THERMODYNAMICS REVIEW. ENERGY ABILITY TO DO WORK UNITS– JOULES (J), WE WILL USE “KJ” CAN BE...
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Transcript of 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 ! !