Problem 55 - General Chemistry

Question

Phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4}\), can be prepared by the reactio of phosphorus(V) oxide, \(\mathrm{P}_{4} \mathrm{O}_{10}\), with water. $$ \frac{1}{4} \mathrm{P}_{4} \mathrm{O}_{10}(s)+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q) ; \Delta H=-96.2 \mathrm{~kJ} $$ What is \(\Delta H\) for the reaction involving \(1 \mathrm{~mol}\) of \(\mathrm{P}_{4} \mathrm{O}_{10}\) ? $$ \mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(a q) $$

Step-by-Step Solution

Answer

\(\Delta H\) for 1 mole of \(\mathrm{P}_4\mathrm{O}_{10}\) is \(-384.8\) kJ.

Step 1: Understand the Given Reaction

The reaction provided in the problem is \( \frac{1}{4} \mathrm{P}_{4} \mathrm{O}_{10}(s)+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q) \) with \( \Delta H = -96.2 \text{ kJ} \). This means that for every 1/4 mole of \( \mathrm{P}_{4} \mathrm{O}_{10} \), 96.2 kJ of energy is released.

Step 2: Calculate Energy for 1 Mole of \(\mathrm{P}_{4} \mathrm{O}_{10}\)

Since the given enthalpy change is for 1/4 mole of \( \mathrm{P}_{4} \mathrm{O}_{10} \), we need to scale it up to find the energy change for 1 mole. We do this by multiplying the enthalpy change by 4 (since there are 4 quarters in a whole). Thus, \[ \Delta H = -96.2 \text{ kJ} \times 4 = -384.8 \text{ kJ} \].

Step 3: Conclusion

The enthalpy change \( \Delta H \) for the reaction involving 1 mole of \( \mathrm{P}_{4} \mathrm{O}_{10} \) is \( -384.8 \text{ kJ} \). This means that when 1 mole of \( \mathrm{P}_{4} \mathrm{O}_{10} \) reacts, it releases 384.8 kJ of energy.

Key Concepts

Enthalpy Change

Enthalpy change, often denoted as \( \Delta H \), is a measure of the total heat content or energy change during a chemical reaction. This concept is important in thermodynamics as it helps us understand whether a reaction is exothermic (releases heat) or endothermic (absorbs heat).
As described in the exercise, the given reaction is exothermic because it has a negative enthalpy change \( \Delta H = -96.2 \text{ kJ} \). This means that energy is being released. For chemical reactions, the enthalpy change can tell us a lot about the energy dynamics involved.
To find out the energy change for a full mole of a reactant, you simply scale the initial enthalpy result according to the reactant's molar requirement in the balanced equation. In this exercise, multiplying the enthalpy change by 4 (since 1/4 mole needs to be scaled to 1 mole) gives us \( -384.8 \text{ kJ} \). This implies that each mole of \( \mathrm{P}_{4} \mathrm{O}_{10} \) releases \( 384.8 \text{ kJ} \) when it reacts with water to form phosphoric acid.
Understanding enthalpy changes is crucial for predicting reaction behavior and energy management in chemical processes.

Chemical Reaction Balancing

Balancing chemical equations is a fundamental skill in chemistry that ensures mass conservation during a chemical reaction. In the given exercise, the initial equation is partially balanced: \( \frac{1}{4} \mathrm{P}_{4} \mathrm{O}_{10}(s) + \frac{3}{2} \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(aq) \). Here, the coefficients ensure that the number of each type of atom is the same on both sides of the reaction.
A balanced equation not only respects the law of conservation of mass but also allows accurate calculations of reactants and products in quantitative analysis. Balancing involves adjusting these coefficients to ensure equality on both sides. This becomes critical in enthalpy change calculations since \( \Delta H \) values depend on the mole amounts determined by the balanced equation.
In this exercise, balancing the entire equation for one full mole: \( \mathrm{P}_{4} \mathrm{O}_{10}(s) + 6 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(aq) \), helps us see the reaction ratio needed for calculating energy changes, highlighting why balancing chemical reactions is key for accurate prediction and computation of enthalpy.

Thermodynamics

Thermodynamics deals with the principles of heat energy transfer and its transformation into chemical energy. It helps us describe how energy moves during a chemical reaction, such as the one in the exercise involving phosphorus pentoxide and water.
Hess's Law is a principle in thermodynamics that states the total enthalpy change for a reaction is the same, regardless of the reaction pathway. This means whether you perform the reaction in one step or multiple intermediate steps, the overall energy change is consistent. This principle underlies the calculation of enthalpy changes by breaking reactions into simpler steps or scaling existing reaction data, as seen in this exercise.
Thermodynamics not only helps predict the energy outcomes of reactions like these but also aids in understanding energy efficiencies and product yields. This understanding can drive advancements in chemical manufacturing, energy production, and environmental management. Therefore, learning how to calculate and interpret the thermodynamic data of chemical reactions can lead to more sustainable practices in chemistry.