Question

Consider the reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) carried out at constant temperature and pressure. If \(\Delta \mathrm{H}\) and \(\Delta \mathrm{U}\) are the enthalpy and internal energy changes for the reaction, which of the following expressions is true? (a) \(\Delta \mathrm{H}=0\) (b) \(\Delta \mathrm{H}=\Delta \mathrm{U}\) (c) \(\Delta \mathrm{H}<\Delta \mathrm{U}\) (d) \(\Delta \mathrm{H}>\Delta \mathrm{U}\)

Short answer

Option (c) \( \Delta \mathrm{H} < \Delta \mathrm{U} \) is true.

Step-by-step solution

Understand the Relationship Between ΔH and ΔU

The relationship between enthalpy change \( \Delta \mathrm{H} \) and internal energy change \( \Delta \mathrm{U} \) can be expressed as:\[ \Delta \mathrm{H} = \Delta \mathrm{U} + \Delta n \cdot R \cdot T \]where \( \Delta n \) is the change in the number of moles of gas, \( R \) is the ideal gas constant, and \( T \) is the temperature. This relationship helps us account for the work done due to volume change at constant temperature and pressure.

Determine Δn for the Reaction

For the given reaction:\[ \mathrm{N}_2 + 3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3 \]Calculate the change in moles of gas \( \Delta n \):- Initial moles of gas = 1 (from \( \mathrm{N}_2 \)) + 3 (from \( \mathrm{H}_2 \)) = 4 moles.- Final moles of gas = 2 (from \( \mathrm{NH}_3 \)).The change in moles of gas is \( \Delta n = 2 - 4 = -2 \).

Compare ΔH and ΔU

Substitute \( \Delta n \) into the relation:\[ \Delta \mathrm{H} = \Delta \mathrm{U} + (-2) \cdot R \cdot T \]This shows that:\( \Delta \mathrm{H} = \Delta \mathrm{U} - 2 \cdot R \cdot T \).Since \( 2 \cdot R \cdot T \) is a positive quantity, \( \Delta \mathrm{H} \) will be less than \( \Delta \mathrm{U} \).

Key concepts

Reaction Thermodynamics

In reaction thermodynamics, understanding the difference between enthalpy (ath\Delta H) and internal energy (ath\Delta U) is crucial. These values tell us about the energy changes during a chemical reaction. The enthalpy change (ath\Delta H) represents the total heat content change in a system at constant pressure. Internal energy change (ath\Delta U), on the other hand, pertains to the sum of potential and kinetic energy changes within the system.To differentiate them, consider that enthalpy also includes the effect of volume work. At constant pressure, if there is a volume change as gas molecules react, this contributes to the enthalpy value. This is why we use the equation: \[\Delta H = \Delta U + \Delta n \cdot R \cdot T\] where:

• \(\Delta n\) is the change in the number of moles of gas
• \(R\) is the ideal gas constant
• \(T\) is the temperature in Kelvin

Hence, in reactions where gas volumes change, \(\Delta H\) usually differs from \(\Delta U\).

Ideal Gas Law

The Ideal Gas Law is a fundamental equation used to estimate the behavior of gases in ideal conditions. It relates the pressure, volume, and temperature of a gas with the amount of substance of the gas (in moles). The equation is expressed as:\[PV = nRT\]where:

• \(P\) is the pressure of the gas
• \(V\) is the volume of the gas
• \(n\) is the number of moles
• \(R\) is the ideal gas constant
• \(T\) is the temperature in Kelvin

For reactions involving gases, like in our original exercise, the Ideal Gas Law helps in understanding how the reaction conditions affect the gaseous reactants and products. It assists in determining \(\Delta n\), the change in moles of gases, which impacts the enthalpy calculation by indicating if there’s expansion or compression of gases in the reaction.

Chemical Equation Balancing

Balancing chemical equations is essential to represent a chemical reaction accurately. Every chemical reaction must obey the law of conservation of mass, meaning the same number of each type of atom must appear on both sides of the equation. To balance a reaction correctly, follow these steps:

• Write down the unbalanced equation.
• Count the number of atoms of each element on both sides.
• Add coefficients in front of compounds to balance one element at a time. It's best to start with the least common elements.
• Ensure the smallest possible whole number coefficients are used.

In our exercise, the reaction \(\mathrm{N}_2 + 3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3\) is already balanced. This balance confirms that two nitrogen atoms and six hydrogen atoms are on each side, reflecting accurate stoichiometry for analyzing thermodynamic properties.