Entropy and Gibbs free energy

2

Exothermic

• The products are lower in energy than the reactants

• Releases energy• Often release heat

3

C + O2 CO2En

ergy

Reactants Products

C + O2

CO2

-395kJ

+ 395 kJ

When will a reaction be exothermicA) When breaking the bonds of the reactants

takes more energy than making the bonds of the products.

B) When breaking the bonds of the reactants takes less energy than making the bonds of the products

C) When you put in energy to break the bondsD) When you get energy by breaking bonds

5

Endothermic

• The products are higher in energy than the reactants

• Absorbs energy• Absorb heat

6

CaCO3 CaO + CO2En

ergy

Reactants Products

CaCO3

CaO + CO2

+176 kJ

CaCO3 + 176 kJ CaO + CO2

7

Heat of Reaction• The heat that is released or absorbed in a

chemical reaction• Equivalent to ΔH• C + O2(g) CO2(g) +393.5 kJ• C + O2(g) CO2(g) ΔH = -393.5 kJ• In thermochemical equation it is important to

say what state• H2(g) + ½ O2 (g) H2O(g) ΔH = -241.8 kJ• H2(g) + ½ O2 (g) H2O(l) ΔH = -285.8 kJ

8

Ener

gy

Reactants Products

Change is downΔH is <0

+ heat

9

Ener

gy

Reactants Products

Change is upΔH is > 0

Reactants + heat

Choose all that apply...

C(s) + 2 S(g) CS2(l) ΔH = 89.3 kJWhich of the following are true?A) This reaction is exothermicB) It could also be written

C(s) + 2 S(g) + 89.3 kJ CS2(l) C) The products have higher energy than the

reactantsD) It would make the water in the calorimeter

colder

11

Heat of Combustion• The heat from the reaction that completely

burns 1 mole of a substance at 25°C and 1 atm• C2H4 + 3 O2 2 CO2 + 2 H2O

• C2H6 + O2 CO2 + H2O

• 2 C2H6 + 7 O2 4 CO2 + 6 H2O

• C2H6 + (7/2) O2 2 CO2 + 3 H2O

• Always exothermic

Heat and phase change• Melting and vaporizing are endothermic

– Breaking things apart• Freezing and condensing are exothermic

– Forming connections

Heat of Fusion• Heat of fusion-ΔHfus- heat to melt one gram

• q = ΔHfus x m• For water 80 cal/g or 334 J/g• Same as heat of solidification• Book uses molar heat of fusion- heat to melt

one mole of solid• q = ΔHfus x n

Calculating Heat

• If there is a temperature change– q = m ΔT C

• If there is a phase change– q = ΔHfus x m or q = ΔHsolid x m

– q = ΔHvap x m or q = ΔHcond x m

• If there is both, do them separately and add.

Example

• Ammonia has a heat of fusion of 332 cal/g. How much heat to melt 15 g of ammonia?

This formula is for all change

• ΔH = ΣΔH°f (products) - ΣΔΗ°f(reactants)

)(reactantsH- (products)H = H of

of

17

Example• CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

ofH CH4 (g) = -74.86 kJofH O2(g) = 0 kJofH CO2(g) = -393.5 kJofH H2O(g) = -241.8 kJ

ΔH= [-393.5 kJ + 2(-241.8 kJ)] - [-74.86 kJ +2 (0 kJ )]

ΔH= -802.2 kJ

Ener

gy

Reactants Products

reactants

products

elements

products)(ofH

reactants)(ofH

exothermic reactants)(products)( of

of HH

Ener

gy

Reactants Products

reactantsproducts

elements

products)(ofH

reactants)(ofH

cendothermi products)(reactants)( of

of HH

Ener

gy

Reaction coordinate

Reactants

Products

Ener

gy

Reaction coordinate

Reactants

Products

Activation Energy - Minimum energy to make the reaction happen – how hard

Ener

gy

Reaction coordinate

Reactants

Products

Activated Complex or Transition State

Activation Energy

• Must be supplied to start the reaction• Low activation energy

– Lots of collision are hard enough– fast reaction

• High Activation energy– Few collisions hard enough– Slow reaction

Ener

gy

Reaction coordinate

Reactants

Products

Ener

gy

Reaction coordinate

Reactants

Products

Activation Energy - Minimum energy to make the reaction happen – how hard

Ener

gy

Reaction coordinate

Reactants

Products

Activated Complex or Transition State

Activation Energy

• Must be supplied to start the reaction• Low activation energy

– Lots of collision are hard enough– fast reaction

• High Activation energy– Few collisions hard enough– Slow reaction

Activation energy

• If reaction is endothermic you must keep supplying heat

• If it is exothermic it releases energy• That energy can be used to supply the

activation energy to those that follow

Ener

gy

Reaction coordinate

Reactants

Products

Overall energy change

Thermodynamics

Will a reaction happen?

Things that Affect Rate• Catalysts- substances that increase the rate of

a reaction without being used up.(enzyme).• Not a reactant nor a product.• Speeds up reaction by giving the reaction a

new path.• The new path has a lower activation energy.• More molecules have this energy.• The reaction goes faster.

Ener

gy

Reaction coordinate

Reactants

Products

Pt surface

HHHH

HH

HH

• Hydrogen bonds to surface of metal.

