Question

Consider the combustion of liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}(l):\) $$ \begin{aligned} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \\ \Delta H=&-726.5 \mathrm{~kJ} \end{aligned} $$ (a) What is the enthalpy change for the reverse reaction? (b) Balance the forward reaction with whole-number coefficients. What is \(\Delta H\) for the reaction represented by this equation? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If the reaction were written to produce \(\mathrm{H}_{2} \mathrm{O}(g)\) instead of \(\mathrm{H}_{2} \mathrm{O}(l)\) would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

Short answer

(a) The enthalpy change for the reverse reaction is \(+726.5 \text{ kJ}\). (b) The balanced reaction and its enthalpy change are the same: \(\Delta H = -726.5 \text{ kJ}\). (c) The forward reaction is thermodynamically favored, as it has a negative enthalpy change. (d) If the product were \(H_2O(g)\) instead of \(H_2O(l)\), the magnitude of \(\Delta H\) would decrease (becoming less negative) because of the additional energy required for the vaporization of water.

Step-by-step solution

(a) Enthalpy Change for Reverse Reaction

To find the enthalpy change for the reverse reaction, we simply change the sign of the given enthalpy change. The enthalpy change of the reverse reaction is the opposite of the forward reaction. In this case, the enthalpy change for the reverse reaction is: \(\Delta H_{\text{reverse}} = +726.5 \text{ kJ}\)

(b) Balancing the Reaction and Finding the Enthalpy Change

The given reaction equation is already balanced with whole-number coefficients: $$ \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ Since the reaction is balanced and the enthalpy change is already provided, the enthalpy change for the balanced reaction is the same: \(\Delta H = -726.5 \text{ kJ}\)

(c) Thermodynamically Favored Reaction

A reaction is thermodynamically favored if it proceeds with a decrease in enthalpy (negative \(\Delta H\)) since it releases energy to the surroundings. In this case, the forward reaction has a negative enthalpy change, which means it is thermodynamically favored over the reverse reaction.

(d) Effect of the Product Being Water Vapor Instead of Liquid Water

If the product were water vapor (\(\mathrm{H}_2\mathrm{O}(g)\)) instead of liquid water (\(\mathrm{H}_2\mathrm{O}(l)\)), we need to take into account the change in the phase of water. The enthalpy change for vaporization of water is positive since energy is required to convert liquid water to water vapor. Therefore, the overall enthalpy change of the reaction would be greater (less negative) than the given value. So, the magnitude of \(\Delta H\) would decrease (becoming less negative).

Key concepts

Combustion Reactions

When we talk about combustion reactions, we're referring to a specific type of chemical reaction where a substance combines with oxygen, releasing energy in the form of heat and light. These reactions are exothermic, meaning they release energy rather than absorb it. That's why your car runs on gasoline and why fires provide warmth.
In our exercise involving methanol,

• The chemical equation illustrates the combustion of methanol with oxygen (\( \text{O}_2 \)).
• The products of this reaction are carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)).
• A notable feature of combustion reactions is that they are often accompanied by a flame, which is a visible indication of the energy being released.

Understanding combustion is crucial not only in chemistry but also in everyday life, as it applies to things like heating, cooking, and energy production.

Thermodynamics

Thermodynamics is the science of energy and how it flows and transforms. One key aspect of thermodynamics in chemical reactions is the concept of enthalpy (\( \Delta H \)), which tells us about the heat change at constant pressure.
For the methanol combustion reaction given, the \( \Delta H = -726.5 \text{ kJ} \) indicates it's an exothermic process — a perfect example of energy being released to the surroundings.

• A negative \( \Delta H \) value suggests that the reaction naturally proceeds towards the forward direction, releasing energy, hence is thermodynamically favored.
• Remember, the change of energy flow is the central point in deciding whether a process will happen on its own or require assistance.

Thermodynamics helps us understand these energy changes and predict the direction and enthusiasm of chemical reactions. It's like having a magical map that guides how energy wants to naturally wander.

Phase Change

Phase change refers to the transformation of a substance from one state of matter to another, such as from liquid (\( \text{l} \)) to gas (\( \text{g} \)). This change plays a significant role in the exercise at hand.
When water changes from liquid to vapor in the methanol combustion problem:

• Energy, known as enthalpy of vaporization, is required to overcome attractions between water molecules — making the process endothermic.
• In the context of the problem, switching water from liquid to vapor will alter the \( \Delta H \) value, rendering it less negative.
• This means less heat is released overall, as some energy is used in this phase transition.

This simple change from liquid to vapor showcases the importance of considering physical states in reactions, as they significantly impact the energy landscape of chemical processes.