Transcript Slide 1

Ch18.1 – Reaction Rates
Reaction Rates explained by collision theory.
-must collide with the correct orientation
-must have enough Kinetic energy to break apart old bonds
when collide.

http://www.saskschools.ca/curr_content/chem30_05/2_kinetics/kinetics2_1.htm
Factors Affecting Reaction Rates
“CANTC”
Concentration – more particles = more collision
Surface Area – more surface area= more places for rxn to occur
Nature of the reactants – some substances are more reactive than others
Temperature – higher temp= more KE
A.E.
Catalyst – Lowers activation energy
(provides an alternate path)
R
P
Factors Affecting Reaction Rates
“CANTC”
Concentration – more particles = more collision
Surface Area – more surface area= more places for rxn to occur
Nature of the reactants – some substances are more reactive than others
Temperature – higher temp= more KE
A.E.
Catalyst – Lowers activation energy
(provides an alternate path)
R
P
Two Types of Catalysts:
1. Heterogeneous catalystA surface for the reaction to take place on.
(Like a work bench) Platinum is the most common.
2. Homogeneous catalyst-get involved in the reaction form intermediate
compounds, but come back out unchanged.
Ex:
H202(l)  H20(l) + 02(g)
The decomposition of hydrogen peroxide is slow.
Iodide ions help speed it up.
Reversible Reactions
2SO2(g) + O2(g)
2SO3(g)
1%
99%
Forward & reverse reactions occurring at same time.
When they reach the same rate, reach chemical equilibrium.
- Doesn’t have to occur in the middle (50%,50%)
Exs) H2CO3
CO2 + H2O
1%
99%
CaCO3
CaO + CO2
99%
1%
At equilibrioum,
Products favored
At equilibrioum,
Reactants favored
Ch18 HW#1 1 - 6
Ch18 HW#1 1 – 6
1. Rate of reaction –
2. 1.0 mol zinc is completely converted to zinc oxide ZnO is 5.3 months.
What is the average rate of conversion in mol/months
3. Refrigerated food vs. room temp food
Ch18 HW#1 1 – 6
1. Rate of reaction – how fast the reaction converts reactants into products
2. 1.0 mol zinc is completely converted to zinc oxide ZnO is 5.3 months.
What is the average rate of conversion in mol/months
3. Refrigerated food vs. room temp food
Ch18 HW#1 1 – 6
1. Rate of reaction – how fast the reaction converts reactants into products
2. 1.0 mol zinc is completely converted to zinc oxide ZnO is 5.3 months.
What is the average rate of conversion in mol/months
1mol
 0.19mol month
5.3months
3. Refrigerated food vs. room temp food
Ch18 HW#1 1 – 6
1. Rate of reaction – how fast the reaction converts reactants into products
2. 1.0 mol zinc is completely converted to zinc oxide ZnO is 5.3 months.
What is the average rate of conversion in mol/months
1mol
 0.19mol month
5.3months
3. Refrigerated food vs. room temp food
Lower temp,
slows rxn rate
Higher temp = higher KE
= higher rate of rxn
4.How does each affect rate of reaction?
a. Temp
b. Concentration
c. Particle size
d. Catalyst
5. Double arrows in equation:
6.How do the amounts of reactants & products change once a reaction
has achieved equilibrium?
4.How does each affect rate of reaction?
a. Temp – higher temp = higher KE = faster rate
b. Concentration – greater concentration = faster rate
c. Particle size – smaller particles =more surface area = faster rate
d. Catalyst – lowers the activation energy = faster rate
5. Double arrows in equation:
6.How do the amounts of reactants & products change once a reaction
has achieved equilibrium?
4.How does each affect rate of reaction?
a. Temp – higher temp = higher KE = faster rate
b. Concentration – greater concentration = faster rate
c. Particle size – smaller particles =more surface area = faster rate
d. Catalyst – lowers the activation energy = faster rate
5. Double arrows in equation:
Represent a reversible reaction.
Ex: N2(g) + 3H2(g)
2NH2(g)
6.How do the amounts of reactants & products change once a reaction
has achieved equilibrium?
4.How does each affect rate of reaction?
a. Temp – higher temp = higher KE = faster rate
b. Concentration – greater concentration = faster rate
c. Particle size – smaller particles =more surface area = faster rate
d. Catalyst – lowers the activation energy = faster rate
5. Double arrows in equation:
Represent a reversible reaction.
Ex: N2(g) + 3H2(g)
2NH2(g)
6.How do the amounts of reactants & products change once a reaction
has achieved equilibrium?
The amounts stay constant. The rate of the forward reaction
equals the rate of the reverse reaction.
Ch18.2 – Factors Affecting Equilibrium
A chemical reaction that has reached equilibrium is a delicate balance.
If it’s disturbed, it will make minute adjustments to restore itself
at a new equilibrium position.
Le Chatelier’s Principle
If stress is applied to a system at equilibrium, the system changes to
relieve the stress.
- Add something
shifts to opposite side
- Remove something
shifts to the side of removal.
- Adding pressure shifts to side with fewer gas particles.
Decreasing pressure shifts to side with more gas.
Ex1) N2(g) + 3H2(g)
2NH3(g) + heat
Remove NH3(g), shifts to ________
Add N2: ________
Add heat: ________
Increase pressure: ________
Ex2) CO2(g) + H2(g) + heat
CO(g) + H2O(g)
Decrease temp: ________
Add heat: ________
Add H2O: ________
Increase pressure: ________
Ex1) N2(g) + 3H2(g)
2NH3(g) + heat
Remove NH3(g), shifts to products
Add N2: products
Add heat: reactants
Increase pressure: products
Ex2) CO2(g) + H2(g) + heat
CO(g) + H2O(g)
Decrease temp: reactants
Add heat: products
Add H2O: reactants
Increase pressure: no affect
Ex3) 2
SO2(g) + O2(g)
2 SO3(g) + heat
Increase SO2 : ________
Increase heat: ________
Add SO3 : ________
Increase pressure : ________
Ch18 HW#2
Lab18.1 – Reaction Rates
- due in 2 days
- Ch18 HW#2 due at beginning of period
Ch18.1,18.2 Review Worksheet
1. CANTC:
2. What do catalysts do to speed up a reaction?
3. Draw catalyst path on energy diagram:
4. 2 types of catalysts:
Ch18.1,18.2 Review Worksheet
1. CANTC:
Concentration
Surface Area
Nature of the reactants
Temp
Catalyst
