Entropy and Gibbs free energy
Exothermic
• The products are lower in energy than the
reactants
• Releases energy
• Often release heat
2
Energy
C + O2  CO2 + 395 kJ
C + O2
-395kJ
CO2
Reactants
3

Products
When will a reaction be exothermic
A) 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 bonds
D) When you get energy by breaking bonds
Endothermic
• The products are higher in energy than the
reactants
• Absorbs energy
• Absorb heat
5
CaCOCaCO
kJ CaO
 CaO
+ 2CO2
+ CO
3 + 176
3
Energy
CaO + CO2
+176 kJ
CaCO3
Reactants
6

Products
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
7
Energy
Change is down
ΔH is <0
Reactants
8

Products + heat
Energy
Change is up
ΔH is > 0
Reactants
+ heat 
Reactants
9
Products
Choose all that apply...
C(s) + 2 S(g)  CS2(l) ΔH = 89.3 kJ
Which of the following are true?
A) This reaction is exothermic
B) It could also be written
C(s) + 2 S(g) + 89.3 kJ  CS2(l)
C) The products have higher energy than the
reactants
D) It would make the water in the calorimeter
colder
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
11
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 = ΔHvap x m or
q = ΔHsolid x m
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)
H = H (products) - H (reactants )
o
f
o
f
Example
• CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
H
H
H
H
o
f
o
f
o
f
o
f
CH4 (g) = -74.86 kJ
O2(g) = 0 kJ
CO2(g) = -393.5 kJ
H2O(g) = -241.8 kJ
 ΔH= [-393.5 kJ + 2(-241.8 kJ)]
- [-74.86 kJ +2 (0 kJ )]
 ΔH= -802.2 kJ
17
Energy
elements
 H
o
f ( reactants)
H
o
f ( products)
reactants
products
H
o
f ( products) 
Reactants

H
o
f ( reactants) exothermic
Products
Energy
elements
 H
o
f ( reactants)
H
products
o
f ( products)
reactants
H
o
f ( reactants) 
Reactants

H
o
f ( products)endothermic
Products
Energy
Reactants
Products
Reaction coordinate
Energy
Activation Energy Minimum energy to
make the reaction
happen – how hard
Reactants
Products
Reaction coordinate
Energy
Activated
Complex or
Transition State
Reactants
Products
Reaction coordinate
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
Energy
Reactants
Products
Reaction coordinate
Energy
Activation Energy Minimum energy to
make the reaction
happen – how hard
Reactants
Products
Reaction coordinate
Energy
Activated
Complex or
Transition State
Reactants
Products
Reaction coordinate
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
Energy
Reactants
Overall energy
change
Products
Reaction coordinate
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.
Energy
Reactants
Products
Reaction coordinate
Catalysts
H H
• Hydrogen bonds to surface
of metal.
• Break H-H bonds
H
H
Pt surface
H H
H H
Catalysts
H
H
H
C
C
H
H H
H
H
Pt surface
Catalysts
• The double bond breaks and bonds to the
catalyst.
H
H
H
C
H
C
H
H
Pt surface
H H
Catalysts
• The hydrogen atoms bond with the carbon
H
H
H
C
H
C
H
H
Pt surface
H H
Catalysts
H
H
H
H
C
C
H
H
H
Pt surface
H
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.
– Negative H.
• 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
– H2O = -285.8 kJ
O2 = 0 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
– H2O = 69.94 J/K
O2 = 205.0 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 = -531.4 kJ
= -531400 J
= 1290 K
ΔS
-412.5 J/K
-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
– H2O = -237.2 kJ
O2 = 0 kJ
S = 0 kJ
• Then Products – Reactants
• ΔG =2 (-237.2) + 2(0)
- 2 (-33.02) - 1(0) = -408.4 kJ