Chemical reactions involve changes in energy

– Breaking bonds requires energy – Forming bonds releases energy

These energy changes can be in the form of heat

– Heat is the flow of chemical energy The study of the changes in energy in

chemical reactions is called thermochemistry.

The energy involved in chemistry is real and generally a measurable value

Chemical reactions involve changes in energy

– Breaking bonds requires energy – Forming bonds releases energy

These energy changes can be in the form of heat

– Heat is the flow of chemical energy The study of the changes in energy in

chemical reactions is called thermochemistry.

The energy involved in chemistry is real and generally a measurable value

INTRO TO THERMOCHEMISTRYINTRO TO THERMOCHEMISTRY

WHAT IS HEAT?WHAT IS HEAT?Hot & cold, are automatically

associated with the words heat and temperature

– Heat & temp are NOT synonyms– The temperature of a substance is

directly related to the energy of its particles, specifically its:

Hot & cold, are automatically associated with the words heat and temperature

– Heat & temp are NOT synonyms– The temperature of a substance is

directly related to the energy of its particles, specifically its:

The Kinetic Energy defines the temperature

– Particles vibrating fast = hot– Particles vibrating slow = cold

The Kinetic Energy defines the temperature

– Particles vibrating fast = hot– Particles vibrating slow = cold

Vibrational energy is transferred from one particle to the next

– One particle collides with the next particle and so on; and so on – down the line

Vibrational energy is transferred from one particle to the next

– One particle collides with the next particle and so on; and so on – down the line

An Ice Cold Spoon A Hot Spoon

2 Hot Spoons2 Hot Spoons

Thermal energy is the total energy of all the particles that make up a substance

– Kinetic energy from vibration of particles– Potential energy from molecular attraction

(within or between the particles) Thermal energy is dependent upon the

amount or mass of material present (KE =½mv2)

Thermal energy is the total energy of all the particles that make up a substance

– Kinetic energy from vibration of particles– Potential energy from molecular attraction

(within or between the particles) Thermal energy is dependent upon the

amount or mass of material present (KE =½mv2)

Thermal energy is also related to the type of material

Thermal energy is also related to the type of material

Different type of materials– May have the same temp, same mass,

but different connectivity– Affected by the potential energy or

the intermolecular forces So it is possible to be at same temp

(same KE) but have very different thermal energies

The different abilities to hold onto or release energy is referred to as the substance’s heat capacity

Different type of materials– May have the same temp, same mass,

but different connectivity– Affected by the potential energy or

the intermolecular forces So it is possible to be at same temp

(same KE) but have very different thermal energies

The different abilities to hold onto or release energy is referred to as the substance’s heat capacity

Thermal energy can be transferred from object to object through direct contact– Molecules collide, transferring energy

from molecule to molecule

Thermal energy can be transferred from object to object through direct contact– Molecules collide, transferring energy

from molecule to molecule

DEFINITION

THE FLOW OF THERMAL ENERGY FROM SOMETHING WITH A

HIGHER TEMP TO SOMETHING WITH A LOWER TEMP

UNITS MEASURED IN JOULES OR CALORIES

TYPES

THROUGH WATER OR AIR = CONVECTION

THROUGH SOLIDS = CONDUCTION

TRANSFERRED ENERGY BY COLLISION WITH PHOTON =

RADIANT ENERGY

HEAT CAPACITYHEAT CAPACITY The measure of how well a material

absorbs or releases heat energy is its heat capacity– It can be thought of as a reservoir to hold heat, how much it holds before it overflows is its capacity

Heat capacity is a physical property unique to a particular material– Water takes 1 calorie of energy to raise temp 1 °C

– Steel takes only 0.1 calorie of energy to raise temp 1 °C

The measure of how well a material absorbs or releases heat energy is its heat capacity– It can be thought of as a reservoir to hold heat, how much it holds before it overflows is its capacity

Heat capacity is a physical property unique to a particular material– Water takes 1 calorie of energy to raise temp 1 °C

– Steel takes only 0.1 calorie of energy to raise temp 1 °C

SPECIFIC HEAT CAPACITYSPECIFIC HEAT CAPACITY The amount of energy it takes to raise

the temp of a standard amount of an object 1°C is that object’s specific heat capacity (Cp)

– The standard amount =1 gram Specific heats can be listed on data

tables– Smaller the specific heat the less

energy it takes the substance to feel hot– Larger the specific heat the more

energy it takes to heat a substance up (bigger the heat reservoir)

