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Current Question (ID: 21498)

Question:

$\text{For the non-stoichiometry reaction } 2\text{A} + \text{B} \rightarrow \text{C} + \text{D}, \text{ the following kinetic data were obtained in three separate experiments (all at 298 K).}$ $\begin{array}{ccc} \text{Initial Concentration (A)} & \text{Initial Concentration (B)} & \text{Initial rate of formation of C (mol L}^{-1} \text{ s}^{-1}) \\ 0.1 \text{ M} & 0.1 \text{ M} & 1.2 \times 10^{-3} \\ 0.1 \text{ M} & 0.2 \text{ M} & 1.2 \times 10^{-3} \\ 0.2 \text{ M} & 0.1 \text{ M} & 2.4 \times 10^{-3} \end{array}$ $\text{The rate law for the formation of C is:}$

Options:

1. $\frac{dc}{dt} = k[\text{A}]^2[\text{B}]$
2. $\frac{dc}{dt} = k[\text{A}][\text{B}]^2$
3. $\frac{dc}{dt} = k[\text{A}]$
4. $\frac{dc}{dt} = k[\text{A}][\text{B}]$

Solution:

$\text{Hint: The order of reaction with respect to A is 1}$ $r = k[\text{A}]^x [\text{B}]^y \text{ (Rate law expression)}$ $1.2 \times 10^{-3} = k[0.1]^x [0.1]^y$ $1.2 \times 10^{-3} = k[0.1]^x [0.2]^y$ $2.4 \times 10^{-3} = k[0.2]^x [0.1]^y$ $\text{Dividing equation (1) by (2) we get } y = 0$ $\text{Dividing equation (1) by (3) we get } x = 1$ $\therefore \text{ rate law } = k[\text{A}]$

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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
}