GENERAL CHEMISTRY 1140

CALORIMETRY

HESS'S LAW

 

CONSIDER THE DESERT

IT GETS VERY HOT IN THE DAY

IT GETS COLD AT NIGHT

BIG TEMPERATURE CHANGE

CONSIDER WISCONSIN IN JULY

IT GETS QUITE WARM IN THE DAY

IT COOLS OFF A LITTLE BIT AT NIGHT

SMALL TEMPERATURE CHANGE

WHY? WHAT IS THE BIG DIFFERENCE BETWEEN THE DESSERT AND WISCONSIN?

SAND vs. WATER!

A BIG DIFFERENCE IN THE ABILITY TO SOAK UP HEAT

 

SPECIFIC HEAT (s) - AMOUNT OF HEAT NEEDED TO RAISE THE TEMPERATURE OF 1 GRAM OF A SUBSTANCE 1 DEGREE C

H₂O = 4.184 J/g^oC

QUARTZ = 0.79 J/g^oC

Al = 0.90 J/g^oC

Cu = 0.38 J/g^oC

Fe = 0.44 J/g^oC

C (GRAPHITE) = 0.72 J/g^oC

C (DIAMOND) = 0.50 J/g^oC

HEAT CAPACITY (C) - AMOUNT OF HEAT NEEDED TO RAISE THE TEMPERATURE OF SOMETHING (APPARATUS, ROOM, ETC.) 1 DEGREE C

FOR A PURE SUBSTANCE: C = m s (m = MASS)

 

HOW DO WE MEASURE HEAT (q)?

IN A CALORIMETER

EITHER CONSTANT VOLUME OR CONSTANT PRESSURE CALORIMETER DEPENDING ON REACTION

CONSTANT VOLUME CALORIMETER

C₆H₁₂O₆(s) + 6 O₂(g) ---> 6 CO₂(g) + 6 H₂O(l)

GASEOUS REACTANTS AND/OR PRODUCTS REQUIRE A CONSTANT VOLUME CALORIMETER SO MASS DOES NOT LEAVE SYSTEM

CONSTANT VOLUME BOMB CALORIMETER

THE SAMPLE AND THE CALORIMETER MAKE UP AN ISOLATED SYSTEM

q_SYSTEM = 0 = q_H2O + q_CALORIMETER + q_REACTION

q_REACTION = -(q_H2O + q_CALORIMETER)

PROBLEM - 1.00 g OF OCTANE, C₈H₁₈, WAS COMPLETELY BURNED IN OXYGEN IN A CONSTANT VOLUME BOMB CALORIMETER WITH A HEAT CAPACITY OF 837 J/^oK AND CONTAINING 1.20 Kg OF WATER. THE TEMPERATURE INITIALLY WAS AT 25.00^oC AND ROSE TO 33.20^oC. CALCULATE THE HEAT EVOLVED PER MOLE OF OCTANE.

 

C₈H₁₈(l) + (25/2)O₂(g) ---> 8CO₂(g) + 9H₂O(l)

 

q_H2O = smDt = 4.184 J/g^o X (1.20 X 10³ g) (33.20^oC - 25.00^oC) = 41.2 X 10³ J

 

q_CAL = 837 J/^o (33.20^o - 25.00^o) = 6.86 X 10³ J

 

q_RXN = - (q_H2O + q_CAL)

 

q_RXN = - (41.2 X 10³ J + 6.86 X 10³ J) = - 48.1 X 10³ J

 

 

CONSTANT PRESSURE CALORIMETER

HCl(aq) + NaOH(aq) ---> NaCl(aq) + H₂O(l)

NO GASEOUS REACTANTS OR PRODUCTS, OPEN TO ATMOSPHERE

 

CONSTANT PRESSURE CALORIMETER

 

 

 

PROBLEM - A 150.0 mL ALIQUOT OF 0.800 M HCl WAS ADDED TO 150.0 mL OF 0.800 M NaOH IN A STYROFOAM CUP - CONSTANT PRESSURE CALORIMETER WITH A HEAT CAPACITY OF 328 J/^o. BOTH SOLUTIONS ORIGINALLY WERE AT 24.32^o. UPON MIXING THE MAXIMUM TEMPERATURE OBTAINED WAS 28.58^o. DETERMINE THE MOLAR HEAT OF NEUTRALIZATION.

FOR DILUTE SOLUTIONS IT IS FAIR TO ASSUME THAT THE DENSITY OF THE DILUTE SOLUTION IS EQUAL TO THE DENSITY OF PURE WATER AND THAT THE SPECIFIC HEAT OF THE DILUTE SOLUTION IS EQUAL TO THE SPECIFIC HEAT OF PURE WATER.

 

q_SOLUTIONS = smDt = 4.184 J/g^o X (300.0 g) (28.58^oC - 24.32^oC) = 5347 J

 

q_CAL = 328 J/^o X (28.58^oC - 24.32^oC) = 1397 J

 

q_RXN = - (q_SOLUTIONS + q_CAL)

 

q_RXN = - (5447 J + 1397 J) = - 6744 J

 

n = 0.1500 L x 0.800 mol/L = 0.120 mol

 

 

THERMODYNAMICS

WE CAN MEASURE DH, AND MANY HAVE!

CAN WE CALCULATE DH FOR NEW SYSTEMS?

