Question

The enthalpy changes of the following reactions can be measured: \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) $$ \Delta_{i} H^{\circ}=-1411.1 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} $$ \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell)+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\ell)\) \(\Delta, H^{\circ}=-1367.5 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}\) (a) Use these values and Hess's law to determine the enthalpy change for the reaction \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell)\) (b) Draw an energy level diagram that shows the relationship between the energy quantities involved in this problem.

Short answer

The enthalpy change for the target reaction is \(-43.6\, \text{kJ/mol-rxn}\).

Step-by-step solution

Identify the Given Reactions and Data

We have two reactions with their enthalpy changes given. The first reaction is the combustion of ethylene (\(\mathrm{C}_{2} \mathrm{H}_{4}\)), with \(\Delta H^{\circ} = -1411.1\, \text{kJ/mol-rxn}\). The second reaction is the combustion of ethanol (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)), with \(\Delta H^{\circ} = -1367.5\, \text{kJ/mol-rxn}\).

Write the Target Reaction

The target reaction for which we need to find the enthalpy change is: \(\mathrm{C}_{2} \mathrm{H}_{4} (\mathrm{g}) + \mathrm{H}_{2} \mathrm{O} (\ell) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} (\ell)\).

Apply Hess's Law

To apply Hess's law, we align the given reactions with the target reaction. We need to reverse the second given reaction to bring \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} (\ell)\) to the reactants side. Therefore, reversing the second reaction involves changing the sign of its enthalpy change yielding: \(\Delta H^{\circ} = +1367.5\, \text{kJ/mol-rxn}\).

Combine Adjusted Reactions

Now, add the first reaction to the reversed second reaction: \(\mathrm{C}_{2} \mathrm{H}_{4} (\mathrm{g}) + 3 \mathrm{O}_2 (\mathrm{g}) \rightarrow 2 \mathrm{CO}_2 (\mathrm{g}) + 2 \mathrm{H}_2 \mathrm{O} (\ell)\) (\(\Delta H^{\circ} = -1411.1\, \text{kJ}\))Reversed second reaction: \(2 \mathrm{CO}_2 (\mathrm{g}) + 3 \mathrm{H}_{2} \mathrm{O} (\ell) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} (\ell) + 3 \mathrm{O}_2 (\mathrm{g})\) (\(\Delta H^{\circ} = +1367.5\, \text{kJ}\))When combined, the intermediate species \(\mathrm{CO}_2\) and \(\mathrm{O}_2\) cancel out, leading to the target reaction: \(\mathrm{C}_{2} \mathrm{H}_{4} (\mathrm{g}) + \mathrm{H}_{2} \mathrm{O} (\ell) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} (\ell)\).

Calculate the Enthalpy Change for the Target Reaction

The total enthalpy change \(\Delta H_{\text{target}}^{\circ}\) for the reaction is the sum of the enthalpies: \ \(\Delta H_{\text{target}}^{\circ} = -1411.1\, \text{kJ} + 1367.5\, \text{kJ} = -43.6\, \text{kJ/mol-rxn}\).

Draw the Energy Level Diagram

Create an energy level diagram starting with \(\mathrm{C}_{2} \mathrm{H}_{4} (\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O} (\ell)\) at a higher energy level on the left. The diagram should show a decrease to \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} (\ell)\) at a lower energy level on the right, with the difference equal to \(-43.6\, \text{kJ}\), demonstrating the exothermic nature of the reaction.

Key concepts

Enthalpy Change

Enthalpy change refers to the heat energy absorbed or released in a chemical reaction at constant pressure. When we talk about reactions, they can either release energy (exothermic) or absorb energy (endothermic). The enthalpy change is denoted as \( \Delta H \). For the reactions in our problem, the given enthalpy changes are negative, indicating they are exothermic.
This means that energy is released into the surroundings, as seen in the combustion of ethylene and ethanol. In calculating the enthalpy change for any target reaction using Hess's Law, we manipulatereverse and sum the given reactions.
Hess's Law states that the total enthalpy change in a chemical reaction is the same, regardless of the route by which the reaction is achieved. Thus, even if intermediate steps involve different reactions, the total enthalpy will only depend on the initial and final states. In practice:

• Reverse the direction of necessary reactions
• Sum up the adjusted enthalpies of the sequence

In our case, the calculated enthalpy change for the target reaction is \(-43.6 \mathrm{kJ/mol-rxn}\), which shows that the reaction releases energy.

Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance combines with oxygen to release energy in the form of heat and light. This process is commonly what happens in burning fuels.
In the problem at hand, both ethylene \( \mathrm{C}_{2} \mathrm{H}_{4} \) and ethanol \( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \) are undergoing combustion with oxygen \( \mathrm{O}_2 \).
When hydrocarbons combust, they typically produce carbon dioxide \( \mathrm{CO}_2 \) and water \( \mathrm{H}_2\mathrm{O} \) as products:

• Ethylene combustion: \( \mathrm{C}_{2} \mathrm{H}_{4} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2} + 2 \mathrm{H}_{2} \mathrm{O} \)
• Ethanol combustion: \( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} + 3 \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2} + 3 \mathrm{H}_{2} \mathrm{O} \)

Through the combustion process, bonds are broken and formed, resulting in the transfer of energy. This is why the enthalpy changes are negative since energy is released overall.

Energy Level Diagram

An energy level diagram is a visual representation that shows the energy changes during a chemical reaction. It helps illustrate whether a reaction is endothermic or exothermic.
In the diagram related to our exercise, we begin with the reactants \( \mathrm{C}_{2} \mathrm{H}_{4} \) and \( \mathrm{H}_{2} \mathrm{O} \) on the left at a higher energy level.
The products, \( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \), are at a lower energy level on the right.
This drop in energy level represents the release of energy, specifically \(-43.6 \mathrm{kJ/mol-rxn}\). The diagram not only shows the enthalpy change, but it also indicates the direction of the energy transfer, with products being more stable than reactants due to the release of energy.
Some useful points to consider when interpreting or drawing such diagrams:

• Upward arrows or lines show energy absorption (endothermic)
• Downward arrows or lines depict energy release (exothermic)
• The length of the arrow correlates to the magnitude of the energy change

In this particular problem, the negative enthalpy change and the downward arrow underline that the target reaction is exothermic.