Prentice-Hall © 2007 General Chemistry: Chapter 20 Slide 1 of 54 Juana Mendenhall, Ph.D. Assistant...

Prentice-Hall © 2007General Chemistry: Chapter 20Slide 1 of 54

Juana Mendenhall, Ph.D.

Assistant Professor

Lecture 2 March 31

Chapter 7: Thermochemistry


Objectives

1. Define work and it’s units.

2. Define, explain, and apply the first law of thermodynamics.

3. Apply the direct and indirect method of the standard enthalpy of formation and reaction.


The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states.

The standard enthalpy of formation of a pure element in its reference state is 0.

Hf°

Standard Enthalpies of Formation


Standard Enthalpy of Formation and Reaction

H is an extensive property. Enthalpy change is directly proportional to the amount of substance

in a system.

aA + bB → cC + dD

Hrxn

= [cHf(C )+ dHf(D)] - [aHf(A) + bHf(B)]

Hrxn

= [nHf(products)] - [mHf(reactants)]

Direct Method Example with C(graphite)


Standard States


Standard Enthalpies of Formation


Indirect Determination of H:Hess’s Law

Hess’s Law states: when reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

N2(g) + O2(g) → 2 NO(g) H = +180.50 kJ

½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ

H changes sign when a process is reversed

NO(g) → ½N2(g) + ½O2(g) H = -90.25 kJ


Hess’s Law

Hess’s law of constant heat summation If a process occurs in stages or steps (even hypothetically), the

enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps.

½N2(g) + O2(g) → NO2(g) H = +33.18 kJ

½N2(g) + ½O2(g) → NO(g) H = +90.25 kJ

NO(g) + ½O2(g) → NO2(g) H = -57.07 kJ


Hess’s Law Schematically


Functions of State

Any property that has a unique value for a specified state of a system is said to be a State Function.

◦ Water at 293.15 K and 1.00 atm is in a specified state.

◦ d = 0.99820 g/mL

◦ This density is a unique function of the state.

◦ It does not matter how the state was established.


7-4 Work

In addition to heat effects chemical reactions may also do work.

Gas formed pushes against the atmosphere.

Volume changes.

Pressure-volume work.

w = Fd [w ] = kg m

s2 = Jm

Work


Pressure Volume Work

w = F d

= (m g)

(m g)

= PV

w = -PextV

A h= A

h


Definition of Terms

State of a system: the values of all relevant macroscopic properties-example: energy, temp., pressure, volume.

State function: ppts that are determined by the state of the systems.

U is a function of state. Not easily measured.

U has a unique value between two states. Is easily measured.


7-5 The First Law of Thermodynamics

Internal Energy, U. Total energy (potential and kinetic) in a system.

•Translational kinetic energy.

•Molecular rotation.

•Bond vibration.

•Intermolecular attractions.

•Chemical bonds.

•Electrons.


First Law of Thermodynamics

A system contains only internal energy. A system does not contain heat or work. These only occur during a change in the system.

Law of Conservation of Energy The energy of an isolated system is constant

U = q + w


First Law of Thermodynamics


Sign conventions for work & heat

Process Sign

Work done by the system on the surroundings

_

Work done on the system by the surroundings

+

Heat absorbed by the system from the surroundings (endothermic process)

+

Heat absorbed by the surroundings from the system (exothermic procees)

_


Example

The work done when a gas is compressed in a

cylinder like that show in the figure is 462 J.

during this process, there is a heat transfer of

128 J from the gas surroundings. Calculate the

energy change for this process.

Prentice-Hall © 2007 General Chemistry: Chapter 20 Slide 1 of 54 Juana Mendenhall, Ph.D. Assistant...

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