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Current Question (ID: 18274)

Question:
$\text{For a reaction, } 2\text{A} + \text{B} \rightarrow \text{C} + \text{D}, \text{ the following observations were recorded:}$ $\begin{array}{|c|c|c|c|} \hline \text{Experiment} & [\text{A}]/\text{mol L}^{-1} & [\text{B}]/\text{mol L}^{-1} & \text{Initial rate of formation of D/mol L}^{-1} \text{ min}^{-1} \\ \hline \text{I} & 0.1 & 0.1 & 6.0 \times 10^{-3} \\ \text{II} & 0.3 & 0.2 & 7.2 \times 10^{-2} \\ \text{III} & 0.3 & 0.4 & 2.88 \times 10^{-1} \\ \text{IV} & 0.4 & 0.1 & 2.40 \times 10^{-2} \\ \hline \end{array}$ $\text{The rate law applicable to the above mentioned reaction would be:}$
Options:
  • 1. $\text{Rate} = k[\text{A}]^2[\text{B}]^3$
  • 2. $\text{Rate} = k[\text{A}][\text{B}]^2$
  • 3. $\text{Rate} = k[\text{A}]^2[\text{B}]$
  • 4. $\text{Rate} = k[\text{A}][\text{B}]$
Solution:
$\text{HINT: The overall bond order of the reaction is 3.}$ $\text{Explanation:}$ $\text{Step 1: Calculate the order of the reaction with respect to } x$ $\text{Let the order of the reaction with respect to A be } x \text{ and with respect to B be } y.$ $\text{Therefore, the rate of the reaction is given by,}$ $\text{Rate} = k[\text{A}]^x[\text{B}]^y$ $6.0 \times 10^{-3} = k[0.1]^x[0.1]^y \quad (i)$ $7.2 \times 10^{-2} = k[0.3]^x[0.2]^y \quad (ii)$ $2.88 \times 10^{-1} = k[0.3]^x[0.4]^y \quad (iii)$ $2.40 \times 10^{-2} = k[0.4]^x[0.1]^y \quad (iv)$ $\text{Dividing equation (iv) by (i), we obtain}$ $\frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}} = \frac{k[0.4]^x[0.1]^y}{k[0.1]^x[0.1]^y}$ $\Rightarrow 4 = \left(\frac{0.4}{0.1}\right)^x$ $\Rightarrow (4)^1 = 4^x$ $\Rightarrow x = 1$ $\text{Step 2:}$ $\text{Calculate the order of the reaction with respect to B}$ $\text{Dividing equation (iii) by (ii), we obtain}$ $\frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = \frac{k[0.3]^x[0.4]^y}{k[0.3]^x[0.2]^y}$ $\Rightarrow 4 = \left(\frac{0.4}{0.2}\right)^y$ $\Rightarrow 4 = 2^y$ $\Rightarrow 2^2 = 2^y$ $\Rightarrow y = 2$ $\text{Therefore, the rate law is}$ $\text{Rate} = k[\text{A}][\text{B}]^2$

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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
}