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Current Question (ID: 18346)

Question:

$\text{Given the following observations:}$ $\begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A] / \text{mol L}^{-1} & [B] / \text{mol L}^{-1} & \text{Initial rate} / \text{mol L}^{-1} \text{min}^{-1} \\ \hline \text{I} & 0.1 & 0.1 & 2.0 \times 10^{-2} \\ \text{II} & X & 0.2 & 4.0 \times 10^{-2} \\ \text{III} & 0.4 & 0.4 & Y \\ \hline \end{array}$ $\text{The reaction between A and B is first-order with respect to A and zero-order}$ $\text{with respect to B. The values of X and Y are, respectively:}$

Options:

1. $X = 0.2 \text{ mol L}^{-1}; Y = 0.08 \text{ mol L}^{-1}(\text{min})^{-1}$
2. $X = 0.02 \text{ mol L}^{-1}; Y = 0.08 \text{ mol L}^{-1}(\text{min})^{-1}$
3. $X = 0.01 \text{ mol L}^{-1}; Y = 0.8 \text{ mol L}^{-1}(\text{min})^{-1}$
4. $X = 0.2 \text{ mol L}^{-1}; Y = 0.8 \text{ mol L}^{-1}(\text{min})^{-1}$

Solution:

$\text{HINT: Rate law } = k([A])^1([B])^0$ $\text{Step 1: Calculate the value of } k \text{ as follows:}$ $\text{The rate of the reaction is given by,}$ $\text{Rate } = k([A])^1([B])^0$ $2.0 \times (10)^{-2} = k \ [0.1]$ $k = \frac{2.0 \times (10)^{-2}}{0.1}$ $k = 0.2 \ (\text{min})^{-1}$ $\text{Step 2:}$ $\text{Calculate the value of } X \text{ as follows:}$ $4.0 \times (10)^{-2} = k([A])^1([B])^0$ $4.0 \times (10)^{-2} = 0.2 [A]$ $\frac{4.0 \times (10)^{-2}}{0.2} = [A]$ $[A] = X = 0.2 \text{ mol L}^{-1}$ $\text{Step 3:}$ $\text{Calculate the value of } Y \text{ as follows:}$ $Y = k([A])^1([B])^0$ $Y = 0.2 \times 0.4$ $Y = 0.08 \text{ mol L}^{-1}(\text{min})^{-1}$

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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
}