Question

Most ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right),\) the starting material for producing polyethylene, comes from petroleum processing. It also occurs naturally as a fruit-ripening hormone and as a component of natural gas. (a) The heat transferred during combustion of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is \(-1411 \mathrm{~kJ} / \mathrm{mol} .\) Write a balanced thermochemical equation. (b) How many grams of \(\mathrm{C}_{2} \mathrm{H}_{4}\) must burn to give \(70.0 \mathrm{~kJ}\) of heat?

Short answer

1.39 g of \(\text{C}_2\text{H}_4\) must burn to release 70.0 kJ of heat, with the balanced equation: \[ \text{C}_2\text{H}_4 + 3 \text{O}_2 \rightarrow 2 \text{CO}_2 + 2 \text{H}_2\text{O} \hspace{0.5cm} \Delta H = -1411 \text{kJ} \].

Step-by-step solution

- Determine the balanced equation

First, write the balanced chemical equation for the combustion of ethylene \(\text{C}_2\text{H}_4\). Combustion involves reacting with oxygen \(O_2\) to form carbon dioxide \(CO_2\) and water \(H_2O\). The balanced equation is: \[ \text{C}_2\text{H}_4 + 3 O_2 \rightarrow 2 \text{CO}_2 + 2 \text{H}_2 \text{O} \].

- Include the heat transfer in the equation

This is a thermochemical equation, which means we need to include the heat transfer. The heat of combustion for ethylene is given as \(-1411\) kJ/mol. This means that when 1 mol of \(\text{C}_2\text{H}_4\) combusts, \(-1411 \, \text{kJ}\) of energy is released:\[ \text{C}_2\text{H}_4 + 3 \text{O}_2 \rightarrow 2 \text{CO}_2 + 2 \text{H}_2\text{O} \hspace{0.5cm} \text{∆}H = -1411 \hspace{0.1cm} \text{kJ} \].

- Compute the molar mass of \(\text{C}_2\text{H}_4\)

Next, find the molar mass of \(\text{C}_2\text{H}_4\). Carbon (C) has an atomic mass of 12.01 g/mol and hydrogen (H) has an atomic mass of 1.01 g/mol: \[ \text{Molar mass of C}_2\text{H}_4 = (2 \times 12.01) + (4 \times 1.01) = 24.02 + 4.04 = 28.06 \hspace{0.1cm} \text{g/mol} \].

- Set up the proportion for the heat transfer

We need to determine how many grams of \(\text{C}_2\text{H}_4\) must burn to give off 70.0 kJ of heat. We know that \(-1411\) kJ of heat is released for every 28.06 grams of ethylene:\[ \frac{-1411 \hspace{0.1cm} \text{kJ}}{28.06 \, \text{g}} = \frac{70.0 \hspace{0.1cm} \text{kJ}}{x \, \text{g}} \].

- Solve for x

To find the mass x, use cross-multiplication in the proportion:\[ x = \frac{70.0 \hspace{0.1cm} \text{kJ} \times 28.06 \hspace{0.1cm} \text{g}}{-1411 \hspace{0.1cm} \text{kJ}} \].Simplify the expression:\[ x = \frac{70.0 \times 28.06}{1411} \, \text{g} = 1.39 \, \text{g} \].

Key concepts

thermochemical equations

Thermochemical equations are a way to express the energy changes during chemical reactions.
These equations combine the chemical equation with the heat of the reaction. By including energy exchanged during the reaction, these equations help us understand the energy efficiency and requirements.
In this exercise, ethylene \(\text{C}_2\text{H}_4\) undergoes combustion, meaning it reacts with oxygen \(\text{O}_2\) to produce carbon dioxide \(\text{CO}_2\) and water \(\text{H}_2\text{O}\). The thermochemical equation here is: \(\text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \hspace{0.5cm} \text{∆H} = -1411 \hspace{0.1cm} \text{kJ}\).
This equation shows not just the reactants and products, but also the heat released during the reaction.

heat of combustion

The heat of combustion is the amount of heat released when a substance undergoes complete combustion.
For ethylene \(\text{C}_2\text{H}_4\), the heat of combustion is \(-1411 \hspace{0.1cm} \text{kJ/mol}\). This means that for every mole of ethylene that combusts, 1411 kilojoules of heat is released.
It's important to note the negative sign, which indicates that the reaction is exothermic (it releases energy). Understanding the heat of combustion helps in applications like fuel efficiency and energy production.

molar mass calculation

Molar mass is a measure of the mass of one mole of a substance. To find the molar mass of ethylene \(\text{C}_2\text{H}_4\), you need to add the atomic masses of all the atoms in the molecule.

The atomic mass of carbon (C) is 12.01 g/mol, and there are 2 carbons: \(\text{2} \times \text{12.01} = 24.02 \text{g/mol}\).
The atomic mass of hydrogen (H) is 1.01 g/mol, and there are 4 hydrogens: \(\text{4} \times \text{1.01} = 4.04 \text{g/mol}\).
Add these together to get the molar mass of \(\text{C}_2\text{H}_4\): \(\text{24.02 + 4.04 = 28.06 \text{g/mol}}\).

stoichiometry

Stoichiometry involves calculating the quantities of reactants and products in chemical reactions. In this exercise, we need to find out how much ethylene is required to produce 70.0 kJ of heat.
We know \(-1411 \hspace{0.1cm} \text{kJ}\) is released for every 28.06 grams of ethylene combusted.

Setting up a ratio, we get:
\(\frac{-1411 \hspace{0.1cm} \text{kJ}}{28.06 \text{g}} = \frac{70.0 \hspace{0.1cm} \text{kJ}}{x \text{g}}\),
where \(x\) is the mass of ethylene needed to produce 70.0 kJ of heat.
Cross-multiplying and solving for \(\text{x}\):
\(\text{x} = \frac{\text{70.0} \times \text{28.06}}{\text{-1411}} = 1.39 \text{g} \).
So, 1.39 grams of ethylene must burn to produce 70.0 kJ of heat.

balanced chemical equations

A balanced chemical equation has an equal number of each type of atom on both sides of the reaction. This ensures mass and atom conservation.
For the combustion of ethylene, we start with the unbalanced equation: \(\text{C}_2\text{H}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\).

Balancing this, we get: \(\text{C}_2\text{H}_4 + \text{3O}_2 \rightarrow \text{2CO}_2 + \text{2H}_2\text{O}\).
This equation indicates that 1 molecule of ethylene reacts with 3 molecules of oxygen to produce 2 molecules of carbon dioxide and 2 molecules of water.
Balancing equations is crucial in stoichiometry for accurate calculations and understanding of chemical reactions.