2. What do catalysts do to speed up a reaction?
3. Draw catalyst path on energy diagram:
4. 2 types of catalysts:
Ch18.1,18.2 Review Worksheet
1. CANTC:
Concentration
Surface Area
Nature of the reactants
Temp
Catalyst
2. What do catalysts do to speed up a reaction?
Provide alternate path/lowers activation energy
3. Draw catalyst path on energy diagram:
4. 2 types of catalysts:
a.e.
R
P
Ch18.1,18.2 Review Worksheet
1. CANTC:
Concentration
Surface Area
Nature of the reactants
Temp
Catalyst
2. What do catalysts do to speed up a reaction?
Provide alternate path/lowers activation energy
3. Draw catalyst path on energy diagram:
4. 2 types of catalysts:
a.e.
R
P
Ch18.1,18.2 Review Worksheet
1. CANTC:
Concentration
Surface Area
Nature of the reactants
Temp
Catalyst
2. What do catalysts do to speed up a reaction?
Provide alternate path/lowers activation energy
3. Draw catalyst path on energy diagram:
4. 2 types of catalysts:
a.e.
R
P
Heterogeneous – provides a place for the
reaction to occur
Homogeneous – becomes part of the reaction mechanism
then gets spit back out.
5.
CaCO3(s) + heat
CaO(s) + CO2(g)
a. CaCO3(s) added: __________
b. Heat added: __________
c. CaO removed: __________
d. Pressure increased: __________
e. Temp decreased: __________
6.
2H2(g) + O2(g)
2H2O(g) + heat
a. H2 is added: __________
b. Heat is added: __________
c. O2 is removed: __________
d. Pressure is increased: __________
e. Temp decreased: __________
5.
CaCO3(s) + heat
CaO(s) + CO2(g)
a. CaCO3(s) added: products
b. Heat added: products
c. CaO removed: products
d. Pressure increased: reactants
e. Temp decreased: products
6.
2H2(g) + O2(g)
2H2O(g) + heat
a. H2 is added: __________
b. Heat is added: __________
c. O2 is removed: __________
d. Pressure is increased: __________
e. Temp decreased: __________
5.
CaCO3(s) + heat
CaO(s) + CO2(g)
a. CaCO3(s) added: products
b. Heat added: products
c. CaO removed: products
d. Pressure increased: reactants
e. Temp decreased: reactants
6.
2H2(g) + O2(g)
2H2O(g) + heat
a. H2 is added: products
b. Heat is added: products
c. O2 is removed: reactants
d. Pressure is increased: products
e. Temp decreased: products
7.
H2(g) + Cl2(g)
2HCl(g) + 93.2kJ
a. H2 is added: __________
b. Heat is added: __________
c. HCl is removed: __________
d. Pressure is increased: __________
e. Temp decreased: __________
8. ____CH2(g) + ____ O2(g)
____O2(g) + ___H2O(g) + 890kJ
a. CO2 is added: __________
b. Heat is added: __________
c. CO2 is removed: __________
d. Pressure is increased: __________
e. Temp decreased: __________
7.
H2(g) + Cl2(g)
2HCl(g) + 93.2kJ
a. H2 is added: products
b. Heat is added: reactants
c. HCl is removed: products
d. Pressure is increased: no effect
e. Temp decreased: products
8. ____CH2(g) + ____ O2(g)
____CO2(g) + ___H2O(g) + 890kJ
a. CO2 is added: __________
b. Heat is added: __________
c. CO2 is removed: __________
d. Pressure is increased: __________
e. Temp decreased: __________
7.
H2(g) + Cl2(g)
2HCl(g) + 93.2kJ
a. H2 is added: products
b. Heat is added: reactants
c. HCl is removed: products
d. Pressure is increased: no effect
e. Temp decreased: products
8.
2 CH2(g) + 3 O2(g)
2 CO2(g) + 2 H2O(g) + 890kJ
a. CO2 is added: __________
b. Heat is added: __________
c. CO2 is removed: __________
d. Pressure is increased: __________
e. Temp decreased: __________
7.
H2(g) + Cl2(g)
2HCl(g) + 93.2kJ
a. H2 is added: products
b. Heat is added: reactants
c. HCl is removed: products
d. Pressure is increased: no effect
e. Temp decreased: products
8.
2 CH2(g) + 3 O2(g)
2 CO2(g) + 2 H2O(g) + 890kJ
a. CO2 is added: reactants
b. Heat is added: reactants
c. CO2 is removed: products
d. Pressure is increased: products
e. Temp decreased: products
9.
Br2(l) + heat
Br2(g)
a. Br2(g) is added: __________
b. Heat is added: __________
c. Br2(g) is removed: __________
d. Pressure is decreased: __________
e. Temp decreased: __________
10. Lab18.1 Alka-seltzer reaction rates:
1. Hot, cold and room temp water affected the rate of reaction, proved:
2. Whole, broken, and powdered tablets affected the rate of reaction,
proved:
3. Cork stopper to test tube reaction of vinegar and baking soda proved:
9.
Br2(l) + heat
Br2(g)
a. Br2(g) is added: reactants
b. Heat is added: products
c. Br2(g) is removed: products
d. Pressure is decreased: products
e. Temp decreased: reactants
10. Lab18.1 Alka-seltzer reaction rates:
1. Hot, cold and room temp water affected the rate of reaction, proved:
2. Whole, broken, and powdered tablets affected the rate of reaction,
proved:
3. Cork stopper to test tube reaction of vinegar and baking soda proved:
9.
Br2(l) + heat
Br2(g)
a. Br2(g) is added: reactants
b. Heat is added: products
c. Br2(g) is removed: products
d. Pressure is decreased: products
e. Temp decreased: reactants
10. Lab18.1 Alka-seltzer reaction rates:
1. Hot, cold and room temp water affected the rate of reaction, proved:
Higher temp = faster rate
2. Whole, broken, and powdered tablets affected the rate of reaction,
proved:
more surface area = faster rate
3. Cork stopper to test tube reaction of vinegar and baking soda proved:
if products have more moles of gas, the cork
creates greater pressure = slower rate
Ch18 HW#2 7 – 9
7.Can a pressure change be used to shift equilibrium every
reversible reaction?
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l) + heat
8. Affect equilibrium: C(s) + H2O(g) + heat
a) lower temp
b) Increase pressure
c) Remove H2
d) Add H2O
e) Catalyst
f) Raise temp and decrease pressure
9. Affect equilibrium: 2SO3(g) + CO2(g) + heat
a) Add CO2
b) Add heat
c) Decrease pressure
d)Remove O2
e) Add catalyst
CO(g) + H2(g)
CS2(g) + 4O2(g)
Ch18 HW#2 7 – 9
7.Can a pressure change be used to shift equilibrium every
reversible reaction?
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l) + heat
Not when there are no gases present, or equal numbers.
8. Affect equilibrium: C(s) + H2O(g) + heat
CO(g) + H2(g)
a) lower temp
b) Increase pressure
c) Remove H2
