thermodynamics review. energy ability to do work units– joules (j), we will use “kj” can be...

THERMODYNAMICS REVIEW

ENERGY

• ABILITY TO DO WORK

• UNITS– JOULES (J), WE WILL USE “KJ”

• CAN BE CONVERTED TO DIFFERENT TYPES

• ENERGY CHANGE RESULTS FROM FORMING AND BREAKING CHEMICAL BONDS IN REACTIONS

SYSTEM VS. SURROUNDINGS

HEAT (Q)

• ENERGY TRANSFER BETWEEN A SYSTEM AND THE SURROUNDINGS DUE TO A TEMPERATURE CHANGE

• TRANSFER IS INSTANTANEOUS FROM HIGH----LOW TEMPERATURE UNTIL THERMAL EQUILIBRIUM

• TEMPERATURE—

• MEASURE OF HEAT, “HOT/COLD”

HEAT (Q) CONTINUED

• KINETIC THEORY OF HEAT

• HEAT INCREASE RESULTING IN TEMPERATURE CHANGE CAUSES AN INCREASE IN THE AVERAGE MOTION OF PARTICLES WITHIN THE SYSTEM.

• INCREASE IN HEAT RESULTS IN

• ENERGY TRANSFER

• INCREASE IN BOTH POTENTIAL AND KINETIC ENERGIES

1ST LAW OF THERMODYNAMICS(CONSERVATION OF ENERGY)• ENERGY CANNOT BE CREATED OR DESTROYED.

• WITH PHYSICAL AND CHEMICAL CHANGES, ENERGY CAN BE TRANSFERRED OR CONVERTED.

• TOTAL ENERGY = ΣENERGY OF ITS COMPONENTS

• ΔU = Q + W , ΔETOTAL = ΔESYS + ΔESURR = 0

ENTHALPY

THERMODYNAMICS 101• FIRST LAW OF THERMODYNAMICS

• ENERGY IS CONSERVED IN A REACTION (IT CANNOT BE CREATED OR DESTROYED)---SOUND FAMILIAR???

• MATH REPRESENTATION: ΔETOTAL = ΔESYS + ΔESURR = 0

• Δ= “CHANGE IN”

• ΔΕ= POSITIVE (+), ENERGY GAINED BY SYSTEM

• ΔΕ= NEGATIVE (-), ENERGY LOST BY SYSTEM

• TOTAL ENERGY = SUM OF THE ENERGY OF EACH PART IN A CHEMICAL REACTION

ENTHALPY (H)• MEASURES 2 THINGS IN A CHEMICAL REACTION:

1) ENERGY CHANGE

2) AMOUNT OF WORK DONE TO OR BY CHEMICAL REACTION

• 2 TYPES OF CHEMICAL REACTIONS:

1)EXOTHERMIC—HEAT RELEASED TO THE SURROUNDINGS, GETTING RID OF HEAT, -ΔΗ

2)ENDOTHERMIC—HEAT ABSORBED FROM SURROUNDINGS, BRINGING HEAT IN, +ΔΗ

**ENTHALPY OF REACTION—HEAT FROM A CHEMICAL REACTION WHICH IS GIVEN OFF OR ABSORBED, UNITS = KJ/MOL

• ENTHALPY OF REACTION

• HEAT FROM A CHEMICAL REACTION WHICH IS GIVEN OFF OR ABSORBED

• AT CONSTANT PRESSURE

• UNITS = KJ/MOL

EXOTHERMIC • TEMPERATURE INCREASE (--ISOLATED SYSTEM)

• HEAT IS RELEASED TO SURROUNDINGS (--OPEN/CLOSED SYSTEM)

• Q = - VALUE

• CHEMICAL THERMAL ENERGY

ENDOTHERMIC

• TEMPERATURE DECREASE (--ISOLATED SYSTEM)

