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Current Question (ID: 21494)

Question:

$\text{For the reaction } 2A + B \rightarrow C, \text{ the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reaction is:}$ $\begin{array}{ccc} [A] \text{ (mol L}^{-1}\text{)} & [B] \text{ (mol L}^{-1}\text{)} & \text{Initial Rate (mol L}^{-1}\text{s}^{-1}\text{)} \\ 0.05 & 0.05 & 0.045 \\ 0.10 & 0.05 & 0.090 \\ 0.20 & 0.10 & 0.72 \end{array}$

Options:

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

Solution:

$\text{Hint: The order of reaction with respect to A is 1 and B is 2.}$ $\text{Step 1:}$ $\text{The general rate law expression is as follows:}$ $\text{Rate} = k[A]^n[B]^m$ $\text{In the rate law, } n \text{ and } m \text{ are the order of A and B respectively.}$ $\text{Step 2:}$ $\text{Calculate the value of } n \text{ and } m \text{ as follows:}$ $0.045 = k[0.05]^n[0.05]^m \quad \ldots(1)$ $0.090 = k[0.20]^n[0.05]^m \quad \ldots(2)$ $\text{Divide equation 1 and 2 as follows:}$ $\frac{0.045}{0.090} = \frac{k[0.05]^n[0.05]^m}{k[0.20]^n[0.05]^m}$ $\frac{1}{2} = \left(\frac{1}{2}\right)^n$ $n = 1$ $0.045 = k[0.05]^n[0.05]^m \quad \ldots(1)$ $0.72 = k[0.20]^n[0.10]^m \quad \ldots(2)$ $\text{Divide equation 1 and 2 as follows:}$ $\frac{0.045}{0.72} = \frac{k[0.05]^n[0.05]^m}{k[0.20]^n[0.10]^m}$ $\frac{0.045}{0.72} \times 4 = \frac{[0.05]^m}{[0.10]^m}$ $\frac{1}{4} \left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^m$ $m = 2$

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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
}