The amount of energy it takes to raise the temp of a standard amount of an object 1°C is that object’s specific heat capacity (Cp)

– The standard amount =1 gram Specific heats can be listed on data

tables– Smaller the specific heat the less

energy it takes the substance to feel hot– Larger the specific heat the more

energy it takes to heat a substance up (bigger the heat reservoir)

SUBSTANCESUBSTANCE SPECIFIC HEAT SPECIFIC HEAT CAPACITY, CCAPACITY, CPP

WATER, HWATER, H22OO 4.184.18J/g°C OR J/g°C OR 11cal/g°Ccal/g°C

ALUMINUM, ALUMINUM, AlAl

.992.992J/g°C J/g°C OR OR .237.237cal/g°Ccal/g°C

TABLE SALT, TABLE SALT, NaClNaCl

.865 .865 J/g°C J/g°C OR OR .207.207cal/g°Ccal/g°C

SILVER, AgSILVER, Ag .235 .235 J/g°C J/g°C OR OR .056.056cal/g°Ccal/g°C

MERCURY, MERCURY, HgHg

.139 .139 J/g°C J/g°C OR OR .033.033cal/g°Ccal/g°C

Specific heats and heat capacities work for gains in heat and in losses in heat– Smaller the specific heat the less

time it takes the substance to cool off– Larger the specific heat the longer

time it takes the substance to cool off Specific heat capacity values are used

to calculate changes in energy for chemical reactions– It’s important for chemists to know

how much energy is needed or produced in chemical reactions

Specific heats and heat capacities work for gains in heat and in losses in heat– Smaller the specific heat the less

time it takes the substance to cool off– Larger the specific heat the longer

time it takes the substance to cool off Specific heat capacity values are used

to calculate changes in energy for chemical reactions– It’s important for chemists to know

how much energy is needed or produced in chemical reactions

SPECIFIC HEAT CAPACITYSPECIFIC HEAT CAPACITY

CHEMICAL REACTIONSCHEMICAL REACTIONS There are 2 types of chemical reactions

– Exothermic reactions reactions in which heat energy is a product

– Endothermic reactions reactions in which heat energy is a reactant

Exothermic reactions typically feel warm as the reaction proceeds– Gives off heat energy, sometimes

quite allot Endothermic reactions typically feel

cooler the longer the reaction proceeds– Absorbs heat energy, sometimes

enough to get very cold

There are 2 types of chemical reactions– Exothermic reactions reactions in

which heat energy is a product– Endothermic reactions reactions in

which heat energy is a reactant Exothermic reactions typically feel warm

as the reaction proceeds– Gives off heat energy, sometimes

quite allot Endothermic reactions typically feel

cooler the longer the reaction proceeds– Absorbs heat energy, sometimes

enough to get very cold

C3H8C3H8 ++ 5O25O2 2043kJ 2043kJ 3CO23CO2 4H2O4H2O++ ++

Exothermic reaction Exothermic reaction

– To a cold camper, the important product here is the heat energy

– To a cold camper, the important product here is the heat energy

NH4NO3+H2O+ 752kJ NH4OH+HNO3

NH4NO3+H2O+ 752kJ NH4OH+HNO3

Endothermic reaction Endothermic reaction

– Similar system as what is found in cold packs

– Similar system as what is found in cold packs

CHANGE IN HEAT ENERGY (ENTHALPY)CHANGE IN HEAT ENERGY (ENTHALPY) The energy used or produced in a

chemical reaction is called the enthalpy of the reaction– Burning a 15 gram piece of paper

produces a particular amount of heat energy or a particular amount of enthalpy

Enthalpy is a value that also contains a component of direction (energy in or energy out)– Heat gained is the out-of

direction; ie exo-

The energy used or produced in a chemical reaction is called the enthalpy of the reaction– Burning a 15 gram piece of paper

produces a particular amount of heat energy or a particular amount of enthalpy

Enthalpy is a value that also contains a component of direction (energy in or energy out)– Heat gained is the out-of

direction; ie exo-

– Heat lost is the into direction; ie endo-

– Heat lost is the into direction; ie endo-

CHANGE IN HEAT ENERGY (ENTHALPY)CHANGE IN HEAT ENERGY (ENTHALPY) The energy used or produced in a chem

rxn is called the enthalpy of the reaction– Burning a 15 gram piece of paper

produces a particular amount of heat energy or a particular amount of enthalpy

Enthalpy is a value that also contains a component of direction (energy in or energy out)– Heat gained is the out-of

direction; ie exo-

The energy used or produced in a chem rxn is called the enthalpy of the reaction– Burning a 15 gram piece of paper

produces a particular amount of heat energy or a particular amount of enthalpy

Enthalpy is a value that also contains a component of direction (energy in or energy out)– Heat gained is the out-of

direction; ie exo-

HEATHEATHEATHEAT HEATHEATHEATHEAT HEATHEATHEATHEAT HEATHEATHEATHEAT

Most common version of enthalpy is when we have a change in enthalpy (H)