• ALL ENERGY GOING INTO REACTION, NOT INTO SURROUNDINGS

• HEAT ABSORBED BY SYSTEM, SURROUNDINGS HAVE TO PUT ENERGY INTO REACTION

• Q = + VALUE

• THERMAL CHEMICAL ENERGY

METHODS FOR CALCULATING ENTHALPY--1) STOICHIOMETRIC CALCULATIONS USING BALANCED CHEMICAL EQUATION

2) CALORIMETRY (LAB BASED METHOD)

3) HESS’S LAW

4) ENTHALPY OF FORMATION

5) BOND ENTHALPIES

**WHICH METHOD IS THE LEAST ACCURATE?

CALORIMETRYHOW DO WE FIND THE CHANGE IN ENERGY/HEAT

TRANSFER THAT OCCURS IN CHEMICAL REACTIONS???

CALORIMETRY • EXPERIMENTALLY “MEASURING” HEAT TRANSFER FOR A CHEMICAL

REACTION OR CHEMICAL COMPOUND

• CALORIMETER

• INSTRUMENT USED TO DETERMINE THE HEAT TRANSFER OF A CHEMICAL REACTION

• DETERMINES HOW MUCH ENERGY IS IN FOOD

• OBSERVING TEMPERATURE CHANGE WITHIN WATER AROUND A REACTION CONTAINER

** ASSUME A CLOSED SYSTEM, ISOLATED CONTAINER

• NO MATTER, NO HEAT/ENERGY LOST

• CONSTANT VOLUME

SPECIFIC HEAT CAPACITY

• AMOUNT OF HEAT REQUIRED TO INCREASE THE TEMPERATURE OF 1G OF A CHEMICAL SUBSTANCE BY 1°C

• UNITS--- J/G°K

• UNIQUE TO EACH CHEMICAL SUBSTANCE

• AL(S) = 0.901J/G°K

• H2O(L) = 4.18 J/G°K

Q = SMΔT

“COFFEE CUP” CALORIMETER (CONT.)

•QCHEMICAL = -

QWATER

•ΔQRXN

•HEAT GAINED/LOST IN EXPERIMENT WITH CALORIMETER

•ΔHRXN

•HEAT GAINED/LOST IN TERMS OF THE BALANCED CHEMICAL EQUATION

EXAMPLE 2: USING THE FOLLOWING DATA, DETERMINE THE METAL’S SPECIFIC HEAT.

• METAL MASS = 25.0G WATER MASS = 20.0G

• TEMPERATURE OF LARGE WATER SAMPLE = 95°C

• INITIAL TEMPERATURE IN CALORIMETER = 24.5°C

• FINAL TEMPERATURE IN CALORIMETER = 47.2°C

• SPECIFIC HEAT OF WATER = 1.00 CAL/G°C OR 4.184 J/G°K (KNOW!!!!)

BOND ENERGY

• ENERGY REQUIRED TO MAKE/BREAK A CHEMICAL BOND

• ENDOTHERMIC REACTIONS

• PRODUCTS HAVE MORE ENERGY THAN REACTANTS

• MORE ENERGY TO BREAK BONDS

• EXOTHERMIC REACTIONS

• REACTANTS HAVE MORE ENERGY THAN PRODUCTS

• MORE ENERGY TO FORM BONDS

BOND ENTHALPY

• FOCUSES ON THE ENERGY/HEAT BETWEEN PRODUCTS AND REACTANTS AS IT RELATES TO CHEMICAL BONDING

• AMOUNT OF ENERGY ABSORBED TO BREAK A CHEMICAL BOND---AMOUNT OF ENERGY RELEASED TO FORM A BOND.

• MULTIPLE CHEMICAL BONDS TAKE MORE ENERGY TO BREAK AND RELEASE MORE ENERGY AT FORMATION

• AMOUNT OF ENERGY ABSORBED = AMOUNT OF ENERGY RELEASED

TO BREAK CHEMICAL BOND TO FORM A CHEMICAL BOND

CALCULATING ΔHRXN. BY BOND ENTHALPIES (4TH METHOD)

• LEAST ACCURATE METHOD

• ΔH = ΣBE (BONDS BROKEN) - ΣBE (BONDS FORMED)

EXAMPLE 1:

• USING AVERAGE BOND ENTHALPY DATA, CALCULATE ΔH FOR THE FOLLOWING REACTION.

