Standard Enthalpy and Enthalpy of Formation

Definition and Concepts

The heat of reaction or enthalpy of reaction (ΔHrxn) is the heat absorbed or released during a chemical reaction. The standard enthalpy of formation (ΔHof) is the enthalpy change when one mole of a substance is formed from its elements in their standard states.

• Standard state: The most stable form of a substance at 1 atm pressure and 25°C.

• Formation reaction: The process of forming 1 mole of a compound from its elements in their standard states.

• ΔHof for elements in their standard state is zero.

Calculating Enthalpy Changes

The standard enthalpy change of a reaction can be calculated using the enthalpies of formation:

• Formula:

• Enthalpy is a state function, meaning its value depends only on the initial and final states, not the path taken.

Example Calculation

For the reaction: NH3(g) + 7/4 O2(g) → NO2(g) + 3/2 H2O(g), using standard enthalpies of formation:

• ΔHof[NH3] = -46 kJ/mol

• ΔHof[NO2] = +34 kJ/mol

• ΔHof[H2O] = -242 kJ/mol

• ΔHof[O2] = 0 kJ/mol

Calculation:

• kJ

Hess's Law

Principle and Explanation

Hess's Law states that the overall enthalpy change for a chemical process is independent of the path taken. This means that the enthalpy change for a reaction is the same whether it occurs in one step or multiple steps.

• The enthalpy change for a reaction can be determined by summing the enthalpy changes of individual steps.

• Enthalpy is a state function, so only the initial and final states matter.

Operational Details

• If you reverse a reaction, the sign of ΔH changes.

• If you multiply the coefficients in a reaction by a constant, multiply ΔH by the same constant.

• When combining reactions, work backwards from the target reaction.

Example: Combining Reactions

Consider the reaction: N2(g) + 2O2(g) → 2NO2(g), ΔH = 68 kJ. This can be carried out in two steps:

• N2(g) + O2(g) → 2NO(g), ΔH1 = 180 kJ

• 2NO(g) + O2(g) → 2NO2(g), ΔH2 = -112 kJ

• Adding the two steps: ΔH = ΔH1 + ΔH2 = 68 kJ

Generalized Hess's Law Formula

• For a process from A to B via intermediates X and Y:

Worked Examples Using Hess's Law

Example: Methane Combustion

• CH4(g) + 2O2(g) → CO2(g) + 2H2O(g), ΔH = -802 kJ

• 2H2O(g) → 2H2O(l), ΔH = -88 kJ

• Combined: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), ΔH = -890 kJ

Example: Formation of Acetaldehyde

• CH3CO(g) + 2O2(g) → 2CO2(g) + H2O(g), ΔH = -981.1 kJ/mol

• CH4(g) + 2O2(g) → CO2(g) + 2H2O(g), ΔH = -802.3 kJ/mol

• Target: 2CH4(g) + 2O2(g) → CH3CO(g) + 3H2O(g)

Example: Formation of HCl

• H2(g) + Cl2(g) → 2HCl(g), ΔH = -185 kJ

• Given reactions and their enthalpy changes, combine and manipulate equations to obtain the target reaction.

Standard Enthalpies of Combustion

Tabulated Values

Standard enthalpies of combustion are important for understanding the energy content of fuels.

Summary of Key Points

• Enthalpy is a state function and can be calculated using standard enthalpies of formation.

• Hess's Law allows calculation of enthalpy changes for complex reactions by combining known reactions.

• Manipulation of equations (reversing, multiplying) affects the sign and magnitude of ΔH.

• Standard enthalpies of combustion are used to compare fuel values.

Additional info:

• Some images and diagrams were omitted as they were not directly relevant to the explanation of the paragraph.

• All equations are provided in LaTeX format for clarity and academic rigor.