TERMOQUIMICA Teoría 8 - Relación transferencia calor volumen constante y presión constante

Thermodynamics in Chemical Reactions

Application of the First Law of Thermodynamics

• The first law of thermodynamics is applied to chemical reactions at constant pressure (qₚ) and constant volume (qᵥ).

• At constant volume, qᵥ equals the change in internal energy of the system.

• At constant pressure, qₚ equals the change in enthalpy, which is a crucial state function related to internal energy.

Understanding Enthalpy and Internal Energy

• Enthalpy (H) can be expressed as H = U + PV, where U is internal energy and PV represents work done by the system.

• The relationship between qₚ and qᵥ can be summarized: ΔH = ΔU + PΔV.

General Relationship Between Heat Transfers

• By substituting qₚ for ΔH and qᵥ for ΔU, we derive a general expression relating both heat transfers: ΔH = qₚ = qᵥ + PΔV.

Specific Cases: Solid and Liquid Reactions

• For reactions involving only solids or liquids, the volume change between reactants and products is negligible.

• If ΔV is close to zero, then PΔV also approaches zero; thus, it follows that qₚ = qᵥ.

Implications for Gaseous Reactions

• In reactions involving gases, there may be significant volume changes depending on the total number of moles before and after the reaction.

• The variation in moles affects volume according to ideal gas laws; more moles lead to greater volumes.

Ideal Gas Law Application

• Using the ideal gas equation (PV=nRT), we can express changes in volume due solely to changes in mole numbers.

• This allows us to substitute terms into our earlier expressions linking heat transfer with mole variations.

Gas Reactions and Moles Calculation

General Expression for Gas Reactions

• The general expression for gas reactions is given by p = q_V + Delta n cdot R cdot T . This accounts for variations in the number of moles, which can be positive, zero, or negative depending on the reaction.

Example: Combustion of Methane

• The combustion of methane ( CH_4 ) involves two moles of oxygen reacting to produce one mole of carbon dioxide and two moles of water. This reaction consists entirely of gaseous components.

• To determine the total moles in reactants and products, sum the stoichiometric coefficients:

• Products: 1 mole CO_2 + 2 moles H_2O = 3 moles.

• Reactants: 1 mole CH_4 + 2 moles O_2 = 3 moles.

Calculating Change in Moles

• The change in the number of moles ( Delta n ) is calculated as:

• Delta n = (moles;of;products) - (moles;of;reactants) = 3 - 3 = 0 .

• Since Delta n = 0, it implies that heat transferred at constant pressure equals heat transferred at constant volume for this specific reaction.

Second Example: Reaction Involving Nitrogen Monoxide

• Consider the reaction between nitrogen monoxide and oxygen to form nitrogen dioxide:

• Adjusted equation: 2 NO + O_2 → 2 NO_2 . All components are gaseous.

• For this reaction:

• Products have a total of 2 moles from nitrogen dioxide.

• Reactants consist of 2 moles from nitrogen monoxide plus one mole from oxygen, totaling to three. Thus, we find:

• Δn = (moles;of;products) - (moles;of;reactants) = 2 - 3 = -1. This indicates a decrease in volume since there are fewer product moles than reactant moles.

Implications on Work Done