• CH4 + 2O2 CO2 + 2H2O ΔH = ?

Bond Average Bond Enthalpy

C-H 413 kJ/mol

O=O 495 kJ/mol

C-O 358 kJ/mol

C=O 799 kJ/mol

O-H 467 kJ/mol

HESS’ LAW

• ENTHALPY CHANGE FOR A CHEMICAL REACTION IS THE SAME WHETHER IT OCCURS IN MULTIPLE STEPS OR ONE STEP

• ΔHRXN = ΣΔHA+B+C (SUM OF ΔH FOR EACH STEP)

• ALLOWS US TO BREAK A CHEMICAL REACTION DOWN INTO MULTIPLE STEPS TO CALCULATE ΔH

• ADD THE ENTHALPIES OF THE STEPS FOR THE ENTHALPY FOR THE OVERALL CHEMICAL REACTION

EXAMPLE 1:

H2O(L) H2O (G) ΔH° = ?

BASED ON THE FOLLOWING:

H2 + ½ O2 H2O(L) ΔH° = -285.83 KJ/MOL

H2 + ½ O2 H2O(G) ΔH° = -241.82 KJ/MOL

ENTHALPY OF FORMATION (ΔHF°)

• ENTHALPY FOR THE REACTION FORMING 1 MOLE OF A CHEMICAL COMPOUND FROM ITS ELEMENTS IN A THERMODYNAMICALLY STABLE STATE.

• ELEMENTS PRESENT IN “MOST THERMODYNAMICALLY STABLE STATE”

• 25°C°, 1ATM

EXAMPLE 5

• ISOPROPYL ALCOHOL (RUBBING ALCOHOL) UNDERGOES A COMBUSTION REACTION

2(CH3)2CHOH + 9O2 6CO2 + 8H2O

ΔH° = -4011 KJ/MOL

CALCULATE THE STANDARD ENTHALPY OF FORMATION FOR ISOPROPYL ALCOHOL.

EXAMPLE 2: • CALCULATE THE ΔH FOR THE FOLLOWING REACTION WHEN

12.8 GRAMS OF HYDROGEN GAS COMBINE WITH EXCESS CHLORINE GAS TO PRODUCE HYDROCHLORIC ACID.

• H2 + CL2 2HCL ΔH = -184.6 KJ

ENTROPY

SPONTANEOUS VS. NONSPONTANEOUS

1)SPONTANEOUS PROCESS

• OCCURS WITHOUT HELP OUTSIDE OF THE SYSTEM, NATURAL

• MANY ARE EXOTHERMIC—FAVORS ENERGY RELEASE TO CREATE AN ENERGY REDUCTION AFTER A CHEMICAL REACTION

• EX. RUSTING IRON WITH O2 AND H2O, COLD COFFEE IN

A MUG

• SOME ARE ENDOTHERMIC

• EX. EVAPORATION OF WATER/BOILING, NACL DISSOLVING IN WATER

SPONTANEOUS VS. NONSPONTANEOUS

2) NONSPONTANEOUS PROCESS

• REQUIRES HELP OUTSIDE SYSTEM TO PERFORM CHEMICAL REACTION, GETS AID FROM ENVIRONMENT

• EX. WATER CANNOT FREEZE AT STANDARD CONDITIONS (25°C, 1ATM), CANNOT BOIL AT 25°C

**CHEMICAL PROCESSES THAT ARE SPONTANEOUS HAVE A NONSPONTANEOUS PROCESS IN REVERSE **

ENTROPY (S)

• MEASURE OF A SYSTEM’S DISORDER

• DISORDER IS MORE FAVORABLE THAN ORDER

• ΔS = S(PRODUCTS) - S(REACTANTS)

