Thermodynamics

WHY SOME REACTIONS HAPPEN AND SOME DON’T !

Copyright Sautter 2003

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CHEMICAL THERMODYNAMICS

• WHAT IS REQUIRED FOR A REACTION TO BE SPONTANEOUS ?

• SPONTANEOUS REFERS TO A REACTION OCCURRING ON ITS OWN WITHOUT ENERGY INPUT. SPONTANEOUS DOES NOT REFER TO REACTION RATE! THE REACTION MAY BE EXTREMELY SLOW BUT AS LONG AS IT CONTINUES TO OCCUR IT IS SAID TO BE SPONTANEOUS.

• UNDERSTANDING CHEMICAL THERMODYNAMICS REQUIRED THE KNOWLEDGE OF THREE FUNDAMENTAL LAWS AND THE NATURE OF HEAT ENERGY.

CHEMICAL THERMODYNAMICS

• THE ENERGY CONTENT OF A SYSTEM CAN BE IN THE FORM OF KINETIC OR POTENTIAL ENERGY. POTENTIAL ENERGY IS FOUND IN THE BONDS THAT JOIN ATOMS TOGETHER TO FORM MOLECULES.

• KINETIC ENERGY IS THE ENERGY OF MOLECULAR MOTIONS. UNDERSTANDING THE KINETIC ENERGY CONTENT OF A SYSTEM REQUIRES US TO RECALL THE KINETIC MOLECULAR THEORY OF MATTER. IT BASIC TENENT IS THAT ALL MOLECULES ARE IN CONSTANT, RANDOM MOTION, EXCEPT AT ABSOLUTE ZERO (-273 CELSIUS).

CHEMICAL THERMODYNAMICS

• MOLECULAR MOTIONS CONSIST OF FOUR TYPES:• (1) TRANSLATIONAL MOTION – THE PLACE TO PLACE

MOTION OF THE MOLECULES THROUGH SPACE.• (2) ROTATIONAL MOTION – MOLECULES ROTATE OR

SPIN AS THEY MOVE THROUGH SPACE• (3) VIBRATIONAL MOTION – THE BONDS BETWEEN

MOLECULES ARE MUCH LIKE SPRINGS AND THE ATOMS THAT ARE BONDED TOGETHER IN MOLECULES MOVE BACK AND FORTH ALONG THE BOND AXIS WITH PARTICULAR VIBRATIONAL FREQUENCIES.

• (4) ELECTRONIC ENERGY – THE ELECTRONS WITHIN THE ATOMS CONTAIN THEIR OWN ENERGIES OF MOTION.

• THE SUM OF ALL THESE ENERGIES FOR THE MOLECULES IN A SYSTEM IS TERMED THE INTERNAL ENERGY (E) OF THE SYSTEM.

CHEMICAL THERMODYNAMICS

• HEAT ENERGY MAY BE ADDED OR REMOVED FROM A CHEMICAL SYSTEM. THOSE SYSTEMS WHICH ABSORB HEAT ENERGY ARE CALLED ENDOTHERMIC. THOSE WHICH RELEASE ENERGY ARE EXOTHERMIC.

• THE FIRST LAW OF THERMODYNAMICS STATES THAT ALL ENERGY TRANSFERS MUST ACCOUNT FOR THE TOTAL AMOUNT OF ENERGY. NO ENERGY MAY BE LOST OR GAINED, IT MAY ONLY BE CHANGED IN FORM. THIS, OF COURSE, IS THE LAW OF CONSERVATION OF ENERGY.

• THE QUANTITY OF HEAT ENERGY ADDED OR REMOVED FROM A SYSTEM CAN BE MEASURED IN JOULES, CALORIES OR BTUs. GENERALLY JOULES AND KILOJOULES ARE THE PREFERRED UNITS OF ENERGY.