NEED A REFERENCE SINCE WE CAN'T MEASURE ABSOLUTE H

THE STANDARD STATE IS ONE ATMOSPHERE (USUALLY 25^oC)

STANDARD ENTHALPY OF FORMATION IS THE HEAT CHANGED WHEN 1 MOLE OF THE COMPOUND IS FORMED FROM THE ELEMENTS IN THEIR STANDARD STATES

DH^o_f

H₂(g) + 1/2 O₂(g) ---> H₂O(l), DH^o_f = -285.8 kJ/mole

H₂(g) + O₂(g) ---> H₂O₂(l), DH^o_f = -187.6 kJ/mole

WHICH REACTION IS MORE EXOTHERMIC?

FORMATION OF WATER

THEREFORE, WE WOULD EXPECT

H₂O₂(l) ---> H₂O(l) + 1/2 O₂(g)

TO BE EXOTHERMIC, BUT BY HOW MUCH?

CONSIDER

C(s) + 1/2 O₂(g) ---> CO(g), DH^o_f = - 110.5 kJ/mole

C(s) + O₂(g) ---> CO₂(g), DH^o_f = - 393.5 kJ/mole

CO(g) + 1/2 O₂(g) ---> CO₂(g), DH^o_f = - 283.0 kJ/mole

NOTE: -283.0 +(-110.5) = -393.5

 

THIS IS A CONSEQUENCE OF THE FIRST LAW OF THERMODYNAMICS

ENERGY CAN BE CONVERTED FROM ONE FORM TO ANOTHER, BUT IT CANNOT BE CREATED OR DESTROYED.

 

HESS' LAW - 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.

WHY?

ENTHALPY IS A STATE FUNCTION, IT DEPENDS ONLY UPON THE STATE AND NOT ON HOW YOU GOT THERE

DH WOULD DEPEND ONLY UPON THE INITIAL STATE AND THE FINAL STATE

 

A ---> B, DH₁

B ---> C, DH₂

C ---> D, DH₃

___________

A + B + C ---> B + C + D

A ---> D, DH_AD = DH₁ + DH₂ + DH₃

 

WHAT IS DH FOR

CaCO₃(S) ---> CaO(S) + CO₂(g)

Ca(s) + 1/2 O₂(g) ---> CaO(s), DH = -635.6 kJ

C(s) + O₂(g) ---> CO₂(g), DH = -393.5 kJ

CaCO₃ ---> Ca(s) + C(s) + 1 1/2 O₂(g), DH = +1206.9 kJ

___________

Ca(s) + 1/2 O₂(g) + C(s) + O₂(g) + CaCO₃ ---> CaO(s) + CO₂(g) + Ca(s) + C(s) + 1 1/2 O₂(g)

DH = (-635.6) + (-393.5) + (+1206.9) = +177.8 kJ

 

WHERE DID THE DH VALUES COME FROM?

HEATS OF FORMATION!

DH^o_RXN = S(DH^o_fPROD) - S(DH^o_fREACT)

TABLES OF HEATS OF FORMATION ABOUND IN THE LITERATURE

IF YOU CAN'T FIND THE HEATS OF FORMATION IN THE LITERATURE, FIND APPROPRIATE REACTIONS, WHICH, WHEN ADDED TOGETHER, GIVE THE DESIRED OVERALL CHANGE.

EVEN MORE COMPLEX SYSTEM CAN BE DETERMINED IF YOU CAN BREAK IT DOWN APPROPRIATELY

 

WHAT IS THE STANDARD HEAT OF FORMATION OF ACETIC ACID, CH₃CO₂H?

HOW IS THIS DETERMINED EXPERIMENTALLY?

HEATS OF COMBUSTION AND HESS' LAW!

2 C(GRAPHITE) + 2 H₂(g) + O₂(g) ---> CH₃CO₂H

 

2 C(GRAPHITE) + 2 O₂(g) ---> 2 CO₂(g), DH = - 571.6 kJ

2 H₂(g) + O₂(g) ---> 2 H₂O(l), DH = - 787.0 kJ

2 CO₂(g) + 2 H₂O(l) ---> CH₃CO₂H + 2 O₂(g), DH = +878.4 kJ

_______________

2 C(GRAPHITE) + 2 O₂(g) + 2 H₂(g) + O₂(g) + 2 CO₂(g) + 2 H₂O(l) ---> 2 CO₂(g) + 2 H₂O(l) + CH₃CO₂H + 2 O₂(g),

DH = (-571.6) +(-787.0) + (+878.4)
= -484.2 kJ

 

HEAT OF SOLUTION - WHEN A SUBSTANCE IS DISSOLVED IN A SOLVENT, HEAT IS EITHER GENERATED OR ABSORBED

NaCl(s) + H₂O(l) ---> Na¹⁺(aq) + Cl^1-(aq)

NaCl(s) ---> Na¹⁺(g) + Cl^1-(g), DH = 788 kJ (LATTICE ENERGY)

Na¹⁺(g) + Cl^1-(g) + H₂O(l) ---> Na¹⁺(aq) + Cl^1-(aq), DH = -784 kJ (HEAT OF HYDRATION)

DH_SOLUTION = +788 +(-784) = +4 kJ (ENDOTHERMIC)

NOTE DH_SOLUTION IS A FUNCTION OF CONCENTRATION
CAN ALSO FIND HEATS OF DILUTION