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Current Question (ID: 21502)

Question:

$\text{Consider the reaction, } 2A + B \rightarrow \text{ Products.}$ $\text{When concentration of B alone was doubled, the half-life did not change.}$ $\text{When the concentration of A alone was doubled, the rate increased by two times.}$ $\text{The unit of rate constant for this reaction is:}$

Options:

1. $\text{L mol}^{-1} \text{s}^{-1}$
2. $\text{no unit}$
3. $\text{mol L}^{-1}\text{s}^{-1}$
4. $\text{s}^{-1}$

Solution:

$\text{Hint: The overall order of the reaction is second order}$ $\text{Step 1:}$ $\text{Let the rate law for given reaction, that is, } 2A + B \rightarrow \text{ Products is-}$ $\text{Rate} = k[A]^n[B]^m$ $\text{When concentration of B gets doubled, the half life did not change.}$ $\text{Hence, half life did not depend on the concentration of B.}$ $\text{For first order reaction, half life did not depend on the concentration of reactant.}$ $\text{Order of B is 1.}$ $\text{When the concentration of A alone was doubled, the rate increased by two times.}$ $\text{Hence, the order of reaction with respect to A is also 1.}$ $\text{Step 2:}$ $\text{Rate} = k[A][B]$ $\text{mol L}^{-1} \text{s}^{-1} = k(\text{mol L}^{-1})^2$ $k = \text{mol}^{-1} \text{L} \text{s}^{-1}$

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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
}