d) Add H2O
e) Catalyst
f) Raise temp and decrease pressure
9. Affect equilibrium: 2SO3(g) + CO2(g) + heat
a) Add CO2
b) Add heat
c) Decrease pressure
d)Remove O2
e) Add catalyst
CS2(g) + 4O2(g)
Ch18 HW#2 7 – 9
7.Can a pressure change be used to shift equilibrium every
reversible reaction?
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l) + heat
Not when there are no gases present, or equal numbers.
8. Affect equilibrium: C(s) + H2O(g) + heat
CO(g) + H2(g)
a) lower temp Reactants
b) Increase pressure
Reactants
c) Remove H2 Products
d) Add H2O
e) Catalyst
f) Raise temp and decrease pressure
9. Affect equilibrium: 2SO3(g) + CO2(g) + heat
a) Add CO2
b) Add heat
c) Decrease pressure
d)Remove O2
e) Add catalyst
CS2(g) + 4O2(g)
Ch18 HW#2 7 – 9
7.Can a pressure change be used to shift equilibrium every
reversible reaction?
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l) + heat
Not when there are no gases present, or equal numbers.
8. Affect equilibrium: C(s) + H2O(g) + heat
CO(g) + H2(g)
a) lower temp Reactants
b) Increase pressure
Reactants
c) Remove H2 Products
d) Add H2O
Products
e) Catalyst
No Change
f) Raise temp and decrease pressure
Both favor products
9. Affect equilibrium: 2SO3(g) + CO2(g) + heat
a) Add CO2
b) Add heat
c) Decrease pressure
d)Remove O2
e) Add catalyst
CS2(g) + 4O2(g)
Ch18 HW#2 7 – 9
7.Can a pressure change be used to shift equilibrium every
reversible reaction?
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l) + heat
Not when there are no gases present, or equal numbers.
8. Affect equilibrium: C(s) + H2O(g) + heat
CO(g) + H2(g)
a) lower temp Reactants
b) Increase pressure
Reactants
c) Remove H2 Products
d) Add H2O
Products
e) Catalyst
No Change
f) Raise temp and decrease pressure
Both favor products
9. Affect equilibrium: 2SO3(g) + CO2(g) + heat
a) Add CO2
Products
b) Add heat
Products
c) Decrease pressure
Products
d)Remove O2
e) Add catalyst
CS2(g) + 4O2(g)
Ch18 HW#2 7 – 9
7.Can a pressure change be used to shift equilibrium every
reversible reaction?
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l) + heat
Not when there are no gases present, or equal numbers.
8. Affect equilibrium: C(s) + H2O(g) + heat
CO(g) + H2(g)
a) lower temp Reactants
b) Increase pressure
Reactants
c) Remove H2 Products
d) Add H2O
Products
e) Catalyst
No Change
f) Raise temp and decrease pressure
Both favor products
9. Affect equilibrium: 2SO3(g) + CO2(g) + heat
a) Add CO2
Products
b) Add heat
Products
c) Decrease pressure
Products
d)Remove O2 Products
e) Add catalyst No Change
CS2(g) + 4O2(g)
Ch18.3 – Spontaneous Reactions
2 things contribute to deciding if a reaction will proceed on its own:
Free Energy & Entropy
Free Energy – energy released from reactions
Gasoline(g) + O2(g)
CO2(g) + H2O(g) + heat
can do work
Exothermic reactions are usually spontaneous
Endothermic reactions are usually non spontaneous
- they can be spontaneous if the amount of disorder created
has a bigger affect than the increase in energy
Entropy (S) – a measure of the amount of disorder in a system.
Natural processes head toward disorder
Natural processes:
Energy: High energy  low energy
Entropy:
Most
order
Solids
↔
Least
energy
Order  Disorder
Liquids
↔
These will oppose
each other.
One will beat the other
Least
order
Gases
Most
energy
Ex) Are the following reactions spontaneous? Predict
a) H2O(l)
b) C(s)
→ H2O(g)
+ O2 (g) → CO2(g) + 393.5 kJ
Ex) Are the following reactions spontaneous? Predict
a) H2O(l) → H2O(g)
more order → more disorder
less energy ← more energy
b) C(s) + O2 (g) → CO2(g) + 393.5
order disorder → disorder
less E more E ← More E (wins)
kJ
Ch18 HW#3 10 – 17
Lab18.2 – Equilibrium
- due in 2 days
- Ch18 HW#3 due at beginning of period
Ch18 HW#3 10 – 17
10) Where does “lost” free energy end up?
11) Does free energy that’s lost as waste heat ever serve a useful function?
12) What’s a spontaneous reaction? Ex from life?
13) Products in one spontaneous reactions are more ordered then reactants.
Is the entropy change favorable or unfavorable?
14) Higher entropy:
a) new pack playing cards or used cards
b) sugar cube dissolved in water or cube?
c) 1g salt crystal or 1g powdered.
15) Are all spontaneous reactions exothermic?
10) Where does “lost” free energy end up?
waste heat energy
11) Does free energy that’s lost as waste heat ever serve a useful function?
12) What’s a spontaneous reaction? Ex from life?
13) Products in one spontaneous reactions are more ordered then reactants.
Is the entropy change favorable or unfavorable?
14) Higher entropy:
a) new pack playing cards or used cards
b) sugar cube dissolved in water or cube?
c) 1g salt crystal or 1g powdered.
15) Are all spontaneous reactions exothermic?
10) Where does “lost” free energy end up?
waste heat energy
11) Does free energy that’s lost as waste heat ever serve a useful function?
Some heat energy can be converted into useful work
12) What’s a spontaneous reaction? Ex from life?
13) Products in one spontaneous reactions are more ordered then reactants.
Is the entropy change favorable or unfavorable?
14) Higher entropy:
a) new pack playing cards or used cards
b) sugar cube dissolved in water or cube?
c) 1g salt crystal or 1g powdered.
15) Are all spontaneous reactions exothermic?
10) Where does “lost” free energy end up?
waste heat energy
11) Does free energy that’s lost as waste heat ever serve a useful function?
Some heat energy can be converted into useful work
12) What’s a spontaneous reaction? Ex from life?
A rxn that proceeds forward on its own
Glucose and oxygen react to form carbon dioxide and water in your body
13) Products in one spontaneous reactions are more ordered then reactants.
Is the entropy change favorable or unfavorable?
reactants  products
If this rxn is spontaneous
less order more order
order doesn’t support this!
14) Higher entropy:
a) new pack playing cards or used cards
b) sugar cube dissolved in water or cube?
c) 1g salt crystal or 1g powdered.
