Slide 1 of 58

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.

Slide 2 of 58

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

Slide 3 of 58

First Law of Thermodynamics

Slide 4 of 58

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.

Slide 5 of 58

Functions of State

U is a function of state. Not easily measured.

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

Slide 6 of 58

Path Dependent Functions

Changes in heat and work are not functions of state. Remember example 7-5, w = -1.24 102 J in a one

step expansion of gas: Consider 2.40 atm to 1.80 atm and finally to 1.20 atm.

Slide 7 of 58

Path Dependent Functions

w = (-1.80 atm)(1.36-1.02)L – (1.30 atm)(2.04-1.36)L

= -0.61 L atm – 0.82 L atm = -1.43 L atm

= -1.44 102 JCompared -1.24 102 J for the two stage process

Slide 8 of 58

7-6 Heats of Reaction: U and H

Reactants → Products

Ui Uf

U = Uf - Ui

= qrxn + w

In a system at constant volume:

U = qrxn + 0 = qrxn = qv

But we live in a constant pressure world!

How does qp relate to qv?

Slide 9 of 58

Heats of Reaction

Slide 10 of 58

Heats of Reaction

qV = qP + w

We know that w = - PV and U = qP, therefore:

U = qP - PV

qP = U + PV

These are all state functions, so define a new function.

Let H = U + PV

Then H = Hf – Hi = U + PV

If we work at constant pressure and temperature:

H = U + PV = qP

Slide 11 of 58

Comparing Heats of Reaction

qP = -566 kJ/mol

= H

PV = P(Vf – Vi)

= RT(nf – ni)

= -2.5 kJ

U = H - PV

= -563.5 kJ/mol

= qV

Slide 12 of 58

Changes of State of Matter

H2O (l) → H2O(g) H = 44.0 kJ at 298 K

Molar enthalpy of vaporization:

Molar enthalpy of fusion:

H2O (s) → H2O(l) H = 6.01 kJ at 273.15 K

Slide 13 of 58

Break the problem into two steps: Raise the temperature of the liquid first then completely vaporize it. The total enthalpy change is the sum of the changes in each step.

Enthalpy Changes Accompanying Changes in States of Matter. Calculate H for the process in which 50.0 g of water is converted from liquid at 10.0°C to vapor at 25.0°C.

= (50.0 g)(4.184 J/g °C)(25.0-10.0)°C + 50.0 g

18.0 g/mol 44.0 kJ/mol

Set up the equation and calculate:

qP = mcH2OT + nHvap

= 3.14 kJ + 122 kJ = 125 kJ

EXAMPLE 7-8

Slide 14 of 58

Standard States and Standard Enthalpy Changes

Define a particular state as a standard state. Standard enthalpy of reaction, H°

The enthalpy change of a reaction in which all reactants and products are in their standard states.

Standard State The pure element or compound at a pressure of

1 bar and at the temperature of interest.

Slide 15 of 58

Enthalpy Diagrams

Slide 16 of 58

7-7 Indirect Determination of H:Hess’s Law

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

substance in a system.

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

Slide 17 of 58

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