The enthalpy absorbed or gained (changed) in a reaction is dependent on the amount of material reacting

– Amount is usually in the form of moles– We can use the coefficient ratios to

energy ratios to calculate how much energy a reaction used or produced

Most common version of enthalpy is when we have a change in enthalpy (H)

The enthalpy absorbed or gained (changed) in a reaction is dependent on the amount of material reacting

– Amount is usually in the form of moles– We can use the coefficient ratios to

energy ratios to calculate how much energy a reaction used or produced

CHANGE IN ENTHALPYCHANGE IN ENTHALPY

(For Example)How much heat will be released if 1.0g of (H2O2) decomposes in a bombardier beetle to produce

a defensive spray of steam

(For Example)How much heat will be released if 1.0g of (H2O2) decomposes in a bombardier beetle to produce

a defensive spray of steam

2H2O2 2H2O + O2 Hº = -190kJ

2H2O2 2H2O + O2 Hº = -190kJ

USING H IN CALCULATIONSUSING H IN CALCULATIONS Chemical reaction equations are

very powerful tools. – Given a reaction equation with an

energy value, We can calculate the amount of energy produced or used for any given amount of reactants.

Chemical reaction equations are very powerful tools. – Given a reaction equation with an

energy value, We can calculate the amount of energy produced or used for any given amount of reactants.

Analyze: we know that if we had 2 mols of H2O2 decomposing we would produce 190kJ of heat, but how much would it be if only 1.0 g of H2O2

Analyze: we know that if we had 2 mols of H2O2 decomposing we would produce 190kJ of heat, but how much would it be if only 1.0 g of H2O2

Therefore: we have to convert our given 1.0 g of H2O2 to moles of H2O2

Therefore: we have to convert our given 1.0 g of H2O2 to moles of H2O2

1.0g H1.0g H22OO22

1molH1molH22OO22

34gH34gH22OO22

2H2O2 2H2O + O2 Hº = -190kJ

2H2O2 2H2O + O2 Hº = -190kJ

= .02941 mol= .02941 mol= .02941 mol= .02941 mol

Molar massMolar massMolar massMolar mass

Therefore: with 2 moles of H2O2 we would produce 190 kJ of energy, but we don’t have 2 moles we only have .02941 mols of H2O2, so how much energy would the bug produce?

Therefore: with 2 moles of H2O2 we would produce 190 kJ of energy, but we don’t have 2 moles we only have .02941 mols of H2O2, so how much energy would the bug produce?

-190kJ-190kJ

2molH2molH22OO22

= -2.8kJ= -2.8kJ= -2.8kJ= -2.8kJ.02941 mol.02941 mol.02941 mol.02941 mol

Reaction equationReaction equationReaction equationReaction equation

2H2O2 2H2O + O2 Hº = -190kJ

2H2O2 2H2O + O2 Hº = -190kJ

How much heat will be released when 4.77 g of ethanol (C2H5OH) react with excess O2 according to the following

equation:

C2H5OH + 3O2 2CO2 + 3H2O Hº=-1366.7kJ

How much heat will be released when 4.77 g of ethanol (C2H5OH) react with excess O2 according to the following

equation:

C2H5OH + 3O2 2CO2 + 3H2O Hº=-1366.7kJ

Example #2Example #2

analyze: we know that if we had 1 mol of ethanol (assuming coefficient of 1 in rxn equation) burning we would produce 1366.7kJ of heat, but how much would it be if only we only had 4.77 g of ethanol?

analyze: we know that if we had 1 mol of ethanol (assuming coefficient of 1 in rxn equation) burning we would produce 1366.7kJ of heat, but how much would it be if only we only had 4.77 g of ethanol?