15) Are all spontaneous reactions exothermic?
10) Where does “lost” free energy end up?
waste heat energy
11) Does free energy that’s lost as waste heat ever serve a useful function?
Some heat energy can be converted into useful work
12) What’s a spontaneous reaction? Ex from life?
A rxn that proceeds forward on its own
Glucose and oxygen react to form carbon dioxide and water in your body
13) Products in one spontaneous reactions are more ordered then reactants.
Is the entropy change favorable or unfavorable?
reactants  products
If this rxn is spontaneous
less order more order
order doesn’t support this!
Therefore entropy is unfavorable
(This rxn MUST have a big energy change.)
14) Higher entropy:
a) new pack playing cards or used cards
b) sugar cube dissolved in water or cube?
c) 1g salt crystal or 1g powdered.
15) Are all spontaneous reactions exothermic?
10) Where does “lost” free energy end up?
waste heat energy
11) Does free energy that’s lost as waste heat ever serve a useful function?
Some heat energy can be converted into useful work
12) What’s a spontaneous reaction? Ex from life?
A rxn that proceeds forward on its own
Glucose and oxygen react to form carbon dioxide and water in your body
13) Products in one spontaneous reactions are more ordered then reactants.
Is the entropy change favorable or unfavorable?
reactants  products
If this rxn is spontaneous
less order more order
order doesn’t support this!
Therefore entropy is unfavorable
(This rxn MUST have a big energy change.)
14) Higher entropy:
a) new pack playing cards or used cards Used
b) sugar cube dissolved in water or cube? Dissolved
c) 1g salt crystal or 1g powdered.
Powder
15) Are all spontaneous reactions exothermic?
10) Where does “lost” free energy end up?
waste heat energy
11) Does free energy that’s lost as waste heat ever serve a useful function?
Some heat energy can be converted into useful work
12) What’s a spontaneous reaction? Ex from life?
A rxn that proceeds forward on its own
Glucose and oxygen react to form carbon dioxide and water in your body
13) Products in one spontaneous reactions are more ordered then reactants.
Is the entropy change favorable or unfavorable?
reactants  products
If this rxn is spontaneous
less order more order
order doesn’t support this!
Therefore entropy is unfavorable
(This rxn MUST have a big energy change.)
14) Higher entropy:
a) new pack playing cards or used cards Used
b) sugar cube dissolved in water or cube? Dissolved
c) 1g salt crystal or 1g powdered.
Powder
15) Are all spontaneous reactions exothermic? Not always,
an endothermic rxn can be spont if big increase in entropy.
16) At normal atm pressure, steam condenses to liquid.
Is entropy change favorable?
17) Guess
a) CaCO3(s)  CaO(s) +CO2(g)
b) 2 H2(g) + O2(g)  2 H2O(l)
c) H2(g) + Cl2(g)  2 HCl(g)
d) CH4(g) + O2(g)  CO2(g) + H2O(g)
e)Br2(l)  Br2(g)
16) At normal atm pressure, steam condenses to liquid.
Is entropy change favorable?
No, liquid has more order.
17) Guess
a) CaCO3(s)  CaO(s) +CO2(g)
b) 2 H2(g) + O2(g)  2 H2O(l)
c) H2(g) + Cl2(g)  2 HCl(g) + 93.2kJ
d) CH4(g) + O2(g)  CO2(g) + H2O(g) + 890kJ
e)Br2(l)  Br2(g)
16) At normal atm pressure, steam condenses to liquid.
Is entropy change favorable?
No, liquid has more order.
17) Guess
a) CaCO3(s)  CaO(s) +CO2(g)
order
order
disorder
low E
low E
high E
b) 2 H2(g) + O2(g)  2 H2O(l)
(spont)
(non-spont)
c) H2(g) + Cl2(g)  2 HCl(g) + 93.2kJ
d) CH4(g) + O2(g)  CO2(g) + H2O(g) + 890kJ
e)Br2(l)  Br2(g)
16) At normal atm pressure, steam condenses to liquid.
Is entropy change favorable?
No, liquid has more order.
17) Guess
a) CaCO3(s)  CaO(s) +CO2(g)
order
order
disorder (spont)
low E
low E
high E
(non-spont)
b) 2 H2(g) + O2(g)  2 H2O(l)
disorder disorder order (non-spont)
high E high E lower E (spont)
c) H2(g) + Cl2(g)  2 HCl(g) + 93.2kJ
d) CH4(g) + O2(g)  CO2(g) + H2O(g) + 890kJ
e)Br2(l)  Br2(g)
16) At normal atm pressure, steam condenses to liquid.
Is entropy change favorable?
No, liquid has more order.
17) Guess
a) CaCO3(s)  CaO(s) +CO2(g)
order
order
disorder (spont)
low E
low E
high E
(non-spont)
b) 2 H2(g) + O2(g)  2 H2O(l)
disorder disorder order (non-spont)
high E high E lower E (spont)
c) H2(g) + Cl2(g)  2 HCl(g) + 93.2kJ (exothermic)
disorder disorder disorder(non-spont)
high E high E lower E (spont)
d) CH4(g) + O2(g)  CO2(g) + H2O(g) + 890kJ
e)Br2(l)  Br2(g)
16) At normal atm pressure, steam condenses to liquid.
Is entropy change favorable?
No, liquid has more order.
17) Guess
a) CaCO3(s)  CaO(s) +CO2(g)
order
order
disorder (spont)
low E
low E
high E
(non-spont)
b) 2 H2(g) + O2(g)  2 H2O(l)
disorder disorder order (non-spont)
high E high E lower E (spont)
c) H2(g) + Cl2(g)  2 HCl(g) + 93.2kJ (exothermic)
disorder disorder disorder(non-spont)
high E high E lower E (spont)
d) CH4(g) + O2(g)  CO2(g) + H2O(g) + 890kJ
disorder disorder disorder (no effect)
high E high E high E high E exothermic (spont)
e)Br2(l)  Br2(g)
17) Guess
a) CaCO3(s)  CaO(s) +CO2(g)
order
order
disorder (spont)
low E
low E
high E
(non-spont)
b) 2 H2(g) + O2(g)  2 H2O(l)
disorder disorder order (non-spont)
high E high E lower E (spont)
c) H2(g) + Cl2(g)  2 HCl(g) + 93.2kJ (exothermic)
disorder disorder disorder(non-spont)
high E high E lower E (spont)
d) CH4(g) + O2(g)  CO2(g) + H2O(g) + 890kJ
disorder disorder disorder (no effect)
high E high E high E high E exothermic (spont)
e)Br2(l)  Br2(g)
order disorder
(spont)
low E high E
(non-spont)
Ch18.4 – Equilibrium Constants
Balance:
H2(g) + I2(g)
HI(g)
Ch18.4 – Equilibrium Constants
H2(g) + I2(g)
At equilibrium:
2%
2HI(g)
98%
Chemists don’t use % to indicate the position of equilibrium,
instead use molarity and ratios.
Ch18.4 – Equilibrium Constants
H2(g) + I2(g)
At equilibrium:
2HI(g)
2%
98%
Chemists don’t use % to indicate the position of equilibrium,
instead use molarity and ratios.
Keq – equilibrium constant
- compares the rate of reverse reaction to rate of forward
Keq
[products]

