Thermodynamics

Thermochemistry

Carol Brown

Saint Mary’s Hall

The 0th Law

Heat flows from hot to cold

The 1st Law

In all changes, energy is conserved

The energy of the universe is constant

∆E=q + w(both q and w are from the system’s

point of view)

P∆V Work

Using the MKS system, determine the units for P∆V.

Calorimetry-Basics

Heat: The total amount of thermal energy contained in a sample of matter– Measured in calories or joules

Temperature: The average kinetic energy of the molecules in a sample of mater.– Measured in Kelvins or degrees Celsius

More basics

Specific heat capacity: The amount of heat it takes to raise one gram of a substance one degree Celsius. – Units: J/g oC

Heat capacity: The amount of heat it takes to raise a system one degree Celsius.– Units--J/oC

Coffee Cup Calorimeter

q = mc∆T

How much heat, in kJ, is necessary to take 20 g of ice at -5o C to steam at 100o C? The specific heat of ice is 2.1 J/g oC; of liquid water is 4.2 J/g oC. The latent heat of fusion is 334 J/g. The latent heat of vaporization is 2268 J/g.

60.7 kJ

Potential Energy Diagrams--Exothermic Reactions

Potential Energy Diagrams--Endothermic Reactions

Important Terms

Enthalpy (heat of reaction)--The amount of heat lost or absorbed during the course of a reaction when the only work done is expansion or contraction at a constant pressure (P∆V). Change in enthalpy is a state function and is symbolized by ∆H.

What drives a reaction?

Enthalpy ∆H

Entropy S

What is a State Function?

Examples of State Functions

EnthalpyEntropyInternal EnergyTemperaturePressure

Thermodynamics

Defining a system

SystemSurroundingsUniverse

Stoichiometric Thermochemistry

When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the molar heat of combustion of methane under these conditions?

Stoichiometric Thermochemistry: Answer

When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the molar heat of combustion of methane under these conditions?

-802 kJ/mol

Stoichiometric Thermochemistry: Your Turn

The oxidation of glucose is described by the following equation:

C6H12O6 + 6O2 --> 6CO2 + 6H2O ∆Ho = -2816 kJ

How much heat in kJ is produced by the oxidation of 1.0 g of glucose?

Stoichiometric Thermochemistry:Your Turn Answer

The oxidation of glucose is described by the following equation:

C6H12O6 + 6O2 --> 6CO2 + 6H2O ∆Ho = -2816 kJ

How much heat in kJ is produced by the oxidation of 1.0 g of glucose?

16 kJ

Another term

Standard molar heat of combustion: The amount of heat released when one mole of a substance is burned in oxygen. The measurements must be taken at standard thermodynamic conditions. i.e. 298 K and 1.00 atm pressure. Symbolized by ∆Ho

comb.

Still another important term

Standard molar heat of formation: The amount of heat lost or absorbed when one mole of product is formed from its elements in their most stable state. Again, the measurements must be taken at thermodynamic standard conditions. Symbolized by ∆Ho

f.

Is this an example of ∆Hocomb,

∆Hof, neither, or both?

C + 1/2 O2 --> CO

CO + 1/2 O2 --> CO2

CH3OH + 3/2 O2 --> CO2 + 2H2O

8 C + 9H2 --> C8H18

CH4 + 2O2 --> CO2 + 2H2O

FeCl2 + 1/2Cl2--> FeCl3

S + O2 --> SO2

Hess’s Law

Since enthalpy is a state function, the change in enthalpy in going from some initial state to some final state is independent of the pathway. This means that in going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in a single step or in a series of steps. This principle is known as Hess’s Law.

Hess’s Law

ΔHreaction= ΔHproducts∑ − ΔHreactants∑

Hess’s Law Problem #1

Given the following data:H2 + 1/2O2 --> H2O(l) ∆Ho=-285.8 kJ

N2O5 + H2O --> 2HNO3 ∆Ho= -76.6 kJ

1/2N2 + 3/2O2 + 1/2H2_--> HNO3 ∆Ho=-174.1 kJ

Calculate ∆Ho for the reaction

2N2 + 5O2 --> 2N2O5

Answer

Given the following data:H2 + 1/2O2 --> H2O(l) ∆Ho=-285.8 kJ

N2O5 + H2O --> 2HNO3 ∆Ho= -76.6 kJ

1/2N2 + 3/2O2 + 1/2H2 -->HNO3 ∆Ho=-174.1 kJ

Calculate ∆Ho for the reaction

2N2 + 5O2 --> 2N2O5

28.4 kJ

Another Hess’s Law Problem

Calculate ∆Ho for the processSb(s) + 5/2 Cl2(g) --> SbCl5(g)

from the following information.

Sb(s) + 3/2 Cl2(g) --> SbCl3(g) ∆Ho -314 kJ

SbCl3(g) + Cl2(g) --> SbCl5(g) ∆Ho -80 kJ

Another Hess’s Law Problem:Answer

Calculate ∆Ho for the processSb(s) + 5/2 Cl2(g) --> SbCl5(g)

from the following information.

Sb(s) + 3/2 Cl2(g) --> SbCl3(g) ∆Ho -314 kJ

SbCl3(g) + Cl2(g) --> SbCl5(g) ∆Ho -80 kJ

Answer: -394 kJ

Spontaneity

The driving force of reactions.

–Enthalpy

–Entropy

Free Energy ∆G

∆G = ∆H - T∆S