Or in general, 'T V

U UU U dV dTV T

For infinitesimal changes,

T V

U UdU dV dTV T

VV

U CT

T

T

UV

Some terms are familiar:

Math for Heat Capacity

Heat capacity at constant volume

Internal pressure at constant temp

Heat Capacity

If dV = 0, then dU = CV(T)dT

However, CV(T) is approximately constant over small temperature changes and above room temperature

so integrate both sides: U = CVT = qv+ w

Since constant volume, w= o and qV=CVT

T V

U UdU dV dTV T

CV(T)T (V)

Internal Energy (5)Many useful, general relationships are derived frommanipulations of partial derivatives, but I will(mercifully) spare you more.Suffice it to say that U is best used for processestaking place at constant volume, with only PV work:Then dU = dqV and U= U2 – U1 = qV

The increase in internal energy of a system in a rigidcontainer is thus equal to the heat qV supplied to it.

We would prefer a different state function forconstant pressure processes: enthalpy.

Enthalpy Defined

At constant V and P, q = U = H

U = q + w

q= qP = U - w, w = -PV

qP = U + PV H

Enthalpy, H U + PV

At Constant P, H = U + PV

dH dU d PV dU PdV VdP

Comparing H and Uat constant P

1. Reactions that do not involve gasesV 0 and H U

H = U + PV

2. Reactions in which ngas = 0V 0 and H U

3. Reactions in which ngas 0 V 0 and H U

A B C D

ConcepTest #1

Which of the following reactions has the largest difference between H and U?

A. NH3 (g) + HCl (g) NH4Cl (s)

B. CO (g) + Cl2 (g) COCl2 (g)

C. ZnS (s) + 3/2 O2 (g) ZnO (s) + SO2 (g)

D. ZnO (s) + CO (g) Zn (s) + CO2 (g)

Comparing H and Uat constant P (2)

1. Reactions that do not involve gasesV 0 and H U

H = U + PV

2. Reactions in which ngas = 0V 0 and H U

3. Reactions for which ngas 0

H U + ngas RT Ideal Gases

Heat Capacity atConstant Volume or Pressure

CP = dqP/dT = (H/T)P = (U/T)P + (PV/T)P

Partial derivative of enthalpy withrespect to T at constant P

CV = dqV/dT = (U/T)V

Partial derivative of internal energy with respect to T at constant V

Ideal Gas: CP = CV + nR

Heat Capacity, C

Molar Heat Capacity (J K-1 mol-1 )Heat needed to raise T of 1 mole by 1 K

q = CmnT

Heat Capacity (J K-1 )Heat needed to raise T of system by 1 K

q = CT

Specific Heat Capacity (J K-1 kg-1 ) Heat needed to raise T of 1 kg by 1 K

q = CSmT

CalorimetryMeasure H and U for Reactions

Isolate sample, bomb, and water

bath from surroundings

Initiate reaction

Heat released causes temperature rise

in system

Calorimetry ProblemFor the complete combustion of 1 mole of ethanol in a bomb calorimeter, 1364.5 kJ mol–1is released at 25 °C. Determine cU and cH.

cU = qV = –1364.5 kJ

cH = cU + nRT = –1364.5 –2.5 = –1367.0 kJ/mol

C2H5OH (l) + 3 O2 (g) 2 CO2 (g) + 3 H2O (l)c

n = -1 At 25 C, RT =8.314 J/ K 298.15 K = 2.5 kJ

Now for cH

Endothermic & Exothermic Processes

H Negative Negative amount of heat absorbed(i.e. heat released by the system)

Exothermic Process

H Positive Positive amount of heat absorbed by the system

Endothermic Process

H = Hfinal - Hinitial

ConcepTest #2

A. H2O (s) H2O (l) H = +

B. CH4 + 2 O2 CO2 + 2 H2O H = –C. H2O (g) H2O (l) H = + D. 2 H (g) H2 (g) H = –

For which of the following reactions is the indicated sign of H incorrect?

Thermochemical Equations

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l)

(a combustion reaction) cH = – 890 kJ

Phases must be specified

H is an extensive property

Sign of H changes when reaction is reversed

Reaction must be balanced

ConcepTest #3

2 CO2 (g) + 4 H2O (l) 2 CH4 (g) + 4 O2 (g)

A. -890 kJ B. -1780 kJ C. +890 kJD. +1780 kJ

What is H for this reaction?

Calculation of rxnH

I. Hess’s LawH of an overall process is the sum of the Hs for the individual steps

II. Use of Standard Enthalpies of Formation

I. Hess’s Law

Calculate H for the reaction:

H2O (solid at 0 °C) H2O (gas at 100 °C)

H2O (s, 0 °C) H2O (l, 0 °C) fusH = 6.0 kJ/mol

H2O (l, 0 °C) H2O (l, 100 °C) CT = 7.5 kJ/mol

H2O (l, 100 °C) H2O (g, 100 °C) vapH = 40.6 kJ/mol

H2O (s, 0 °C) H2O (g, 100 °C) rxnH = 54.1 kJ/mol

II. Standard Enthalpy of Formation

The standard enthalpy of formation ( fH°)of a compound is the enthalpy change for the formation of one mole of compound from the elements in their standard state.

Designated by superscript o: H°

For example, CO2:

C (graphite) + O2 (g) CO2 (g)

rxnH° = -393.5 kJ/mol Table 2.7

f H° CO2 (g) = -393.5 kJ/mol