CHEMICAL THERMODYNAMICS• HOW CAN HEAT ENERGY BE MEASURED?• MEASURING HEAT REQUIRES THREE FUNDAMENTAL

QUANTITIES.• (1) THE AMOUNT OF SUBSTANCE HEATED (MASS IN

GRAMS OR KILOGRAMS)• (2) THE KIND OF SUBSTANCE HEATED (THIS IS

MEASURED USING THE SPECIFIC HEAT OF THE SUBSTANCE)

• (3) THE TEMPERATURE CHANGE OF THE SUBSTANCE (IN DEGREES CELSIUS OR KELVIN)

• THE EQUATION RELATING THESE VARIABLES IS: Q = M x C x T WHERE Q = HEAT LOST OF GAINED, M = MASS OF SUBSTANCE HEATED OR COOLED, C = THE SPECFIC HEAT OF THE SUBSTANCE T = THE TEMPERATURE CHANGE OF THE SUBSTANCE

CHEMICAL THERMODYNAMICS

• A COMMON DEVICE USED TO MEASURE THE HEAT OF REACTION IS A BOMB CALORIMETER. IN THIS APPARATUS, A REACTION IS CARRIED OUT IN A CHAMBER IMMERSED IN A CONTAINER OF WATER. HEAT FLOW OUT OF THE REACTION CHAMBER AND INTO THE WATER. BASED ON THE TEMPERATURE CHANGE OF THE WATER AND ASSOCIATED EQUIPMENT THE HEAT OF REACTION ( E) CAN BE CALCULATED.

• A BOMB CALORIMETER IS A CONSTANT VOLUME MEASUREMENT MEANING THAT ALL RELEASED HEAT IS USED TO RAISE THE INTERNAL ENERGY OF THE SYSTEM AND NONE IS USED TO DO WORK.

• IN OTHER THERMODYNAMIC SYSTEM CONSTANT PRESSURE PROCESSES OCCUR. HERE THE WORK ASSOCIATED WITH VOLUME CHANGES OF THE GASES MUST BE CONSIDERED.

IGNITERPOWER SUPPLY

THERMOMETER

STIRRER

REACTIONCHAMBERINSULATED

CONTAINER

WATER CHAMBER

BOMB CALORIMETER(CONSTANT VOLUME)

HEAT FLOWS OUT OF THE REACTION CHAMBERINTO THE SURROUNDING WATER

CHEMICAL THERMODYNAMICS• CALORIMETRIC CALCULATIONS :When 0.1567 moles of ammonia react with HCl gas, in a calorimeter with

water equivalent of 200 grams (the equipment’s heat storing capacity equals that or 200 grams of water) the temperature of the 2000 grams of water in the calorimeter increases from 26.13 to 29.27 0C. What is E for the reaction ?

• Solution: Q = M x C x T, mass = (2000 + 200) grams C* for water = 4.18 joules / gram x degree Celsius T = (29.27 – 26.13) = 3.14 0C Q (heat lost from reaction) = 2200 x 4.18 x 3.14 = 28875 j E (heat of reaction) = 28.875 KJ / 0.1567 moles = 184.4 KJ/moleSince heat is lost by the reaction it is exothermic and E = - 184.4 KJ/m*specific heat varies depending on the substance. C for water is a value

required for many problems and is 4.18 joules / gram x degree Celsius

CONSTANT VOLUME VS CONSTANT PRESSURE SYSTEMS

• IN CONSTANT VOLUME SYSTEMS ALL ADDED OR REMOVED HEAT ENERGY ALTERS THE INTERNAL ENERGY OF THE SYSTEM AND NONE IS USED TO DO WORK.

• WORK FROM GASEOUS EXPANSION OR COMPRESSION IS CALCULATED BY MULTILPYING PRESSURE TIMES THE RESULTING VOLUME CHANGE OF THE SYSTEM (W = P x V). IN A CONSTANT VOLUME SYSTEM SUCH AS THE BOMB CALORIMETER, VOLUME IS ZERO AND THEREFORE THE WORK DONE IS ZERO.

• THE FIRST LAW OF THERMODYNAMICS IN EQUATION FORM IS: E = Q + W, WHERE E IS THE INTERNAL ENERGY CHANGE, Q IS HEAT ADDED OR REMOVED AND W IN WORK DONE BY OR ON THE SYSTEM. IF W = 0 THEN E = Q AND ALL HEAT ADDED OR REMOVED RESULTS IN A INTERNAL ENERGY CHANGES.

