Problem 110 - General Chemistry

Question

Ethylene glycol, \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\), is used as antifreeze. It is produced from ethylene oxide, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\), by the reaction $$ \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}(l) $$ Use Hess's law to obtain the enthalpy change for this reaction from the following enthalpy changes: $$ \begin{gathered} 2 \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H=-2612.2 \mathrm{~kJ} \\ \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}(l)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) ; \\ \Delta H=-1189.8 \mathrm{~kJ} \end{gathered} $$

Step-by-Step Solution

Answer

The enthalpy change is \(-116.3~\mathrm{kJ}\).

Step 1: Understand the Given Enthalpy Changes

We are given two reactions with their corresponding enthalpy changes. The first reaction involves combustion of ethylene oxide: \[ 2 \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}(g) + 5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{CO}_{2}(g) + 4 \mathrm{H}_{2} \mathrm{O}(l); \quad \Delta H=-2612.2~\mathrm{kJ} \] The second reaction is the combustion of ethylene glycol: \[ \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}(l) + \frac{5}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + 3 \mathrm{H}_{2} \mathrm{O}(l); \quad \Delta H=-1189.8~\mathrm{kJ} \]

Step 2: Write Enthalpy Change for Ethylene Oxide Formation

Since the first reaction gives the combustion of 2 moles of ethylene oxide, we must alter it to reflect 1 mole, allowing us to extract its enthalpy of formation. Divide the entire reaction and enthalpy change by 2: \[ \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}(g) + \frac{5}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(l); \quad \Delta H=-1306.1~\mathrm{kJ} \]

Step 3: Reverse Ethylene Glycol Equation

To use Hess's Law, we need to reverse the given combustion reaction of ethylene glycol, since we're interested in forming ethylene glycol from carbon dioxide and water. Change the sign of the enthalpy to reflect this reversal: \[ 2 \mathrm{CO}_{2}(g) + 3 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}(l) + \frac{5}{2} \mathrm{O}_{2}(g); \quad \Delta H=1189.8~\mathrm{kJ} \]

Step 4: Apply Hess's Law to Find Overall Enthalpy Change

Using Hess's Law, we can find the enthalpy change of forming ethylene glycol from ethylene oxide and water by summing up the rearranged reactions. The desired reaction is: \[ \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}(g) + \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}(l) \] For achieving this using intermediate steps from previous equations:1. Formation of ethylene oxide from its components: \(-1306.1~\mathrm{kJ}\) 2. Formation of ethylene glycol from its components: \(+1189.8~\mathrm{kJ}\) Sum them to find the enthalpy change: \[ \Delta H = (-1306.1) + 1189.8 = -116.3~\mathrm{kJ} \]

Step 5: Conclusion

The enthalpy change for the reaction of ethylene oxide with water to form ethylene glycol is \( \Delta H = -116.3~\mathrm{kJ} \).

Key Concepts

Enthalpy Change

Enthalpy change is a key concept in chemical thermodynamics that represents the heat absorbed or released during a chemical reaction at constant pressure. It is symbolized by \( \Delta H \), where a negative value indicates an exothermic reaction, and a positive value indicates an endothermic reaction. In Hess's law, the overall enthalpy change for a reaction is the sum of the enthalpy changes for a series of intermediate steps leading to the final reaction.
Imagine you have several smaller reactions that make up a larger one. If each small step has a known \( \Delta H \), you can add them all up to find the total enthalpy change of the overall reaction. This makes use of the state function nature of enthalpy, meaning it depends only on the initial and final states of the reaction, not the path or steps taken.
In our exercise, we found the \( \Delta H \) for the formation of ethylene glycol from ethylene oxide by considering the combustion reactions of these compounds. Through reversing and adjusting these reactions, the cumulative \( \Delta H \) was calculated as \(-116.3\, \text{kJ}\).

Combustion Reactions

Combustion reactions are chemical processes where a substance combines with oxygen, releasing energy in the form of light and heat. They are a type of redox reaction and are highly exothermic, contributing to our understanding of chemical energy and enthalpy changes.
When compounds like hydrocarbons or alcohols undergo combustion, they produce carbon dioxide and water as byproducts. The reactions involved are typically straightforward and predictable, making them ideal for studying enthalpy changes.
In the given exercise, both ethylene oxide and ethylene glycol undergo combustion. These reactions help us understand the enthalpy changes for forming or breaking compounds. By breaking down these combustion reactions to manageable steps, Hess's law is effectively applied to calculate the overall energy change in the formation of ethylene glycol from ethylene oxide.

Chemical Thermodynamics

Chemical thermodynamics is the field of chemistry that studies the energy changes accompanying chemical reactions. It is central to understanding how reactions proceed, how energy is stored, and how it transforms. Central concepts include enthalpy, entropy, and the Gibbs free energy.
The principles of chemical thermodynamics dictate whether a reaction is spontaneous, the conditions under which equilibrium is reached, and the size of the reaction's driving force. Understanding these thermodynamic properties allows chemists to predict how reactions behave under different conditions.
Hess's law falls under this umbrella. It showcases that the total enthalpy change in a complex reaction will be the same, regardless of the multiple intermediary steps involved. This principle supports the conservation of energy, affirming that energy cannot be created or destroyed. In the given exercise, by using Hess's law, we calculated the enthalpy change involved in the reaction, which is integral to predicting whether the ethylene oxide and water reaction results in the formation of ethylene glycol.