• ΔS IS (+) WITH INCREASED DISORDER

• STATE FUNCTION

• ONLY DEPENDENT ON INITIAL AND FINAL STATES OF A REACTION

• EX. EVAPORATION, DISSOLVING, DIRTY HOUSE

THERMODYNAMIC LAWS

1ST LAW OF THERMODYNAMICS

• ENERGY CANNOT BE CREATED OR DESTROYED

2ND LAW OF THERMODYNAMICS

• THE ENTROPY OF THE UNIVERSE IS ALWAYS INCREASING.

• NATURALLY FAVORS A DISORDERED STATE

WHEN DOES A SYSTEM BECOME MORE DISORDERED

FROM A CHEMICAL REACTION? (ΔS > 0)

1)MELTING

2)VAPORIZATION

3)MORE PARTICLES PRESENT IN THE PRODUCTS THAN THE REACTANTS

• 4C3H5N3O9 (L) 6N2 (G) + 12CO2 (G) + 10H2O (G) + O2 (G)

4)SOLUTION FORMATION WITH LIQUIDS AND SOLIDS

5)ADDITION OF HEAT

3RD LAW OF THERMODYNAMICS

THE ENTROPY (ΔS) OF A PERFECT CRYSTAL IS 0 AT A TEMPERATURE OF ABSOLUTE ZERO (0°K).

• NO PARTICLE MOTION AT ALL IN CRYSTAL STRUCTURE

• ALL MOTION STOPS

HOW DO WE DETERMINE IF A CHEMICAL REACTION IS

SPONTANEOUS?1)CHANGE IN ENTROPY (ΔS)

2)GIBBS FREE ENERGY (ΔG)

CHANGE IN ENTROPY (ΔS)

• FOR A CHEMICAL REACTION TO BE SPONTANEOUS (ΔST > 0),

THERE MUST BE AN INCREASE IN SYSTEM’S ENTROPY (ΔSSYS>

0) AND THE REACTION MUST BE EXOTHERMIC (ΔSSURR > 0).

• EXOTHERMIC REACTIONS ARE FAVORED, NOT ENDOTHERMIC REACTIONS.

• EXOTHERMIC (ΔH < 0, ΔS > 0)

• ENDOTHERMIC (ΔH > 0, ΔS < 0)

• ΔST = ΔSSYS + ΔSSURR

• IF ΔST > 0, THEN THE CHEMICAL REACTION IS SPONTANEOUS

EXAMPLE 1:

WILL ENTROPY INCREASE OR DECREASE FOR THE FOLLOWING?

a)N2 (G) + 3H2 (G) 2NH3 (G)

b)2KCLO3 (S) 2KCL (S) + 3O2 (G)

c)CO(G) + H2O(G) CO2 (G) + H2 (G)

d)C12H22O11 (S) C12H22O11

HOW DO WE CALCULATE THE ENTROPY CHANGE (ΔS) IN A CHEMICAL REACTION?

• SAME METHOD AS USING THE ENTHALPIES OF FORMATION TO CALCULATE ΔH AND USE THE SAME TABLE.

• AA + BB CC + DD

ΔS° =[C (ΔS°C) + D(ΔS°D)] - [A (ΔS°A) + B (ΔS°B)]

EXAMPLE 2: CALCULATE ΔS° FOR THE FOLLOWING REACTION AT 25°C….

4HCL(G) + O2 (G) 2CL2 (G) + 2H2O (G)

FREE ENERGY AND EQUILIBRIUM

• AT EQUILIBRIUM, ΔG = 0, SO REACTION QUOTIENT (Q) = EQUILIBRIUM CONSTANT (K)

• AT EQUILIBRIUM

• ΔG° = - RTLNK

• ENABLES THE REACTION’S EQUILIBRIUM CONSTANT (K) TO BE CALCULATED FROM THE CHANGE IN FREE ENERGY (ΔG°)

ΔG = ΔG° + RTLNQ

• THE MAGNITUDE OF ΔG° INDICATES HOW FAR THE CHEMICAL REACTION IN ITS STANDARD STATE IS FROM EQUILIBRIUM.