• Break H-H bonds

Catalysts

Pt surface

HH

HH

Catalysts

C HH CHH

Pt surface

HH

HH

Catalysts

C HH CHH

• The double bond breaks and bonds to the catalyst.

Pt surface

HH

HH

Catalysts

C HH CHH

• The hydrogen atoms bond with the carbon

Pt surface

H

Catalysts

C HH CHH

H HH

Energy

• Substances tend react to achieve the lowest energy state.

• Most chemical reactions are exothermic.• Doesn’t work for things like ice melting.• An ice cube must absorb heat to melt, but it

melts anyway. Why?

Entropy• The degree of randomness or disorder.• Better – number of ways things can be

arranged• S• The First Law of Thermodynamics - The energy

of the universe is constant.• The Second Law of Thermodynamics -The

entropy of the universe increases in any change.

• Drop a box of marbles.• Watch your room for a week.

Entropy

Entropy of a solid

Entropy of a

liquid

Entropy of a gas

• A solid has an orderly arrangement.• A liquid has the molecules next to each other

but isn’t orderly• A gas has molecules moving all over the

place.

Entropy increases when...

• Reactions of solids produce gases or liquids, or liquids produce gases.

• A substance is divided into parts -so reactions with more products than reactants have an increase in entropy.

• The temperature is raised -because the random motion of the molecules is increased.

• a substance is dissolved.

Entropy calculations

• There are tables of standard entropy (pg 407).• Standard entropy is the entropy at 25ºC and 1

atm pressure.• Abbreviated Sº, measure in J/K.• The change in entropy for a reaction is• ΔSº= ΣSº(Products)-ΣSº(Reactants).• Calculate ΔSº for this reaction • CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

• Calculate ΔSº for this reaction • CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

• For CH4 Sº = 186.2 J/K-mol

• For O2 Sº = 205.0 J/K-mol

• For CO2 Sº= 213.6 J/K-mol

• For H2O(g) Sº = 188.7 J/K-mol

Spontaneity

Will the reaction happen, and how can we make it?

Spontaneous reaction

• Reactions that will happen.• Nonspontaneous reactions don’t.• Even if they do happen, we can’t say how fast.• Two factors influence.• Enthalpy (heat) and entropy(disorder).

Two Factors• Exothermic reactions tend to be spontaneous.

– NegativeH.• Reactions where the entropy of the products

is greater than reactants tend to be spontaneous.– Positive ΔS.

• A change with positive ΔS and negative ΔH is always spontaneous.

• A change with negative ΔS and positive ΔH is never spontaneous.

Gibbs Free Energy

• The energy free to do work is the change in Gibbs free energy.

• ΔGº = ΔHº - TΔSº (T must be in Kelvin)• All spontaneous reactions release free energy. • So ΔG <0 for a spontaneous reaction. ΔG is

negative

Problems

• Using the information on page 407 and pg 190 determine if the following changes are spontaneous at 25ºC.

• 2H2S(g) + O2(g) 2H2O(l) + S(rhombic)

2H2S(g) + O2(g) 2H2O(l) + 2S

• We find ΔHf° for each component

– H2S = -20.1 kJ O2 = 0 kJ

– H2O = -285.8 kJ S = 0 kJ

• Then Products – Reactants• ΔH =2 (-285.8 kJ) + 2(0 kJ)

- 2 (-20.1 kJ) - 1(0 kJ) = -531.4 kJ

2H2S(g) + O2(g) 2H2O(l) + 2 S

• we find S for each component– H2S = 205.6 J/K O2 = 205.0 J/K

– H2O = 69.94 J/K S = 31.9 J/K

• Then Products – Reactants• ΔS= 2 (69.94 J/K) + 2(31.9 J/K)

- 2(205.6 J/K) - 205 J/K = -412.5 J/K

2H2S(g) + O2(g) 2H2O(l) + 2 S• ΔG = ΔH – T ΔS• G = -531.4 kJ - 298K (-412.5 J/K)• G = -531.4 kJ - -123000 J• ΔG = -531.4 kJ - -123 kJ• ΔG = -408.4 kJ• Spontaneous• Exergonic- it releases free energy.• At what temperature does it become

spontaneous?

Spontaneous

• It becomes spontaneous when ΔG = 0• That’s where it changes from positive to negative.• Using 0 = ΔH – T ΔS and solving for T• 0 - ΔH = - T ΔS • - ΔH = -T

ΔS• T = ΔH =

ΔS = 1290 K -531.4 kJ-412.5 J/K

= -531400 J -412.5 J/K

There’s Another Way• There are tables of standard free energies of

formation compounds.(pg 414)• ΔGºf is the free energy change in making a

compound from its elements at 25º C and 1 atm.

• for an element ΔGºf = 0

• Look them up.• ΔGº= ΔGºf(products) - ΔGºf(reactants)

2H2S(g) + O2(g) 2H2O(l) + 2S

• From we find ΔGf° for each component

– H2S = -33.02 kJ O2 = 0 kJ

– H2O = -237.2 kJ S = 0 kJ

• Then Products – Reactants• ΔG =2 (-237.2) + 2(0)

- 2 (-33.02) - 1(0) = -408.4 kJ