Question

When a certain amount of ethylene was combusted \(6226 \mathrm{~kJ}\) heat was evolved. If heat of combustion of ethylene is \(1411 \mathrm{~kJ}\). The volume of \(\mathrm{O}_{2}\) (at NTP) that entered into the reaction is (a) \(296.5 \mathrm{ml}\) (b) \(296.5 \mathrm{~L}\) (c) \(6226 \times 22.4 \mathrm{~L}\) (d) \(22.4 \mathrm{~L}\).

Short answer

The volume of \(\text{O}_2\) that entered into the reaction is \( 296.5 \text{ L} \). (Option b)

Step-by-step solution

Write the Combustion Reaction for Ethylene

The combustion reaction for ethylene (\( \text{C}_2\text{H}_4 \)) with oxygen (\( \text{O}_2 \)) is:\[ \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \]This shows that 1 mole of ethylene reacts with 3 moles of oxygen.

Calculate Moles of Ethylene Combusted

Given the heat of combustion of ethylene is \( 1411 \text{ kJ/mol} \) and the total heat evolved is \( 6226 \text{ kJ} \). Calculate the moles of ethylene combusted using the formula: \[\text{Moles of ethylene} = \frac{\text{Heat evolved}}{\text{Heat of combustion}} = \frac{6226}{1411} \approx 4.41 \text{ moles} \]This means around \( 4.41 \text{ moles} \) of ethylene were combusted.

Calculate Moles of Oxygen Used

From the balanced chemical equation, 1 mole of ethylene requires 3 moles of oxygen for complete combustion. Therefore, \[\text{Moles of } \text{O}_2 = 4.41 \times 3 = 13.23 \text{ moles} \]There are approximately \( 13.23\text{ moles} \) of oxygen used in the reaction.

Calculate Volume of Oxygen at NTP

At NTP, 1 mole of any gas occupies a volume of \( 22.4 \text{ L} \). Calculate the volume of oxygen using the formula: \[\text{Volume of } \text{O}_2 = 13.23 \times 22.4 = 296.352 \text{ L} \]This means about \( 296.352 \text{ L} \) of oxygen entered into the reaction.

Key concepts

Understanding Chemical Equations

A chemical equation is a symbolic representation of a chemical reaction. It shows the substances involved, known as reactants, on one side and the substances produced, known as products, on the other side. For the combustion of ethylene, the chemical equation is:
\[ \text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \]
This equation indicates that one molecule of ethylene (\( \text{C}_2\text{H}_4 \)) reacts with three molecules of oxygen (\( \text{O}_2 \)) to produce two molecules each of carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)).
Chemical equations must be balanced, which means the number of each type of atom on both sides must be equal. This reflects the conservation of mass, as no atoms are lost or gained in the reaction. Balancing a chemical equation requires adjusting coefficients, which are numbers placed in front of compounds to ensure this equality. When studying combustion reactions like this, focus on recognizing the reactants and predicting the products to understand how substances transform.

Mastering Mole Calculations

Mole calculations are central to understanding chemical reactions quantitatively. The mole is a fundamental unit in chemistry that represents a specific number of particles, usually atoms or molecules. One mole contains Avogadro's number of particles, which is approximately \( 6.022 \times 10^{23} \).
To determine how many moles of a substance are involved in a reaction, you often use known values such as the heat of combustion. For instance, knowing ethylene's heat of combustion is \( 1411 \text{ kJ/mol} \), and \( 6226 \text{ kJ} \) of heat is evolved, you can calculate the moles of ethylene combusted with:\[\text{Moles of ethylene} = \frac{\text{Heat evolved}}{\text{Heat of combustion}} = \frac{6226}{1411} \approx 4.41 \text{ moles}\]Understanding this calculation helps in determining the extent of a reaction by linking the energy involved to the amount of reactants consumed. Moreover, using the coefficients from the balanced equation, you can relate moles of different substances involved, gaining insight into the entire reaction process.

Determining Gas Volume at NTP

When dealing with gases in chemical reactions, it is crucial to be familiar with normal temperature and pressure (NTP). At NTP, which is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm, one mole of any ideal gas occupies a volume of \( 22.4 \text{ L} \).
This property allows us to easily convert between moles and volume for gases. For the reaction involving ethylene and oxygen, since the calculation showed that approximately \( 13.23 \text{ moles} \) of \( \text{O}_2 \) participated in the reaction, the volume can be calculated as:\[\text{Volume of } \text{O}_2 = 13.23 \times 22.4 = 296.352 \text{ L}\]Such calculations are powerful tools in chemistry, helping to predict the quantities of gases needed or produced in reactions under standard conditions. This information is crucial for practical applications, from industrial processes to laboratory experiments, ensuring efficient use of resources.