C2H5OH + 3O2 2CO2 + 3H2O Hº=-1366.7kJ

C2H5OH + 3O2 2CO2 + 3H2O Hº=-1366.7kJ

4.77g C4.77g C22HH55OHOH1mol C1mol C22HH55OHOH

46g C46g C22HH55OHOH= .1037 mol= .1037 mol= .1037 mol= .1037 mol

Therefore: with 1 mole of C2H5OH we would produce 1366.7 kJ of energy, but we don’t have 1 mole we only have .1037 mols of C2H5OH, so how much energy would the reaction produce?

Therefore: with 1 mole of C2H5OH we would produce 1366.7 kJ of energy, but we don’t have 1 mole we only have .1037 mols of C2H5OH, so how much energy would the reaction produce?

-1366.7kJ-1366.7kJ

1mol C1mol C22HH55OHOH= -142 kJ= -142 kJ= -142 kJ= -142 kJ.1037 .1037

molmol.1037 .1037

molmol

H =H = FINAL TEMP – INITIAL TEMPFINAL TEMP – INITIAL TEMP

SPECIFICHEAT

SPECIFICHEATMASSMASS

We can also track energy changes due to temp changes, using H=mCT:

We can also track energy changes due to temp changes, using H=mCT:

If the temp difference is positive– The reaction is exothermic because the

final temp is greater than the initial temp

– So the enthalpy is positive

If the temp difference is positive– The reaction is exothermic because the

final temp is greater than the initial temp

– So the enthalpy is positive if the temp change is negative– makes the enthalpy negative– the reaction absorbed heat into the

system, so it’s endothermic

if the temp change is negative– makes the enthalpy negative– the reaction absorbed heat into the

system, so it’s endothermic

If you drink 4 glasses of ice water at 0°C, how much heat energy is

transferred as this water is brought to body temperature?

Each glass contains 250 g of water & body temperature is 37°C.

If you drink 4 glasses of ice water at 0°C, how much heat energy is

transferred as this water is brought to body temperature?

Each glass contains 250 g of water & body temperature is 37°C.

mass of 4 glasses of water:– m = 4 x 250g = 1000g H2O

change in water temp:– Tf – Ti = 37°C - 0°C

specific heat of water:– C = 4.18 J/g•C° (from previous slide)

mass of 4 glasses of water:– m = 4 x 250g = 1000g H2O

change in water temp:– Tf – Ti = 37°C - 0°C

specific heat of water:– C = 4.18 J/g•C° (from previous slide)

H=mCTH=mCT H=(1000g)(4.18J/g•°C)(37°C)

H=(1000g)(4.18J/g•°C)(37°C)

H= 160,000JH= 160,000J

Enthalpy is dependent on the conditions of the reaction– It’s important to have a standard set of

conditions– This allow us to compare the affect of

temperatures, pressures, etc., has on different substances

Chemists have defined a standard set of conditions– Standard Temperature = 298K or 25°C– Standard Pressure = 1atm or 760mmHg

Enthalpy produced in a reaction under standard conditions is the standard enthalpy (H°)

Enthalpy is dependent on the conditions of the reaction– It’s important to have a standard set of

conditions– This allow us to compare the affect of

temperatures, pressures, etc., has on different substances

Chemists have defined a standard set of conditions– Standard Temperature = 298K or 25°C– Standard Pressure = 1atm or 760mmHg

Enthalpy produced in a reaction under standard conditions is the standard enthalpy (H°)

Standard enthalpies can be found on tables of data measured as standard enthalpies of formations

Standard enthalpies of formations are measured values for the energy to form chemical compounds (Hf°)

– H2 gas & O2 gas can be ignited to produce H2O and a bunch of energy

– The amount of energy produced by the reaction is 285kJ for every mol of water produced

Standard enthalpies can be found on tables of data measured as standard enthalpies of formations

Standard enthalpies of formations are measured values for the energy to form chemical compounds (Hf°)

– H2 gas & O2 gas can be ignited to produce H2O and a bunch of energy

– The amount of energy produced by the reaction is 285kJ for every mol of water produced