[reactants]
The brackets represent molarity.
Each substance’s molarity is raised
to the power of the reaction coefficient.
Ex1) Write Keq for H2(g) + I2(g)
2HI(g)
Ex 2) At 10oC, a 1 liter container has 0.0045 mol N2O4 in equilibrium
with 0.030 mol NO2
a) Write an equation:
b) Write Keq:
c) Calc Keq:
Ex 2) At 10oC, a 1 liter container has 0.0045 mol N2O4 in equilibrium
with 0.030 mol NO2
a) Write an equation:
b) Write Keq:
c) Calc Keq:
K eq
N2O4
[ NO2 ] 2

[ N 2 O4 ]
2NO2
K eq
[0.03]2

 0.2
[0.0045]
Keq is unique for every reaction, and for every temp.
If temp goes up, usually Keq goes up.
Keq > 1 products favored
Keq < 1 reactants favored
Ex 3) At a certain temp, 0.500M H2, 0.500M N2 are
in equilibrium with 2.500M NH3. Calc Keq
Ex 4) Given Keq = 100, [NH3] = 1.25M, [H2] = 1.75M, Find [N2]
Ch18 HW#4 WS
Ch18 HW#4
1. Write the equilibrium expression
for the oxidation of hydrogen to
form water vapor.
2H2(g) + O2(g)
2H2O(g)
2. Write the equilibrium expression
for the formation of nitrosyl bromide.
2NO(g) + Br2(g)
2NOBr(g)
3. Write the equilibrium expression
for the following reaction
NO(g) + O3(g)
O2(g) + NO2(g)
Ch18 HW#4
1. Write the equilibrium expression
for the oxidation of hydrogen to
form water vapor.
2H2(g) + O2(g)
K eq 
2H2O(g)
2. Write the equilibrium expression
for the formation of nitrosyl bromide.
2NO(g) + Br2(g)
2NOBr(g)
3. Write the equilibrium expression
for the following reaction
NO(g) + O3(g)
O2(g) + NO2(g)
[ H 2 O]2
[ H 2 ]2 [O2 ]
Ch18 HW#4
1. Write the equilibrium expression
for the oxidation of hydrogen to
form water vapor.
2H2(g) + O2(g)
K eq 
2H2O(g)
2. Write the equilibrium expression
for the formation of nitrosyl bromide.
2NO(g) + Br2(g)
2NOBr(g)
[ H 2 ]2 [O2 ]
K eq 
3. Write the equilibrium expression
for the following reaction
NO(g) + O3(g)
[ H 2 O]2
O2(g) + NO2(g)
[ NOBr]2
[ NO]2 [ Br2 ]
Ch18 HW#4
1. Write the equilibrium expression
for the oxidation of hydrogen to
form water vapor.
2H2(g) + O2(g)
K eq 
2H2O(g)
2. Write the equilibrium expression
for the formation of nitrosyl bromide.
2NO(g) + Br2(g)
2NOBr(g)
K eq
O2(g) + NO2(g)
[ NO2 ][O2 ]

[ NO][O3 ]
[ H 2 ]2 [O2 ]
K eq 
3. Write the equilibrium expression
for the following reaction
NO(g) + O3(g)
[ H 2 O]2
[ NOBr]2
[ NO]2 [ Br2 ]
4. Write the equilibrium expression
for the following reaction
CH4(g) + CI2(g)
CH3CI(g) + HCI(g)
5. Write the equilibrium expression
for the following reaction
CH4(g) + H20(g)
CO(g) + 3H2(g)
6. Write the equilibrium expression
for the following reaction
CO(g) + 2H2(g)
CH3OH(g)
7. Same:
2C2H6(g) + 7O2(g)
4CO2(g) + 6H2O(g)
4. Write the equilibrium expression
for the following reaction
CH4(g) + CI2(g)
CH3CI(g) + HCI(g)
K eq
5. Write the equilibrium expression
for the following reaction
CH4(g) + H20(g)
CO(g) + 3H2(g)
6. Write the equilibrium expression
for the following reaction
CO(g) + 2H2(g)
CH3OH(g)
7. Same:
2C2H6(g) + 7O2(g)
4CO2(g) + 6H2O(g)
[CH 3Cl ][ HCl]

[CH 4 ][Cl2 ]
4. Write the equilibrium expression
for the following reaction
CH4(g) + CI2(g)
CH3CI(g) + HCI(g)
5. Write the equilibrium expression
for the following reaction
CH4(g) + H20(g)
CO(g) + 3H2(g)
K eq
[CH 3Cl ][ HCl]

[CH 4 ][Cl2 ]
K eq
[CO][ H 2 ]3

[CH 4 ][ H 2O]
6. Write the equilibrium expression
for the following reaction
CO(g) + 2H2(g)
CH3OH(g)
7. Same:
2C2H6(g) + 7O2(g)
4CO2(g) + 6H2O(g)
4. Write the equilibrium expression
for the following reaction
CH4(g) + CI2(g)
CH3CI(g) + HCI(g)
5. Write the equilibrium expression
for the following reaction
CH4(g) + H20(g)
CO(g) + 3H2(g)
6. Write the equilibrium expression
for the following reaction
CO(g) + 2H2(g)
K eq
[CH 3Cl ][ HCl]

[CH 4 ][Cl2 ]
K eq
[CO][ H 2 ]3

[CH 4 ][ H 2O]
K eq 
CH3OH(g)
7. Same:
2C2H6(g) + 7O2(g)
4CO2(g) + 6H2O(g)
[CH 3OH ]
[CO][ H 2 ]2
4. Write the equilibrium expression
for the following reaction
CH4(g) + CI2(g)
CH3CI(g) + HCI(g)
5. Write the equilibrium expression
for the following reaction
CH4(g) + H20(g)
CO(g) + 3H2(g)
6. Write the equilibrium expression
for the following reaction
CO(g) + 2H2(g)
K eq
[CH 3Cl ][ HCl]

[CH 4 ][Cl2 ]
K eq
[CO][ H 2 ]3

[CH 4 ][ H 2O]
K eq 
CH3OH(g)
[CH 3OH ]
[CO][ H 2 ]2
7. Same:
2C2H6(g) + 7O2(g)
4CO2(g) + 6H2O(g)
K eq 
[CO2 ]4 [ H 2O]6
[C2 H 6 ]2 [O2 ]7
8. Write the equilibrium expression
for the following reaction, and solve for Keq
SnO2(s) + 2CO(g)
(.25)
(.10)
Sn(s) + 2CO2(g)
(.75) (.60)
(Which side of this reaction is favored?)
9. Write the equilibrium expression
for the following reaction, and solve for Keq
C(s) + CO2(g)
(1.3) (2.25)
2CO(g)
(.5)
Who’s favored?
10. Write the equilibrium expression
for the following reaction, and solve for Keq
FeO(s) + CO(g)
(1)
(2)
Who’s favored?
Fe(s) + CO2(g)
(.5) (.5)
8. Write the equilibrium expression
for the following reaction, and solve for Keq K eq
SnO2(s) + 2CO(g)
(.25)
(.10)
Sn(s) + 2CO2(g)
(.75) (.60)
(Which side of this reaction is favored?)
K eq 
products favored
9. Write the equilibrium expression
for the following reaction, and solve for Keq
C(s) + CO2(g)
(1.3) (2.25)
2CO(g)
(.5)
Who’s favored?
10. Write the equilibrium expression
for the following reaction, and solve for Keq
FeO(s) + CO(g)
(1)
(2)
Who’s favored?
Fe(s) + CO2(g)
(.5) (.5)