CONSTANT VOLUME VS CONSTANT PRESSURE SYSTEMS

• CONSTANT PRESSURE SYSTEMS OCCUR COMMONLY IN LAB SITUATIONS WHERE REACTIONS ARE OPEN TO THE ATMOSPHERE (USUALLY ABOUT 1 ATM.)

• WHEN GASES EXPAND OR CONTRACT, THE WORK TERM IS EQUAL TO P x V.

• WHEN HEAT IS ADDED A SYSTEM, Q IS POSITIVE. WHEN HEAT IS RELEASED, Q IS NEGATIVE.

• WHEN WORK IS DONE BY A SYSTEM (GAS EXPANDS AND WORK LEAVES THE SYSTEM), W IS NEGATIVE.

WHEN WORK IS DONE ON A SYSTEM (SYSTEM IS COMPRESSED AND WORK IS ADDED TO THE SYSTEM), W IS POSITIVE.

• WHEN ENERGY IS RELEASED BY A REACTION, E = POSITIVE. WHEN ENERGY IS ASBORBED BY A REACTION, E = NEGATIVE.

CONSTANT PRESSUREE = q + w = q + P V

VOLUMECHANGE

(V)CONSTANTPRESSURE

AS GASEXPANDS

ADDED HEAT (+q)

W = - P VWORK DONE

BY THESYSTEM

MOVEABLEPISTON

CONSTANT VOLUME VS CONSTANT PRESSURE SYSTEMS

• SUMMARY:

HEAT ADDED, Q = +, HEAT LOST, Q = -WORK DONE ON SYSTEM, W = +

WORK DONE BY THE SYSTEM, W = -INTERNAL ENERGY INCREASES, E = +

INTERNAL ENERGY DECREASES, E = -• FOR CONSTANT PRESSURE REACTIONS THE HEAT THAT IS

GAINED OR LOST BY THE REACTION IS CALLED ENTHALPY CHANGE ( H)

AND H = E + W OR H = E + P V • IN MOST ALL REACTIONS AT ATMOSPHERIC PRESSURE,

THE WORK FACTOR (P V) IS VERY SMALL COMPARED TO THE INTERNAL ENERGY CHANGE ( E) AND THEREFORE IN MOST CASES H AND E ARE PRACTICALLY THE SAME VALUE.

CONSTANT VOLUME VS CONSTANT PRESSURE SYSTEMS

• For the decomposition of CaCO3 to CaO and CO2 what is the difference between E and H value at 25 0C and 1 atm.?

CaCO3(S) CaO(S) + CO2 (g)

• SOLUTION: H CAN BE CALCULATED USING HESSES LAW

H RXN = HPRODUCTS - HREACTANTS

H RXN = ((-635.5) + (-394)) - (-1207) = + 571 KJ H = E + P V, P V = n RT, THEREFORE H = E + n RT AND E = H - n RT n = MOLES OF GASEOUS PRODUCTS – MOLES OF GASEOUS

REACTANTS (n = 1 –0) = 1 E = (+571) - ((1) x 0.00831 x (25 + 273)) = +571 – 2.48 = 569 KJ

THE DIFFERENCE BETWEEN H AND E IS ONLY 2.48 K OR ONLY 0.43 %

CAUSES OF SPONTANEOUS CHANGE

• FROM AN ENERGY STANDPOINT, REACTIONS ALWAYS TEND TO MOVE IN THE DIRECTION OF LOWEST ENERGY THAT IS EXOTHERMIC (ENERGY RELEASING) REACTONS ARE FAVORED. HOWEVER, THERE ARE NUMEROUS INSTANCES OF SPONTANEOUS ENDOTHERMIC (ENERGY ABSORBING) REACTIONS. HOW CAN THIS BE EXPLAINED?

• THE ANSWER LIES IN THE SECOND LAW OF THERMODYNAMICS, A CONCEPT KNOWN AS ENTROPY.

• ENTROPY IS ALSO REFERRED TO AS “RANDOMNESS” OR “DISORDER”. THE SECOND LAW STATES THAT ALL PROCESSES HAVE A NATURAL TENDENCY TO PROGRESS TO STATES OF INCREASED ENTROPY OR DISORDER. IN THIS PRINCIPLE LIES THE ANSWER TO SPONTANEOUS ENDOTHERMIC REACTIONS.

