Thermochemistry

Chapter 17:3

Pages 520-525

A. Heat of FusionA. Heat of Fusion

Molar Heat of Fusion (Hfus)

• Heat absorbed by one mole of a solid substance when it melts to a liquid at a constant temperature

Molar Heat of Solidification (Hsolid)

• Heat lost by one mole of a liquid substance when it solidifies at a constant temperature

Hfus = - Hsolid

A. Heat of FusionA. Heat of Fusion

Molar Heat of Fusion (Hfus)

H2O(s) H2O(l) Hfus = 6.02 kJ/mol

Molar Heat of Solidification (Hsolid)

H2O(l) H2O(s) Hsolid = -6.02 kJ/mol

B. Heating CurvesB. Heating Curves

Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

B. Heating CurvesB. Heating Curves

Temperature Change• change in KE (molecular motion) • depends on heat capacity

Phase Change• change in PE (molecular arrangement)• temp remains constant

C. Heat of VaporizationC. Heat of Vaporization

Molar Heat of Vaporization (Hvap)• energy required to boil 1 mole of a

substance at its b.p.Hvap for water = 40.7 kJ/mol• H2O(l) H2O(g) Hvap = 40.7 kJ/mol

• usually larger than Hfus…why?

EX: sweating, steam burns

• How much heat energy is required to melt 25 grams of ice at 0.000oC to liquid water at a temperature of 0.000oC?

25 g H2O 1 mol H2O

18.02 g H2O

= 8.4 kJ

6.02 kJ

1 mol H2O

ice

D. Practice ProblemsD. Practice Problems

D. Practice ProblemsD. Practice Problems

• How much heat energy is required to change 500.0 grams of liquid water at 100.0oC to steam at 100.0oC?

= 1129 kJ

500.0 g H2O 1 mol H2O

18.02 g H2O

40.7 kJ

1 mol H2O

steam

D. Practice ProblemsD. Practice Problems

• How many kJ are absorbed when 0.46g of C2H5Cl vaporizes at its normal boiling point? The molar Hvap is 26.4 kJ/mol.

= 0.19 kJ

0.46 g C2H5Cl 1 mol C2H5Cl

64.52 g C2H5Cl

26.4 kJ

1 mol C2H5Cl

E. Heat Calculations with State ChangesE. Heat Calculations with State Changes

q = mCΔTCH2O(l) = 4.184 J/goC

QTC(l)

ΔHfus = 6.02

kJ/mol

QPC(f)

ΔHvap = 40.7

kJ/molQPC(v)

q = mCΔTCH2O(s) = 2.03 J/goC

QTC(s)

QTC(v)

q = mCΔTCH2O(g) = 2.02 J/goC

See handout “How to

Calculate Heat in Changes of

State”

E. Heat Calculations with State ChangesE. Heat Calculations with State Changes

q = mCΔTCH2O(l) = 4.184 J/goC

QTC(l)

ΔHvap = 40.7

kJ/molQPC(v) QTC(v)

q = mCΔTCH2O(g) = 2.02 J/goC

• How much heat energy is required to change 100.0 grams of liquid water at 25.0oC to steam at 120.0oC?

E. Heat Calculations with State ChangesE. Heat Calculations with State Changes

QTC(l) = (100.0g)(4.184J/goC)(100-25oC)= 31380 J = 31.38 kJ

QPC(v) =

QTC(v) = (100.0g)(2.02J/goC)(120-100oC)= 4040 J = 40.40 kJ

q = mCΔTCH2O(l) = 4.184 J/goC

QTC(l)

ΔHvap = 40.7 kJ/molQPC(v)

QTC(v) q = mCΔTCH2O(g) = 2.02 J/goC

= 225.9 kJ

100.0 g H2O 1 mol H2O

18.02 g H2O

40.7 kJ

1 mol H2O

+

+

= 298 kJ

E. Heat of SolutionE. Heat of Solution

During the formation of a solution, heat is either released or absorbed

Enthalpy change caused by dissolution of 1 mol of a substance is the molar heat of solution Hsoln

Examples: hot packs, cold packs

E. Heat of SolutionE. Heat of Solution

NaOH(s) → Na+(aq) + OH-(aq)Hsoln = -445.1 kJ/mol

NH4NO3(s) → NH4+(aq) + NO3

-(aq)

Hsoln = 25.7 kJ/mol

F. Practice ProblemsF. Practice Problems

• How much heat (in kJ) is released when 20.0 g of NaOH(s) is dissolved in water? The molar Hsoln is -445.1 kJ/mol.

= 223 kJ

20.0 g NaOH 1 mol NaOH

40.00 g NaOH

-445.1 kJ

1 mol NaOH