[ Sn][CO2 ]2
[ SnO2 ][CO]2
[.75][.60]2
[.25][.10]
2
 108
8. Write the equilibrium expression
for the following reaction, and solve for Keq K eq
SnO2(s) + 2CO(g)
(.25)
(.10)
Sn(s) + 2CO2(g)
(.75) (.60)
K eq 
(Which side of this reaction is favored?)
products favored
9. Write the equilibrium expression
for the following reaction, and solve for Keq
C(s) + CO2(g)
(1.3) (2.25)
2CO(g)
(.5)
K eq 
Who’s favored?
reactants favored K
eq

10. Write the equilibrium expression
for the following reaction, and solve for Keq
FeO(s) + CO(g)
(1)
(2)
Who’s favored?
Fe(s) + CO2(g)
(.5) (.5)

[ Sn][CO2 ]2
[ SnO2 ][CO]2
[.75][.60]2
[.25][.10]
2
 108
[CO]2
[C ][CO2 ]
2
[.5]2
[1.3][2.25]
2
 0.085
9. Write the equilibrium expression K eq
for the following reaction, and solve for Keq
C(s) + CO2(g)
(1.3) (2.25)
2CO(g)
(.5)
Who’s favored?

K eq 
reactants favored
[CO]2
[C ][CO2 ]2
[.5]2
[1.3][2.25]
10. Write the equilibrium expression
for the following reaction, and solve for Keq
FeO(s) + CO(g)
(1)
(2)
Fe(s) + CO2(g)
(.5) (.5)
K eq
Who’s favored?
reactants favored
K eq
2
 0.085
[ Fe][CO2 ]

[ FeO][CO]
[.5][.5]