ENTROPY – A NATURAL TENDENCY FOR ALL CHANGE

AN UNLIKELY EVENT

SPONTANEOUSDECREASE IN

ENTROPY

A VERY LIKELY EVENT

SPONTANEOUSINCREASE IN

ENTROPY

COINS SWEPTFROM A TABLE

ENTROPY AND THE NATURAL DIRECTION OF CHANGE

• THE NATURAL DIRECTION OF ALL CHANGE IS TOWARDS INCREASING ENTROPY. THE STATE OF MATTER WITH THE GREATEST ENTROPY IS GAS. INTERMEDIATE ENTROPY IS THAT OF LIQUIDS AND THE LEAST DISORDERED STATE IS SOLID (THE KINETIC MOLECULAR THEORY JUSTIFIES THESE OBSERVATIONS).

• THE ONLY SUBSTANCE HAVING A PERFECTLY ORDERED STATE (ZERO ENTROPY) IS A PERFECT CRYSTAL AT ABSOLUTE ZERO (THIS IS THE THIRD LAW OF THERMODYNAMICS). THESE CONDITIONS CAN NEVER BE MET, THEREFORE ALL SUBSTANCES HAVE SOME INNATE ENTROPY OR RANDOMNESS.

• FROM AN ENTROPY STANDPOINT, REACTIONS TEND TO PROGRESS TOWARDS THE FORMATION OF GASES FROM LIQUIDS AND SOLIDS AND ALSO TEND TOWARDS DECOMPOSITION, (FORMING MORE MOLECULES FROM FEWER) THEREBY BEING CAPABLE OF AN EVEN GREATER DISORDERED CONDITION.

MEASURING ENTROPY• ENTROPY IS SYMBOLIZED WITH S. ENTROPY CHANGE

CAN BE DEFINED AS THE HEAT ABSORBED BY A SYSTEM DIVIDED BY THE KELVIN TEMPERATURE AT WHICH THE HEAT IS ADDED.

S = Q / T (JOULES/ KELVIN)• THE CHANGE IN ENTROPY FOR A CHEMICAL REACTION

CAN BE CALCULATED BY:

S = SPRODUCTS - S REACTANTS

• ENTROPY VALUES FOR REACTANTS AND PRODUCTS CAN BE OBTAINED FROM THERMODYNAMIC TABLES.

• AVAILABLE ENTROPY VALUES ARE REFERRED AS STANDARD VALUES (THOSE MEASURED AT 25 0C AND 1 ATM GAS PRESSURES AND 1 MOLAR SOLUTION CONCENTRATIONS. THE SYMBOL FOR THESE CONDITIONS IS S0, THE SUPERSCRIPT 0 INDICATING STANDARD STATE CONDITIONS.)

CALCULATING S0 FOR A REACTION

• Problem: What is the standard entropy change for urea reacting with water to form CO2 and ammonia? CO(NH2)2(aq) + H2O(l) CO2(g) + 2 NH3(g)

• S0 CO(NH2)2(aq) = 175.8 J/ mol K

S0 H2O(l) = 69.96 J/ mol K

S0 CO2(g) = 213.6 J/ mol K

S0 NH3(g) = 192.5 J/ mol K S0 = (213.6 + 2(192.5)) – (173.8 + 69.96) = 354.8 J/K• SINCE THE REACTION PRODUCED THREE GAS

MOLECULES FROM ONE AQUEOUS MOLECULE AND ONE LIQUID MOLECULE, WE EXPECT A LARGE INCREASE IN ENTROPY AND +354.8 J/K IS A LARGE ENTROPY INCREASE!

ENTHALPY & ENTROPY AND REACTION SPONTANEITY

• WHEN SYSTEMS REACT AND ENTROPY INCREASES (A NATURAL TENDENCY BASED ON THE SECOND LAW) ENERGY IS ASBORBED. ACCORDING TO THE FIRST LAW REACTIONS TEND TOWARDS LOWEST ENERGY STATES, THAT THEY TEND TO RELEASE ENERGY. HOW CAN WE TELL WHAT WILL REALLY OCCUR WHEN THESE TWO OPPOSING TENDENCIES DRIVE A REACTION?

