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Current Question (ID: 21489)

Question:

$\text{The results given in the below table were obtained during kinetic studies of the following reaction:}$ $2\text{A} + \text{B} \rightarrow \text{C} + \text{D}$ $\begin{array}{|c|c|c|c|} \hline \text{Experiment} & [\text{A}]/\text{mol L}^{-1} & [\text{B}]/\text{mol L}^{-1} & \text{Initial rate/mol L}^{-1} \text{ min}^{-1} \\ \hline \text{I} & 0.1 & 0.1 & 6.00 \times 10^{-3} \\ \text{II} & 0.1 & 0.2 & 2.40 \times 10^{-2} \\ \text{III} & 0.2 & 0.1 & 1.20 \times 10^{-2} \\ \text{IV} & X & 0.2 & 7.20 \times 10^{-2} \\ \text{V} & 0.3 & Y & 2.88 \times 10^{-1} \\ \hline \end{array}$ $\text{X and Y in the given table are respectively:}$

Options:

1. $0.3, 0.4$
2. $0.4, 0.3$
3. $0.4, 0.4$
4. $0.3, 0.3$

Solution:

$\text{Hint: The order of reaction with respect to A is 1 and with respect to B is 2}$ $\text{Step 1:}$ $\text{From rate law}$ $r = -\frac{1}{2} \frac{d[\text{A}]}{dt} = -\frac{d[\text{B}]}{dt} = K[\text{A}]^x[\text{B}]^y$ $6 \times 10^{-3} = K(0.1)^x (0.1)^y \quad \cdots \quad (1)$ $2.4 \times 10^{-2} = K(0.1)^x (0.2)^y \quad \cdots \quad (2)$ $1.2 \times 10^{-2} = K(0.2)^x (0.1)^y \quad \cdots \quad (3)$ $(3) \div (1) \Rightarrow x = 1$ $(2) \div (3) \Rightarrow y = 2$ $\text{So, order with respect to A = 1}$ $\text{Step 2:}$ $\text{Order with respect to B = 2}$ $(4) \div (3)$ $\left( \frac{x}{0.2} \right) \times \left( \frac{0.2}{0.1} \right)^2 = \frac{7.2 \times 10^{-2}}{1.2 \times 10^{-2}}$ $x = \frac{6 \times 0.2}{4}$ $x = 0.3\text{M}$ $(5) \div (4)$ $\left( \frac{y}{0.2} \right)^2 = \frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}}$

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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
}