H2 + ½02 H2O

H2 + ½02 H2O

Hf°=-285.8kJ/mol

Hf°=-285.8kJ/mol

STANDARD ENTHALPIES OF FORMATION

SYMBOLSYMBOL FORMULASFORMULAS HHff°°kJ/kJ/molmol

AlClAlCl33(s)(s) Al + 3/2ClAl + 3/2Cl22 AlClAlCl33

-705.6-705.6

AlAl22OO33(s)(s) 2Al + 3/2O2Al + 3/2O22 AlAl22OO33

-1676.0-1676.0

COCO22(g)(g) C + OC + O22 CO CO22 -393.5-393.5

HH22O(g)O(g) HH22 + 1/2O + 1/2O22 HH22OO

-241.8-241.8

CC33HH88(g)(g) 3C + 4H3C + 4H22 C C33HH88 -104.7-104.7

CALORIMETRYCALORIMETRY Calorimetry is the process of measuring

heat energy – Measured using a device called a

calorimeter– Uses the heat absorbed by H2O to

measure the heat given off by a reaction or an object

The amount of heat soaked up by the water is equal to the amount of heat released by the reaction

Calorimetry is the process of measuring heat energy – Measured using a device called a

calorimeter– Uses the heat absorbed by H2O to

measure the heat given off by a reaction or an object

The amount of heat soaked up by the water is equal to the amount of heat released by the reaction

HSYS=-HSURHSYS=-HSUR

Hsys is the system or what is taking place in the main chamber (reaction etc.) And

Hsur is the surroundings which is generally water.

Hsys is the system or what is taking place in the main chamber (reaction etc.) And

Hsur is the surroundings which is generally water.

A COFFEE CUPCALORIMETERA COFFEE CUPCALORIMETER

A BOMB CALORIMETER

A BOMB CALORIMETERUSED WHEN TRYING

TO FIND THE AMOUNTOF HEAT PRODUCED BYBURNING SOMETHING.

USED WHEN TRYINGTO FIND THE AMOUNT

OF HEAT PRODUCED BYBURNING SOMETHING.

USED FOR A REACTIONIN WATER,

OR JUST A TRANSFEROF HEAT.

USED FOR A REACTIONIN WATER,

OR JUST A TRANSFEROF HEAT.

HSYSHSYS== -HSUR-HSUR

+ SIGN MEANS

HEAT WASABSORBED BY THE REACTION

+ SIGN MEANS

HEAT WASABSORBED BY THE REACTION

- SIGN MEANSHEAT WAS

RELEASED BYWATER

- SIGN MEANSHEAT WAS

RELEASED BYWATER

With calorimetry we use the sign of what happens to the water– When the water loses heat into the

system it obtains a (-) sign

With calorimetry we use the sign of what happens to the water– When the water loses heat into the

system it obtains a (-) sign

CALORIMETRYCALORIMETRY

HEATHEATHEATHEATHEATHEATHEATHEAT

HEATHEATHEATHEATHEATHEATHEATHEAT

You calculate the amount of heat absorbed by the water (using H= mCT)

Which leads to the amount of heat given off by the reaction– you know the mass of the water (by

weighing it)– you know the specific heat for water

(found on a table)– and you can measure the change in the

temp of water (using a thermometer)

You calculate the amount of heat absorbed by the water (using H= mCT)

Which leads to the amount of heat given off by the reaction– you know the mass of the water (by

weighing it)– you know the specific heat for water

(found on a table)– and you can measure the change in the

temp of water (using a thermometer)

CALORIMETRYCALORIMETRY

A block of Al that weighs 72.0g is heated to 100°C is dropped in a

calorimeter containing 120. mL of water at 16.6°C.

the H2O’s temp rises to 27°C.

A block of Al that weighs 72.0g is heated to 100°C is dropped in a

calorimeter containing 120. mL of water at 16.6°C.

the H2O’s temp rises to 27°C.- mass of Al = 72g

- Tinitial of Al = 100°C

- Tfinal of Al = 27°C

- CAl = .992J/g°C (from table)

- mass of Al = 72g

- Tinitial of Al = 100°C

- Tfinal of Al = 27°C

- CAl = .992J/g°C (from table)

HSYSHSYS

H =H = 72g72g .992J/g°C.992J/g°C 27°C-100°C27°C-100°C

HH == -5214J-5214J

We can do the same calc with the water information:

– Mass of H2O= 120g– Tinitial of H2O= 16.6°C– Tfinal of H2O = 27°C– CH2O= 4.18J/g°C (from table)

We can do the same calc with the water information:

– Mass of H2O= 120g– Tinitial of H2O= 16.6°C– Tfinal of H2O = 27°C– CH2O= 4.18J/g°C (from table)

HSURHSUR

HH==

5216J5216J

Equal but opposite, means that since the Al decreased in

temperature, it released heat causing the H2O to increase in

temperature.