 0.125
[1][2]
11. For the reaction 2ICl(g)
I2(g) +Cl2(g), Keq= 0.11
At a particular time, the following concentrations are measured:
[ICl] = 2.5M, [I2] = 2.0M, [Cl2] = 1.2M.
Is this reaction at equilibrium?
If not, in which direction will the reaction proceed?
11. For the reaction 2ICl(g)
I2(g) +Cl2(g), Keq= 0.11
At a particular time, the following concentrations are measured:
[ICl] = 2.5M, [I2] = 2.0M, [Cl2] = 1.2M.
Is this reaction at equilibrium?
If not, in which direction will the reaction proceed?
K eq 
K eq 
[ I 2 ][Cl2 ]
[ ICl ]2
[1.2][2]
[2.5]
2
 0.384
Keq is too high, so reaction will proceed backwards towards the
reactants, until it lowers to 0.11.
Ch18.5 Solving Keq‘s
Ex1) 1 mol of H2 & 1 mol of I2 are placed in a 1 L sealed flask, and begin
reacting, at 450oC. At equilibrium, 1.56 mol of hydrogen iodide is
present. Calc Keq
H2(g) + I2(g)  2HI(g)
Ex2) 1 mol of I2 gas is made to react with 1 mol of Cl2 gas inside a sealed
1 L container. At a certain temp, Keq for the rxn is 0.84. What are
the equilibrium concentrations?
I2 + Cl2
2 ICl
Ex3) In the oxidation of hydrogen, 1 mol of oxygen is placed in a sealed 1L
flask with 1 mol of hydrogen gas. At a certain temp, Keq for this
reaction is 20.4. how much water vapor is produced?
Ch18 HW#5 18-22
Ch18 HW#5 18-22
18. Keq for: N2(g) + 3H2(g) ↔ 2NH3(g)
19. Cale Keq
N2(g) + 3H2(g) ↔ 2NH3(g)
[.25] [.15]
[.10]
20. Keq for 2HI(g) ↔ H2(g) + I2(g)
Ch18 HW#5 18-22
18. Keq for: N2(g) + 3H2(g) ↔ 2NH3(g)
[NH3]2
Keq =
[N2] [H2]3
19. Cale Keq
N2(g) + 3H2(g) ↔ 2NH3(g)
[.25] [.15]
[.10]
20. Keq for 2HI(g) ↔ H2(g) + I2(g)
Ch18 HW#5 18-22
18. Keq for: N2(g) + 3H2(g) ↔ 2NH3(g)
[NH3]2
Keq =
[N2] [H2]3
N2(g) + 3H2(g) ↔ 2NH3(g)
[.25] [.15]
[.10]
19. Cale Keq
[.1]2
Keq =
= 11.9
[.25] [.15]3
20. Keq for 2HI(g) ↔ H2(g) + I2(g)
Ch18 HW#5 18-22
18. Keq for: N2(g) + 3H2(g) ↔ 2NH3(g)
[NH3]2
Keq =
[N2] [H2]3
N2(g) + 3H2(g) ↔ 2NH3(g)
[.25] [.15]
[.10]
19. Cale Keq
[.1]2
Keq =
= 11.9
[.25] [.15]3
20. Keq for 2HI(g) ↔ H2(g) + I2(g)
[H2] [I2]
Keq =
[HI]2
21.Products or reactants?
a) Keq = 1 x 102
b) Keq = 0.003
c) Keq = 3.5
d) Keq = 6 x 10-4
22. How many moles of I & HI at equilibrium?
2HI(g) ↔ 1H2(g) + 1I2(g)
Start:
[X]
[0]
[0]
Equilibrium: [x – 1.0] [.5]
[.5]
21.Products or reactants?
a) Keq = 1 x 102
b) Keq = 0.003
c) Keq = 3.5
d) Keq = 6 x 10-4
P
R
P
R
22. How many moles of I & HI at equilibrium?
2HI(g) ↔ 1H2(g) + 1I2(g)
Start:
[X]
[0]
[0]
Equilibrium: [x – 1.0] [.5]
[.5]
21.Products or reactants?
a) Keq = 1 x 102
b) Keq = 0.003
c) Keq = 3.5
d) Keq = 6 x 10-4
P
R
P
R
22. How many moles of I & HI at equilibrium?
2HI(g) ↔ 1H2(g) + 1I2(g)
Start:
[X]
[0]
[0]
Equilibrium: [x – 1.0] [.5]
[.5]
[H2] [I2]
Keq=
[HI]2
[.5] [.5]
(.02)=
[x – 1.0]2
Ch18.6 – Entropy Calculations
Entropy (S) – measure of disorder
Units:
Table of Standard Entropies ΔSo (@ 25oC, 101.3 kPa)
- on HW sheet
Ex1) Calculate the standard entropy change, ΔSo , that
occurs when 1 mol H2O(g) is condensed to 1 mol H2O(l) @ STP.
H2O(g)  H2O(l)
- Do you think there’s an increase or decrease in disorder?
Ch18.6 – Entropy Calculations
Entropy (S) – measure of disorder
Units:
Table of Standard Entropies ΔSo (@ 25oC, 101.3 kPa)
- on HW sheet
Ex1) Calculate the standard entropy change, ΔSo , that
occurs when 1 mol H2O(g) is condensed to 1 mol H2O(l) @ STP.
H2O(g)  H2O(l)
Reactants
Products
(1mol)(188.7J/K·mol) (1mol)(69.94J/K·mol)
188.7 J/K
69.94 J/K
ΔSo = Soproducts – Soreactants
Δ S = 69.94J/K – 188.7J/K = -118J/K
ΔSo = (+) increase in disorder (more chaos)
ΔSo = (-) decrease in disorder (more ordered)
Ex2) What is the standard change in entropy:
2 NO(g) + O2(g)  2NO2(g)
Ex2) What is the standard change in entropy:
2 NO(g) + O2(g)  2NO2(g)
ΔS = prod – react
480 – 626.2 = -145.2 J/K
Dec in disorder
Ch18 HW#6 23,24
Ch18 HW#6 23,24
23. How does the magnitude of the standard entropy of an element
in gas state compare to that of the liquid state?
Ch18 HW#6 23,24
23. How does the magnitude of the standard entropy of an element
in gas state compare to that of the liquid state?
Gases always have higher entropies than their liquids.
(More disorder)
Ch18 HW#6 24
24a)
CaCO3(s) ↔ CaO(s) +CO2(g)
(1mol)(88.7J/K·mol) (1mol)(39.75J/K·mol)+(1mol)(213.6J/K·mol
= 88.7J/K
= 253.35 J/K
b)
2H2(g)+ O2(g)→ 2H2O(l)
(2mol)(130.6J/K·mol)+(1mol)(205J/K·mol)
466.2J/K
c)
(2mol)(69.94J/K·mol)
139.88J/K
H2(g)+ Cl2(g)→ 2HCl(g)
(1mol)(130.6J/K·mol)+(1mol)(223J/K·mol) (2mol)(186.7J/K·mol)
353.6J/K
373.4J/k
Ch18 HW#6 24
24a)
CaCO3(s) ↔ CaO(s) +CO2(g)
(1mol)(88.7J/K·mol) (1mol)(39.75J/K·mol)+(1mol)(213.6J/K·mol
= 88.7J/K
= 253.35 J/K
ΔS = prod – react
= 253.35 – 88.7 = +164.65 J/K (Inc)
b)
2H2(g)+ O2(g)→ 2H2O(l)
(2mol)(130.6J/K·mol)+(1mol)(205J/K·mol)
466.2J/K
c)
(2mol)(69.94J/K·mol)
139.88J/K
H2(g)+ Cl2(g)→ 2HCl(g)
(1mol)(130.6J/K·mol)+(1mol)(223J/K·mol) (2mol)(186.7J/K·mol)
353.6J/K
373.4J/k
Ch18 HW#6 24
24a)
CaCO3(s) ↔ CaO(s) +CO2(g)
(1mol)(88.7J/K·mol) (1mol)(39.75J/K·mol)+(1mol)(213.6J/K·mol
= 88.7J/K
= 253.35 J/K
ΔS = prod – react
= 253.35 – 88.7 = +164.65 J/K (Inc)
b)
2H2(g)+ O2(g)→ 2H2O(l)
(2mol)(130.6J/K·mol)+(1mol)(205J/K·mol) (2mol)(69.94J/K·mol)
466.2J/K
139.88J/K
ΔS = 139.88 – 466.2
= -326.3 J/K (Dec)
c)
H2(g)+ Cl2(g)→ 2HCl(g)
(1mol)(130.6J/K·mol)+(1mol)(223J/K·mol) (2mol)(186.7J/K·mol)
353.6J/K
373.4J/k
Ch18 HW#6 24
24a)
CaCO3(s) ↔ CaO(s) +CO2(g)
(1mol)(88.7J/K·mol) (1mol)(39.75J/K·mol)+(1mol)(213.6J/K·mol
= 88.7J/K
= 253.35 J/K
ΔS = prod – react
= 253.35 – 88.7 = +164.65 J/K (Inc)
b)
2H2(g)+ O2(g)→ 2H2O(l)
(2mol)(130.6J/K·mol)+(1mol)(205J/K·mol) (2mol)(69.94J/K·mol)
466.2J/K
139.88J/K
ΔS = 139.88 – 466.2
= -326.3 J/K (Dec)
c)
H2(g)+ Cl2(g)→ 2HCl(g)
(1mol)(130.6J/K·mol)+(1mol)(223J/K·mol) (2mol)(186.7J/K·mol)
353.6J/K
373.4J/k
ΔS = 373.4 – 353.6
= +19.8 J/K (Inc)
d)
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
(1)(186.2)+(2)(205)
596J/K
e)
(1)(213.6)+(2)(188.7)
591J/K
Br2(l)  Br2(g)
(1)(152.3) (1)(245)
152.3J/K
245J/K
d)
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
(1)(186.2)+(2)(205)
596J/K
(1)(213.6)+(2)(188.7)
591J/K
ΔS = 591 – 596 = -5 (dec)
e)
Br2(l)  Br2(g)
(1)(152.3) (1)(245)
152.3J/K
245J/K
d)
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
(1)(186.2)+(2)(205)
596J/K
(1)(213.6)+(2)(188.7)
591J/K
ΔS = 591 – 596 = -5 (dec)
e)
Br2(l)  Br2(g)
(1)(152.3) (1)(245)
152.3J/K
245J/K
ΔS = prod – react = 245 – 152.3 = +92.7J/K (Inc)
Ch18.7 – Free Energy Calculations
- Gibb’s free energy, ΔG, is the max amount of energy
available in a chemical rxn to do useful work.
- Use ΔG to decide if a reaction is spontaneous
ΔG = (-) spontaneous
ΔG = (+) non-spontaneous
- ΔG = 0 for all free elements
- ΔG values given on HW sheet & test
- Two ways to solve ΔG problems:
1. ΔG = products – reactants
2. ΔG = ΔH – T·ΔS
Entropy
Heat absorbed Temp
or released
(298K)
Ex1) Use ΔG values “given” to determine if spontaneous:
C(s) + O2(g)  CO2(g)
Ex2) Same for
Na(s) + Cl2(g)  NaCl(s)
Ex3) Use ΔH & S to determine if rxn is spontaneous:
C(s)
O2(g)
Standard Heats of Formation, Hf0
0 kJ/mol
Standard Entropies, S0
5.694J/mol.K 205J/mol.K
0 kJ/mol
C(s) + O2(g)  CO2(g)
CO2(g)
-393.5kJ/mol
213.6J/mol.K
Ex4) Use ΔH & S to determine if rxn is spontaneous
C2H6(g)
O2(g)
CO2(g)
-393 kJ/mol
-286kJ/mol
63.8 J/mol.K
80.7 J/mol.K
Hf0
-84.7kJ/mol
0 kJ/mol
S0
68.2 J/mol.K
64.1 J/mol.K
H2O(g)
2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6H2O(g)
Ch18 HW#7
Ch18 HW#7
25. Calculate the standard free energy change ΔG for the reaction between
iron (III) oxide and carbon (graphite). Refer to the table of free energies.
Standard Gibbs Free
Energies of Formation (Gf)
Fe2O3(s) C(s) (graphite)
-741.0
0.0
Fe(s) CO2(g)
0.0 -394.4
2Fe2O3(s)+3C(s)4Fe(s)+3CO2(g)
26. A reaction is endothermic (positive ΔH) and has a positive entropy.
Would this reaction more likely be spontaneous at high or low
temperatures? Explain.
27. A reaction has ΔS of –122 J/(K x mol) and a ΔH of –78 kJ/mol.
Is this reaction spontaneous at 285˚C?
28. Calculate the standard free energy change ΔG for the reaction
between nitrogen and hydrogen, given the ΔH and ΔS values.
N2(g)
ΔHfº (kJ/mol)
0
ΔSº (J/K.mol)
191.5
+
3H2(g)
0
130.6

