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Current Question (ID: 18269)

Question:
$\text{The rate constant of a particular reaction has the dimension of frequency. The order of the reaction is:}$
Options:
  • 1. $\text{Zero.}$
  • 2. $\text{First.}$
  • 3. $\text{Second.}$
  • 4. $\text{Fractional.}$
Solution:
$\text{HINT: Units of rate constant}$ $\text{Explanation:}$ $\text{The number of periods or cycles per second is called frequency. The SI unit of frequency is s}^{-1}.$ $\text{The rate constant unit for the first order is same as frequency.}$ $\text{Find the unit of the rate constant for the first order is as follows:}$ $R = k[A]^1$ $\frac{R}{A} = k$ $k = \frac{\text{mol}}{\text{litre}} \times \text{time} \frac{\text{mol}}{\text{litre}} = \frac{1}{\text{time}} = \text{s}^{-1}$

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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
}