Equal but opposite, means that since the Al decreased in

temperature, it released heat causing the H2O to increase in

temperature.

HH==

120g120g 4.18J/g°C 4.18J/g°C 27°C-16.6°C27°C-16.6°C

ThermochemistryThermochemistry

System - the part of the universe that is being studied

Surroundings - the rest of the universe

Boundary - the separation between the the system and surroundings

Systems may be: (1) open; (2) closed; or (3) isolated (adiabatic)

ThermochemistryThermochemistry

Laws of Thermodynamics

Zero Law - Two objects in contact will have the same temperature

1st Law - The energy of the universe is constant

2nd Law - The entropy of the universe is expanding

3rd Law - The entropy of a perfect crystalline substance is 0 at absolute zero (0 K)

The First Law of Thermodynamics

The total energy and mass of a system plus its surroundings remains constant.

∆E = q + w

Energy is measured in units of:Calorie (cal) or Joule* (j) where

1 calorie = 4.184 joules

*The SI unit is the Joule

Heat of reaction is measured in a calorimeter.

A bomb calorimeter An ordinary calorimeter constant volume

E (internal energy)constant pressure

H (enthalpy)

∆E ∆H

The enthalpy change of a reaction (∆H):

the heat absorbed or released during a reaction at constant

pressure.

A reaction that releases heat is exothermic, ∆H is negative.

A reaction that absorbs heat is endothermic, ∆H is positive.

The first law does not predict the direction

of a process!

The Second Law of Thermodynamics

Chemical and physical changes will take place in a direction to produce maximum

disorder in the system + surroundings.

The Second Law of Thermodynamics

Disorder orrandomness

Order

Entropy (S) is a quantitative measure of disorder.

The more positive is the value of S, the greater is the disorder.

(∆Ssystem + ∆Ssurroundings) > 0 for a spontaneous process

(∆Ssystem + ∆Ssurroundings) > 0 for a spontaneous process

Hard to measure ∆Ssystem + surroundings

Free Energy (G)

∆G° = ∆H° - T∆S°

∆G°: the change in free energy of a system ∆H°: the change in enthalpy of the system∆S°: the change in entropy of the system T: absolute temperature (in Kelvin units, K, oC+273)

For a spontaneous process, ∆G < 0.

More free energy(great work capacity)

Less free energy(less work capacity)

Spontaneous change

∆G < 0

battery

(1)∆G < 0: exothermic (exergonic) (there is a net loss of energy).

The “A to B” reaction is favorable and spontaneous.

(2)∆G > 0: endothermic (endergonic) (there is a net gain of energy)

The “A to B” reaction is unfavorable and not spontaneous.

(3)∆G = 0: the reaction is at equilibrium.

A B

∆G = GB - GA

The battery is dead!

∆G tells us how far the system is from equilibrium

∆G << 0 ∆G = 0 ∆G > 0∆G < 0

A B

A B∆G = GB-GA

The relationship of ∆G to Keq and the concentration of reactants and products

A + B C + D

R is the gas constant (1.987 cal/mol.degree). T is the absolute temperature (oK)

P=1M, R=1M (standard condition)

∆Go = - RT ln Keq = - 2.303 RT log Keq

∆Go is the standard free energy change

At the standard temperature of 298oK (25oC):

∆Go = -1.36 log Keq

(1) Keq >1, then ∆Go < 0

(2) Keq< 1, then ∆Go > 0

(3) Keq= 1, then ∆Go = 0

At pH 7, [H+] is 10-7M, the symbol ∆Go´ is used.

∆Go´ = -1.36 log K´eq

∆Go´

K´eq --------Kcal/mol kJ/mol

10-5 6.82 28.5310-4 5.46 22.8410-3 4.09 17.1110-2 2.73 11.4210-1 1.36 5.69 1 0 0

10 -1.36 -5.69102 -2.73 -11.42

103 -4.09 -17.11104 -5.46 -22.84

∆G = ∆Go + RT ln P

R

1. ∆G is predictive .

2. ∆G depends on Keq, and the concentrations of P and R.

Sometimes, ∆Go > 0, but if P/R is very small, then ∆G < 0.

Summary:

∆H, ∆S and ∆G

∆G = ∆H - T∆S

∆G < 0, R P

RP

∆G0

∆G0 = -RTlnKeq Keq>1, ∆G0<0

RP

∆G = ∆Go + RT ln PR

∆G < 0, R P