Problem 73 - Chemistry

Question

Consider the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \quad \Delta H=-92.6 \mathrm{~kJ} / \mathrm{mol}\) When \(2 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) react with \(6 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) to form \(4 \mathrm{~mol}\) of \(\mathrm{NH}_{3}\) at 1 atm and a certain temperature, there is a decrease in volume equal to \(98 \mathrm{~L}\). Calculate \(\Delta U\) for this reaction. (The conversion factor is \(1 \mathrm{~L} \cdot \mathrm{atm}=101.3 \mathrm{~J} .\)

Step-by-Step Solution

Answer

\(\Delta U = -175.2626\, \mathrm{kJ}\) for the reaction.

Step 1: Understand the Reaction and Given Data

The reaction is \( \mathrm{N}_2(g) + 3\mathrm{H}_2(g) \rightarrow 2\mathrm{NH}_3(g) \) with \( \Delta H = -92.6\, \mathrm{kJ/mol} \). \( 2\, \mathrm{mol} \) of \( \mathrm{N}_2 \) reacts with \( 6\, \mathrm{mol} \) of \( \mathrm{H}_2 \) to form \( 4\, \mathrm{mol} \) of \( \mathrm{NH}_3 \). The pressure is \( 1\, \mathrm{atm} \), and there is a decrease in volume of \( 98\, \mathrm{L} \). We need to calculate \( \Delta U \) which is the change in internal energy.

Step 2: Use the Helmoltz Free Energy Equation

The relationship between \( \Delta U \) (change in internal energy) and \( \Delta H \) (change in enthalpy) is given by the equation:\[ \Delta U = \Delta H - P\Delta V \]where \( P \) is the pressure, and \( \Delta V \) is the change in volume. Given \( \Delta H = -92.6\, \mathrm{kJ/mol} \), \( P = 1\, \mathrm{atm} \), and \( \Delta V = -98\, \mathrm{L} \).

Step 3: Calculate the Product of PΔV

To convert the volume change from liters to joules, we use the conversion factor:\[ P\Delta V = \Delta V \times 1\, \mathrm{atm} = -98\, \mathrm{L} \times 101.3\, \mathrm{J/L} = -9937.4\, \mathrm{J} \]Convert kilojoules to joules:\[ \Delta H = -92.6 \times 2 = -185.2\, \mathrm{kJ} = -185200\, \mathrm{J} \]

Step 4: Calculate ΔU Using the Values Found

Substitute the values for \( \Delta H \) and \( P\Delta V \) into the equation:\[ \Delta U = -185200\, \mathrm{J} - (-9937.4\, \mathrm{J}) = -185200\, \mathrm{J} + 9937.4\, \mathrm{J} \]\( \Delta U = -175262.6\, \mathrm{J} \) (or equivalently \(-175.2626\, \mathrm{kJ}\) by converting back to kilojoules).

Step 5: Write Down the Final Answer

The calculated change in internal energy for the reaction, \( \Delta U \), is \(-175.2626\, \mathrm{kJ}\).

Key Concepts

Enthalpy Change

In thermodynamics, the enthalpy change (\( \Delta H \)) is a critical concept representing the heat absorbed or released during a reaction at constant pressure. When considering the reaction between nitrogen (\( \mathrm{N}_{2} \)) and hydrogen (\( \mathrm{H}_{2} \)) to produce ammonia (\( \mathrm{NH}_{3} \)), the given enthalpy change is \(-92.6 \mathrm{kJ} / \mathrm{mol}\). This indicates that the reaction is exothermic, meaning it releases heat. The enthalpy change helps us understand the energy dynamics of a reaction:

• Reactions with negative \( \Delta H \) values release energy.
• Reactions with positive \( \Delta H \) values absorb energy.

For our exercise, calculating the total change for the entire reaction involves multiplying the per mole enthalpy change by the number of moles of nitrogen reacting, leading to \(-185.2 \mathrm{kJ}\) for 2 moles.

Volume Work

Volume work is an essential concept in understanding reactions involving gases. It refers to the work done by or against the system when the volume changes under constant external pressure. The formula used to calculate work done by the system is \(-P\Delta V\), where \( P \) is the pressure, and \( \Delta V \) is the change in volume. In the context of the nitrogen-hydrogen reaction:

• The change in volume, \( \Delta V \), was given as \(-98 \mathrm{L}\).
• Using the conversion factor \(1 \mathrm{L} \cdot \mathrm{atm} = 101.3 \mathrm{J}\), the energy change due to volume work can be calculated as \(-9937.4 \mathrm{J}\)

This value is essential when calculating the total change in internal energy, as it represents the energy required to compress the gaseous products.

Helmholtz Free Energy

Helmholtz free energy (\( F \)) is a wider thermodynamic quantity that relates to the energy available to do work at constant temperature and volume. Unlike enthalpy, which deals with constant pressure, Helmholtz free energy is more focused on constant volume systems. However, our main focus here is on the change in internal energy derived using a similar-covered concept: \[ \Delta U = \Delta H - P\Delta V \]This equation depicts the relationship between internal energy (\( \Delta U \)), enthalpy change (\( \Delta H \)), and the work done due to volume change. In our exercise, this formula helps bridge the gap between enthalpy change and the actual internal energy change considering work done on or by the gas.

Nitrogen-Hydrogen Reaction

The nitrogen-hydrogen reaction is an essential chemical process that produces ammonia, a significant compound for fertilizers. The reaction \( \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g) \) is both a synthesis and an economically valuable interaction. Let's break down some key aspects:

• The balanced equation illustrates the stoichiometry of the reaction, with 1 mole of nitrogen reacting with 3 moles of hydrogen to form 2 moles of ammonia.
• The reaction decreases volume, typical for reactions producing fewer moles of gas from more moles of reactants.
• The process is exothermic, as seen by the negative enthalpy change, making it energetically favorable under specific conditions.

This reaction's understanding is crucial as it encapsulates important concepts like stoichiometry, energy flow, and the economics of chemical synthesis.