• THE ANSWER IS FOUND IN THE GIBBS - HELMHOTZ EQUATION. IT RELATES THE ENTHALPY FACTOR AND THE ENTROPY FACTOR, BOTH WHICH DRIVE REACTION SPONTANIETY. RECONCILING THESE DRIVING FORCES RESULTS IN A TERM CALLED GIBBS FREE ENERGY CHANGE (G)

• THIS TERM ALLOWS US TO DECIDE ON THE NET EFFECT OF ENTROPY AND ENTHALPY ON A CHEMICAL REACTION AND PREDICT ITS SPONTANEITY.

GIBBS FREE ENERGY & REACTION SPONTANIETY

• THE GIBBS – HELMHOTZ EQUATION IS:• G = H - T S • G = GIBBS FREE ENERGY (KJ/MOLE) H = ENTHALPY CHANGE (KJ/MOLE) S = ENTROPY CHANGE (JOULES/MOLE) T = KELVIN TEMPERATURE (ALWAYS +)• IF S FOR A REACTION IS (+) THE REACTION IS FAVORED BY

ENTROPY INCREASING.• IF H IS (–) THE REACTION IS FAVORED BY ENTHALPY

DECREASING.• G = (-) - (+) (+) = (-) , SINCE BOTH ENTROPY AND ENTHALPY

DRIVE THE REACTION TOWARD SPONTANEITY THE REACTION MUST BE SPONTANEOUS FORWARD AND THUS WE CAN CONCLUDE THAT A NEGATIVE G INDICATES A SPONTANEOUS FORWARD REACTION.

GIBBS FREE ENERGY & REACTION SPONTANIETY

• IF THE SIGN OF H = + THE REACTION IS NOT FAVORED BY ENTHALPY CHANGES.

• IF THE SIGN OF S = - THE REACTION IS NOT FAVORED BY ENTROPY CHANGES.

• THE REACTION CANNOT BE SPONTANEOUS SINCE BOTH ENTROPY AND ENTHALPY OPPOSE REACTION SPONTANEITY.

• G = (+) - (+) (-) = (-) , A NEGATIVE G THEREFORE INDICATES A NONSPONTANEOUS REACTION FORWARD (SPONTANEOUS IN REVERSE).

• IF G = 0 THEN THE REACTION IS EQUALLY SPONTANEOUS BOTH FORWARD AND REVERSE AND THEREFORE IS AT EQUILIBRIUM.

SUMMARY OF G INDICATORS

• G = + (FORWARD REACTION IS SPONTANEOUS)• G = - (FORWARD REACTION IS NONSPONTANEOUS)• G = 0 ( REACTION IS AT EQUILIBRIUM)• G0 MAY BE CALCULATED IN TWO WAYS. (1) G = H - T S

* (2) G0 = G0 PRODUCTS - G0 REACTANTS

* G0 AS WELL AS H0 AND S0 VALUES CAN BE OBTAINED FROM STANDARD THERMODYNAMIC TABLES. ALSO AN IMPORTANT NOTE IS THAT H AND S VALUE ARE RELATIVELY TEMPERATURE INDEPENDENT MEANING THAT THEY ARE CONSIDERED TO BETHE SAME AT ALL TEMPERATURES. G HOWEVER VARIES GREATLY WITH TEMPERATURE SINCE TEMPERATURE IS USED IN ITS CALCULATION (G = H - T S )

USING G VALUES• PROBLEM: Is the reaction:

N2 (g) + O2 (g) 2 NO (g) Spontaneous at 25 0C with 1 atm pressures?

Given data: H0 for NO(g) = 21.57 Kcal/mole

S0 for NO(g) = 50.347 cal/K, S0 for N2(g) =45.77 cal/K

S0 for O2(g) = 49.003 cal/K

SOLUTION: G0 = H0 - T S0

S0 = SPRODUCTS - S REACTANTS

S0 = (2 x 50.347) – (45.77 + 49.003) = 5.92 cal/K H0 = HPRODUCTS - HREACTANTS

H0 = (2 x 21.57) – ( 0 + 0) = 43.14 Kcal/mole G0 = 43.14 - (298)(0.00592) = + 41.38 Kcal/mole G0 is + and the reaction is not spontaneous at standard conditions!

