Q42 / 180 · COMEDK 2024 (Morning Shift) · Chemistry
Propane in presence of $\mathrm{O}_2$ gas undergoes complete combustion to produce $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$. The required $\mathrm{O}_2$ for this combustion reaction was produced by the electrolysis of water. For what duration of time had water been electrolysed by passing $200 \mathrm{~A}$ current so that Oxygen gas produced could completely burn $44 \mathrm{~g}$ of Propane?
- Correct answerA2.68 hours
- B1.98 hours
- C3.86 hours
- D1.34 hours
Solution
The balanced chemical equation for the combustion of propane is:
$\text{C}_{3}\text{H}_{8} + 5\text{O}_{2} \rightarrow 3\text{CO}_{2} + 4\text{H}_{2}\text{O}$
The molar mass of propane ($\text{C}_{3}\text{H}_{8}$) is $3 \times 12 + 8 \times 1 = 44 \text{ g/mol}$.
Moles of propane = $\dfrac{44 \text{ g}}{44 \text{ g/mol}} = 1 \text{ mol}$.
From the stoichiometry, $1 \text{ mol}$ of propane requires $5 \text{ mol}$ of $\text{O}_{2}$.
The electrolysis of water is given by:
$2\text{H}_{2}\text{O} \rightarrow 2\text{H}_{2} + \text{O}_{2}$
For $1 \text{ mol}$ of $\text{O}_{2}$ produced, $4 \text{ moles}$ of electrons are required (since each oxygen atom changes oxidation state from $-2$ to $0$, and there are $2$ oxygen atoms per $\text{O}_{2}$ molecule).
Total moles of electrons required for $5 \text{ mol}$ of $\text{O}_{2}$ = $5 \times 4 = 20 \text{ mol}$.
Using Faraday's law of electrolysis, $Q = n \times F = I \times t$, where $F \approx 96500 \text{ C/mol}$.
$20 \times 96500 = 200 \times t$
$t = \dfrac{20 \times 96500}{200} = 9650 \text{ seconds}$.
Converting to hours: $t = \dfrac{9650}{3600} \approx 2.68 \text{ hours}$.
Answer: 2.68 hours
$\text{C}_{3}\text{H}_{8} + 5\text{O}_{2} \rightarrow 3\text{CO}_{2} + 4\text{H}_{2}\text{O}$
The molar mass of propane ($\text{C}_{3}\text{H}_{8}$) is $3 \times 12 + 8 \times 1 = 44 \text{ g/mol}$.
Moles of propane = $\dfrac{44 \text{ g}}{44 \text{ g/mol}} = 1 \text{ mol}$.
From the stoichiometry, $1 \text{ mol}$ of propane requires $5 \text{ mol}$ of $\text{O}_{2}$.
The electrolysis of water is given by:
$2\text{H}_{2}\text{O} \rightarrow 2\text{H}_{2} + \text{O}_{2}$
For $1 \text{ mol}$ of $\text{O}_{2}$ produced, $4 \text{ moles}$ of electrons are required (since each oxygen atom changes oxidation state from $-2$ to $0$, and there are $2$ oxygen atoms per $\text{O}_{2}$ molecule).
Total moles of electrons required for $5 \text{ mol}$ of $\text{O}_{2}$ = $5 \times 4 = 20 \text{ mol}$.
Using Faraday's law of electrolysis, $Q = n \times F = I \times t$, where $F \approx 96500 \text{ C/mol}$.
$20 \times 96500 = 200 \times t$
$t = \dfrac{20 \times 96500}{200} = 9650 \text{ seconds}$.
Converting to hours: $t = \dfrac{9650}{3600} \approx 2.68 \text{ hours}$.
Answer: 2.68 hours
Hence, the correct answer is option: 'A'.