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Current Question (ID: 18355)

Question:

$\text{The rate of a reaction increases 4-fold when concentration of reactant is increased 16 times.}$ $\text{If the rate of reaction is } 4 \times 10^{-6} \text{ mol L}^{-1}\text{s}^{-1} \text{ when concentration of the reactant is } 4 \times 10^{-4} \text{ mol L}^{-1}, \text{ the rate constant of the reaction will be:}$

Options:

1. $2 \times 10^{-4} \text{ m}^{1/2} \text{ L}^{-1/2} \text{ s}^{-1}$
2. $1 \times 10^{-2} \text{ s}^{-1}$
3. $4 \times 10^{-4} \text{ mol}^{-1/2} \text{ L}^{-1/2} \text{ s}^{-1}$
4. $25 \text{ mol}^{-1} \text{ L min}^{-1}$

Solution:

$\text{(1) HINT: Rate } \propto \sqrt{\text{concentration}}$ $\text{Explanation:}$ $\text{The rate of a reaction increases 4-fold when concentration of reactant is increased 16 times. Thus,}$ $\text{Rate } \propto \sqrt{\text{Concentration}}, \text{ Rate } = k \sqrt{\text{Concentration}}$ $k = \frac{\text{rate}}{(\text{conc.})^{1/2}} = \frac{4 \times 10^{-6}}{(4 \times 10^{-4})^{1/2}} = \frac{4 \times 10^{-6}}{2 \times 10^{-2}}$ $= 2 \times 10^{-4} \text{ mole}^{1/2} \text{ L}^{-1/2} \text{ s}^{-1}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}