Chapter 5: Problem 29
Consider the reaction $$\begin{aligned}2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow & 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \\ \Delta H=&+483.6 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$$ at a certain temperature. If the increase in volume is 32.7 \(\mathrm{L}\) against an external pressure of \(1.00 \mathrm{~atm},\) calculate \(\Delta U\) for this reaction. \((1 \mathrm{~L} \cdot \mathrm{atm}=101.3 \mathrm{~J})\)
Short Answer
Expert verified
\( \Delta U \) is 480.29 kJ.
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Enthalpy Change
In thermodynamics, **enthalpy change** (\( \Delta H \) ) represents the heat absorbed or released during a reaction at constant pressure. For the given exercise, the reaction involves the conversion of two moles of water vapor into hydrogen and oxygen gases. The enthalpy change for this process is \( +483.6 \, \text{kJ/mol} \). This positive value indicates that the reaction is endothermic, meaning it absorbs heat from the surroundings. It is crucial to convert enthalpy values consistently into comparable units, such as converting \( \text{kJ} \text{ to } \text{J} \) , to facilitate accurate calculations like determining internal energy changes. Understanding enthalpy aids in predicting whether a reaction requires or releases energy, which is a fundamental aspect of chemical thermodynamics.
Internal Energy Change
**Internal energy change** (\( \Delta U \)) is a key concept in understanding the total energy variation within a system as it undergoes a chemical reaction. For this exercise, the internal energy change can be calculated using the formula:\[ \Delta U = \Delta H - W \]where \( \Delta H \) represents the enthalpy change, and \( W \) is the work done by the system. In our example, after converting \( \Delta H \) to joules, we have \( 483600 \, \text{J} \). The work calculated was \( 3313.71 \, \text{J} \). Plugging these into the formula gives us:\[ \Delta U = 483600 \, \text{J} - 3313.71 \, \text{J} = 480286.29 \, \text{J} \]Internal energy change helps in understanding how energy balance shifts between different forms during a reaction, conveying changes not only involving heat but also other types of work like mechanical.
Work-Energy Principle
The **work-energy principle** is pivotal in analyzing how energy allows work to be done in a system. In the context of chemical reactions, when the volume of gas changes,work is done by the system against an external pressure. This exercise involves an increase in volume,equal to \( 32.7 \, \text{L} \) under \( 1.00 \, \text{atm} \)of pressure. Using the formula for work done by a gaseous system, \( W = P \Delta V \), we substitute our values to find:\( W = 32.7 \, \text{L} \cdot \text{atm} \).Converting this into Joules gives:\( W = 3313.71 \, \text{J} \).Understanding this principle is fundamental because it illustrates how mechanical work affects energy change computations in chemical processes, revealing the relationship between volume change and energy transfer.
Conversion of Units
The **conversion of units** is a crucial step in accurately solving thermodynamic problems. This exercise exemplifies the necessity of converting units to maintain consistency throughout the calculations.The reaction's enthalpy change is initially given in \( \text{kJ} \) per mole,but calculations require \( \Delta H \) to be in \( \text{J} \) for uniformity with the work component, expressed in Joules. We convert \( 483.6 \, \text{kJ/mol} \) to \( 483600 \, \text{J} \).Similarly, the work done, calculated in \( \text{L} \cdot \text{atm} \), is converted using \( 1 \, \text{L} \cdot \text{atm} = 101.3 \, \text{J} \) to determine the \( 3313.71 \, \text{J} \) value. This ensures that the final calculation of \( \Delta U \)in Joules and subsequently in \( \text{kJ} \) is precise, underscoring the importance of using consistent units to avoid errors in chemical calculations.
