \text{ mol L}^{-1}\text{s}^{-1}$

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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
}