Chemical Thermodynamics

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CHEMICAL THERMODYNAMICS WHY SOME REACTIONS HAPPEN AND SOME DON’T ! Copyright Sautter 2003

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The next slide is a quick promo for my books after which the presentation will begin Thanks for your patience! Walt S. Wsautter@optonline.net More stuff at: www.wsautter.com

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CHEMICAL THERMODYNAMICS: 

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: 

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: 

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: 

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: 

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: 

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.

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IGNITER POWER SUPPLY THERMOMETER STIRRER REACTION CHAMBER INSULATED CONTAINER WATER CHAMBER BOMB CALORIMETER (CONSTANT VOLUME) HEAT FLOWS OUT OF THE REACTION CHAMBER INTO THE SURROUNDING WATER

CHEMICAL THERMODYNAMICS: 

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 0 C. 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 0 C 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/mole Since 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: 

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 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. *

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CONSTANT PRESSURE  E = q + w = q + P  V VOLUME CHANGE (  V) CONSTANT PRESSURE AS GAS EXPANDS ADDED HEAT (+q) W = - P  V WORK DONE BY THE SYSTEM MOVEABLE PISTON *

CONSTANT VOLUME VS CONSTANT PRESSURE SYSTEMS: 

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: 

CONSTANT VOLUME VS CONSTANT PRESSURE SYSTEMS For the decomposition of CaCO 3 to CaO and CO 2 what is the difference between  E and  H value at 25 0 C and 1 atm.? CaCO 3(S)  CaO (S) + CO 2 (g) SOLUTION:  H CAN BE CALCULATED USING HESSES LAW  H RXN =   H PRODUCTS -   H REACTANTS  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: 

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.

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ENTROPY – A NATURAL TENDENCY FOR ALL CHANGE AN UNLIKELY EVENT SPONTANEOUS DECREASE IN ENTROPY A VERY LIKELY EVENT SPONTANEOUS INCREASE IN ENTROPY COINS SWEPT FROM A TABLE

ENTROPY AND THE NATURAL DIRECTION OF CHANGE: 

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: 

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 =  S PRODUCTS -  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 0 C AND 1 ATM GAS PRESSURES AND 1 MOLAR SOLUTION CONCENTRATIONS. THE SYMBOL FOR THESE CONDITIONS IS S 0 , THE SUPERSCRIPT 0 INDICATING STANDARD STATE CONDITIONS.)

CALCULATING  S0 FOR A REACTION: 

CALCULATING  S 0 FOR A REACTION Problem: What is the standard entropy change for urea reacting with water to form CO 2 and ammonia? CO(NH 2 ) 2(aq) + H 2 O (l)  CO 2(g) + 2 NH 3(g) S 0 CO(NH 2 ) 2(aq) = 175.8 J/ mol K S 0 H 2 O (l) = 69.96 J/ mol K S 0 CO 2(g) = 213.6 J/ mol K S 0 NH 3(g) = 192.5 J/ mol K  S 0 = (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: 

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: 

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: 

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: 

SUMMARY OF  G INDICATORS  G = + (FORWARD REACTION IS SPONTANEOUS)  G = - (FORWARD REACTION IS NONSPONTANEOUS)  G = 0 ( REACTION IS AT EQUILIBRIUM)  G 0 MAY BE CALCULATED IN TWO WAYS. (1)  G =  H - T  S * (2)  G 0 =   G 0 PRODUCTS -   G 0 REACTANTS *  G 0 AS WELL AS  H 0 AND S 0 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: 

USING  G VALUES PROBLEM : Is the reaction: N 2 (g) + O 2 (g)  2 NO (g) Spontaneous at 25 0 C with 1 atm pressures? Given data:  H 0 for NO (g) = 21.57 Kcal/mole S 0 for NO (g) = 50.347 cal/K, S 0 for N 2(g) =45.77 cal/K S 0 for O 2(g) = 49.003 cal/K SOLUTION :  G 0 =  H 0 - T  S 0  S 0 =  S PRODUCTS -  S REACTANTS S 0 = (2 x 50.347) – (45.77 + 49.003) = 5.92 cal/K H 0 =   H PRODUCTS -   H REACTANTS H 0 = (2 x 21.57) – ( 0 + 0) = 43.14 Kcal/mole G 0 = 43.14 - (298)(0.00592) = + 41.38 Kcal/mole G 0 is + and the reaction is not spontaneous at standard conditions!

DECIDING ON REACTION SPONTANEITY WITHOUT CALCULATION: 

DECIDING ON REACTION SPONTANEITY WITHOUT CALCULATION PROBLEM: If  H = -46.9 Kcal for the reaction : 2 H 2 O 2 (l)  2 H 2 O (l) + O 2 (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: 

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 * T K  S  G RXN TYPE - + + - SPONT AT ALL TEMPS + + - + NON SPONT AT ALL TEMPS - + - + / - TEMP , SPONT + + + + / - TEMP , SPONT

FREE ENERGY AND EQUILIBRIUM: 

FREE ENERGY AND EQUILIBRIUM A E B G F R E E E N E R G Y N 2 O 4(g)  2 NO 2(g) MOLES OF N 2 O 4 REACTED 0 .5 1.0 (  G) SLOPE = 0 (Equilibrium) Starting with 1 mole of N 2 O 4 A = ALL N 2 O 4 E = EQUILIBRIUM B =ALL NO 2

FREE ENERGY AND EQUILIBRIUM: 

FREE ENERGY AND EQUILIBRIUM The equation which relates  G and equilibrium as derived from the previous graph:  G =  G 0 + RT ln Q  G = free energy at any point during reaction  G 0 = 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 = K e therefore: 0 =  G 0 + RT ln K e  G 0 = - RT ln K e *

FREE ENERGY AND EQUILIBRIUM: 

FREE ENERGY AND EQUILIBRIUM Problem: Find K p for the reaction: 2 SO 2(g) + O 2(g)  2 SO 3(g) ,  G 0 (SO 2(g) ) = -300KJ/mol,  G 0 (SO 3(g) ) = -370 KJ/mol Solution:  G 0 =   G 0 PRODUCTS -   G 0 REACTANTS  G 0 = (2 x –370) - (2 x – 300) = -140 KJ  G 0 = - RT ln K p, , ln K p = -  G 0 / RT ln K p = -( -140 x 10 3 )/ (8.31 x 298) = 56.55 K p = 3.6 x 10 24 * *A large negative  G 0 corresponds to at large equilibrium constant. Both indicate a highly spontaneous reaction in the forward direction. *

SUMMARY OF EQUATIONS: 

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 =   H PRODUCTS -   H REACTANTS

SUMMARY OF EQUATIONS: 

SUMMARY OF EQUATIONS (7) Entropy Change for a Reaction  S =  S PRODUCTS -  S REACTANTS (8) Gibbs Helmhotz Equation  G =  H - T  S (9) Standard Free Energy Calculation  G 0 =   G 0 PRODUCTS -   G 0 REACTANTS (10) Standard Free Energy from the Gibbs Equation  G 0 =  H 0 - T  S 0 (11) Free Energy during a Reaction  G =  G 0 + RT ln Q (12) Standard Free Energy and the Equilibrium Constant  G 0 = - RT ln K e

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THE END