2NH3(g)
– 46.19
192.5
29. 2SO3(g)  2SO2(g) + O2
Calculate Keq for this reaction if the equilibrium concentrations are:
[SO2] = 0.42M, [O2] = 0.072M, [SO3] = 0.072M
30. In each of the following pairs choose the system with the higher entropy.
a. ___ A heap of loose stamps
or
___ in an album
b. ___ Ice cubes in their tray
or
___ ice cubes in a bucket
c. ___ 10 mL of water at 100˚C
or
___ 10 mL of steam at 100˚C
d. ___ the people watching the parade
or
___ a parade
31. Explain how the equilibrium position of this reaction is affected by
the following changes:
4HCl(g) + O2(g) ⇄ 2Cl2(g) + 2H2O(g)
a.
b.
c.
d.
Add Cl2
Remove O2
Increase pressure
Use a catalyst
__________
__________
__________
__________
Ch18 HW#8
32. When gently warmed, the element iodine will sublime: I2(s)  I2(g)
Is this process accompanied by an increase or decrease in entropy?
__________
33. Is there an increase or decrease in entropy when air is cooled
and liquefied (changed from a gas to a liquid)?
__________
34. Is the degree of disorder increasing or decreasing in these reactions?
a. H2(g) + Br2(l)  2HBr(g)
__________
b. CuSO4 * 5H2O(s)  CuSO4(s) + 5H2O(g)
__________
c. 2XeO3(s)  2Xe(g) + 3O2(g)
__________
35. Classify each of these systems as (A)always spontaneous,
(N) never spontaneous, or (D) depends on the relative magnitude
of the heat and entropy changes.
a. Entropy decreases, heat is released
__________
b. Entropy decreases, heat is absorbed
__________
c. Entropy increases, heat is absorbed
__________
d. Entropy increases, heat is released
__________
36. Calculate the standard entropy change associated with each reaction.
Refer to the table of standard entropies.
Standard Entropies (S0) H2O2(l) H2O(l) O2(g) H2O(g) Cl2(g) HCl(g)
92
69.94 205.0 188.7 223.0 186.7
a. 2H2O2(l)  2H2O(l) + O2(g)
37a. Calculate the change in free energy, ΔG, for the following reaction,
given the following enthalpies and entropies, to determine if the reaction
is spontaneous.
2H2O(g) + 2Cl2(g)  4HCl(g) + O2(g)
H2O(g)
Standard entropy,S0 (J/K.mol)
188.7
Standard enthalpy,ΔH0 (kJ/mol) -241.8
Cl2(g)
223.0
0
HCl(g)
186.7
-92.3
O2(g)
205.0
0
37b. Given the following enthalpy and free energy values, calculate
the free energy, ΔS, for the following reaction:
Mg3N2(c) + H2O(l)  Mg(OH)2(c) + NH3(g)
↑
ΔG
___ Mg3N2(c) + ___ H2O(l) 
ΔH
↓
ΔHf0
ΔG
Mg3N2(c)
-461
-422
H2O(l)
-286
-237
|
|
___ Mg(OH)2(c) + ___ NH3(g)
|
|
Mg(OH)2(c) 2NH3(g)
-925
-46.1
-834
-16.5
Ch18 Review
1. The industrial production of ammonia is described by this reversible reaction.
N2(g) + 3H2(g)  2NH3(g) + 92 kJ
What effect do the following changes have on the equilibrium position?
a. addition of heat
b. increase in pressure
c. addition of catalyst
d. removal of heat
e. removal of NH3
2. Predict the direction of the entropy change in each of the following reactions:
a. CaCO3(s)  CaO(s) + CO2(g)
b. NH3(g) + HC1(g)  NH4C1(s)
c. 2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g)
d. CaO(s) + CO2(g)  CaCO3(s)
3. Write the expression for the equilibrium constant for each of the following:
a. 2HBr(g)  H2(g) + Br2(g)
b. 2SO3(g)  2SO2(g) + O2(g)
c. CO2(g) + H2(g)  CO(g) + H2O(g)
d. 4NH3(g) + 5O2(g)  6H2O(g) + 4NO(g)
4. At 750C this reaction reaches equilibrium in a 1-L container:
H2(g) + CO2(g)  H2O(g) + CO(g)
An analysis of the equilibrium mixture gives the following results:
hydrogen 0.053 mol; carbon dioxide 0.053 mol; water 0.047 mol;
carbon monoxide 0.047 mol. Calculate Keq for the reaction.
5. Comment on the favorability of product formation in each of these reactions:
a. H2(g) + F2(g)  2HF(g)
Keq = 1x1013
b. SO2(g) + NO2(g)  NO(g) + SO3(g)
Keq = 1x102
c. 2H2O(g)  2H2(g) + O2(g)
Keq = 6x10-28
6. The standard entropies are given below for some substance at 25C.
KBrO3(s), S0 = 149.2 J/K.Mol KBr(s), S0 = 96.4 J/K.Mol
O2(g), S0 = 205.0 J/K.Mol Calculate ΔS0 for this reaction:
KBrO3(s)  KBr(s) + 3/2 O2(g)
7. From the data, calculate the standard free energy change for
the following reaction and say whether the reaction is spontaneous
or nonspontaneous:
Standard Gibbs Free Energies
of Formation (Gf) (in kJ/mol)
H2S(g)
H2(g)
S(s)
-33.02
0.0
0.0
H2S(g)  H2(g) + S(s)