Import Question JSON

Current Question (ID: 18264)

Question:

$\text{The correct expression for the rate of reaction given below is:}$ $5\text{Br}^- (\text{aq}) + \text{BrO}_3^- (\text{aq}) + 6\text{H}^+ (\text{aq}) \rightarrow 3\text{Br}_2 (\text{aq}) + 3\text{H}_2\text{O} (\text{l})$

Options:

1. $\frac{\Delta [\text{Br}^-]}{\Delta t} = 5 \frac{\Delta [\text{H}^+]}{\Delta t}$
2. $\frac{\Delta [\text{Br}^-]}{\Delta t} = \frac{6}{5} \frac{\Delta [\text{H}^+]}{\Delta t}$
3. $\frac{\Delta [\text{Br}^-]}{\Delta t} = \frac{5}{6} \frac{\Delta [\text{H}^+]}{\Delta t}$
4. $\frac{\Delta [\text{Br}^-]}{\Delta t} = 6 \frac{\Delta [\text{H}^+]}{\Delta t}$

Solution:

$\text{HINT: Rate of reaction =} \frac{1}{\text{stoichiometric coefficient of reactant}} \times \frac{d[\text{concentration of reactant}]}{dT}$ $\text{Explanation:}$ $\text{The chemical reaction is as follows:}$ $5\text{Br}^- (\text{aq}) + \text{BrO}_3^- (\text{aq}) + 6\text{H}^+ (\text{aq}) \rightarrow 3\text{Br}_2 (\text{aq}) + 3\text{H}_2\text{O} (\text{l})$ $\text{Rate law expression for the above equation can be written as}$ $-\frac{1}{5} \frac{\Delta [\text{Br}^-]}{\Delta t} = -\frac{\Delta [\text{BrO}_3^-]}{\Delta t} = -\frac{1}{6} \frac{\Delta [\text{H}^+]}{\Delta t} = +\frac{1}{3} \frac{\Delta [\text{Br}_2]}{\Delta t}$ $\Rightarrow \frac{\Delta [\text{Br}^-]}{\Delta t} = -\frac{\Delta [\text{BrO}_3^-]}{\Delta t} = -\frac{5}{6} \frac{\Delta [\text{H}^+]}{\Delta t}$ $\Rightarrow \frac{\Delta [\text{Br}^-]}{\Delta t} = \frac{5}{6} \frac{\Delta [\text{H}^+]}{\Delta t}$

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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
}