• ΔG° = 0 , EQUILIBRIUM

• ΔG°= LARGE VALUE, FAR FROM EQUILIBRIUM

• ΔG° = SMALL VALUE, CLOSE TO EQUILIBRIUM

• THE SIGN (+, - ) INDICATES WHICH DIRECTION THE REACTION NEEDS TO SHIFT TO ACHIEVE EQUILIBRIUM

• POSITIVE (+) -------- SHIFT TO LEFT, NO REACTION

• NEGATIVE (-) -------- SHIFT TO RIGHT, REACTION GOES TO COMPLETION

WHAT IS THE RELATIONSHIP BETWEEN FREE ENERGY(ΔG) AND K?

GIBBS FREE ENERGY

CHANGE IN GIBBS FREE ENERGY (ΔG)

•ΔG = ΔH – TΔS• RELATES ENTHALPY AND ENTROPY TO DETERMINE

WHICH HAS MORE IMPORTANCE IN DETERMINING WHETHER A REACTION IS SPONTANEOUS

• COMBINES ENERGY TRANSFER AS HEAT (ΔH) AND ENERGY RELEASED TO CONTRIBUTE TO DISORDER (ΔS)

CHANGE IN GIBBS FREE ENERGY (ΔG)

•ΔG = ΔH – TΔS

• ΔG < 0 , SPONTANEOUS REACTION

• ENERGY AVAILABLE TO DO WORK

• ΔG > 0, NONSPONTANEOUS REACTION

• ENERGY DEFICIENCY, NO LEFTOVER ENERGY AND NOT ENOUGH ENERGY FOR REACTION

HOW CAN WE APPLY THE GIBBS EQUATION TO DETERMINE SPONTANEITY OF

REACTION?

ΔH ΔS ΔG Result

- + - Spontaneous(all temperatures)

+ - + Nonspontaneous (all temperatures)

- - - Spontaneous (low temperatures)

+ + + Nonspontaneous(low temperatures)

ΔG = ΔH – TΔS

TWO PATHS TO CALCULATING ΔG

1) ΔG = ΔH – TΔS

• DETERMINE ΔH. WHAT METHODS CAN WE USE?

• DETERMINE ΔS.

• THEN CALCULATE ΔG

TWO PATHS TO CALCULATING ΔG

2) USE STANDARD FREE ENERGY OF FORMATION (ΔGF °) VALUES TO

DETERMINE ΔG

• STANDARD FREE ENERGY OF FORMATION (ΔGF °) --- ΔG° FOR

THE FORMATION OF 1 MOLE OF A CHEMICAL COMPOUND IN ITS STANDARD STATE.

• ΔGF ° FOR ELEMENT FORMATION IN THEIR MOST STABLE STATE

= 0.

• AA + BB CC + DD

ΔG° =[C (ΔGF°)C + D(ΔGF°)D] - [A (ΔGF°)A + B

(ΔGF°)B ]

EXAMPLE 2:

A) FIND ΔG FOR A CHEMICAL REACTION GIVEN ΔH = -218 KJ AND ΔS = -765 J/K AT 32°C.

B) AT WHAT TEMPERATURE DOES THIS REACTION BECOME SPONTANEOUS? ASSUME ONLY TEMPERATURE CHANGES.

EXAMPLE 3:

CALCULATE ΔG°RXN UNDER STANDARD CONDITIONS FOR

THE FOLLOWING REACTION USING ΔGF° VALUES.

FE2O3 (S) + 2AL(S) 2FE(S) + AL2O3 (S)

•A SPONTANEOUS REACTION IS NOT NECESSARILY FAST!!!!• REACTION RATE INVOLVES KINETICS ! !