DECIDING ON REACTION SPONTANEITY WITHOUT CALCULATION

• PROBLEM: If H = -46.9 Kcal for the reaction:

2 H2O2 (l) 2 H2O (l) + O2 (g) is the reaction spontaneous?

SOLUTION: Since H is temperature independent the reaction is exothermic at all temperatures and is favored by lower energy (First Law).

Two liquid molecules convert to two liquid and a gas molecule and is therefore favored by increasing entropy(Second Law)

Both lower enthalpy (H is negative) and increasing entropy (S is positive) favor the forward reaction and the reaction is sure to be spontaneous at all temperatures.

DECIDING ON REACTION SPONTANEITY WITHOUT CALCULATION

• THE SIGNS OF H, S AND G ARE CAN BE USED TO DECIDE ON THE SPONTANEITY OF REACTIONS AS WELL AS THE TEMPERATURE DEPENDENCE OF A REACTION.(* KELVIN TEMPERATURE IS ALWAYS +)

• H *TK S G RXN TYPE

• - + + - SPONT AT ALL TEMPS • + + - + NON SPONT AT ALL TEMPS • - + - + / - TEMP , SPONT • + + + + / - TEMP , SPONT

FREE ENERGY AND EQUILIBRIUM

A

E

BG

FREE

ENERGY

N2O4(g) 2 NO2(g)

MOLES OF N2O4 REACTED

0 .5 1.0

( G) SLOPE = 0 (Equilibrium)

Starting with 1 mole of N2O4

A = ALL N2O4

E = EQUILIBRIUMB =ALL NO2

FREE ENERGY AND EQUILIBRIUM

• The equation which relates G and equilibrium as derived from the previous graph: G = G0 + RT ln Q

G = free energy at any point during reaction

G0 = standard free energy (298 K, 1atm, 1molar)

R = 8.31 joules / mole K

T = Kelvin temperature

Q = [products] / [reactants] at a point during reaction

• At equilibrium, G =0 and Q = Ke therefore:

0 = G0 + RT ln Ke

G0 = - RT ln Ke

FREE ENERGY AND EQUILIBRIUM• Problem: Find Kp for the reaction:

2 SO2(g) + O2(g) 2 SO3(g) ,

G0 (SO2(g)) = -300KJ/mol,

G0 (SO3(g)) = -370 KJ/mol• Solution: G0 = G0 PRODUCTS - G0 REACTANTS

G0 = (2 x –370) - (2 x – 300) = -140 KJ

G0 = - RT ln Kp,, ln Kp = - G0 / RT

ln Kp = -( -140 x 103)/ (8.31 x 298) = 56.55

Kp = 3.6 x 1024 *

• *A large negative G0 corresponds to at large equilibrium constant. Both indicate a highly spontaneous reaction in the forward direction.

SUMMARY OF EQUATIONS

• (1) Calculation of Heat Content

Q = M x C x T

• (2) First Law of Thermodynamics

E = Q + W

• (3) Constant Pressure (Enthalpy Change)

H = E + P V

• (4) Constant Pressure (Enthalpy Change of a Reaction)

H = E + n RT

• (5) Constant Volume (Internal Energy Change)

E = Q

• (6) Hesses Law

H RXN = HPRODUCTS - HREACTANTS

SUMMARY OF EQUATIONS

• (7) Entropy Change for a Reaction

S = SPRODUCTS - S REACTANTS

• (8) Gibbs Helmhotz Equation G = H - T S • (9) Standard Free Energy Calculation

G0 = G0 PRODUCTS - G0 REACTANTS

• (10) Standard Free Energy from the Gibbs Equation G0 = H0 - T S0

• (11) Free Energy during a Reaction G = G0 + RT ln Q• (12) Standard Free Energy and the Equilibrium Constant

G0 = - RT ln Ke