ch18.1 – reaction rates reaction rates explained by collision theory

Ch18.1 – Reaction RatesReaction 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 occurNature of the reactants – some substances are more reactive than othersTemperature – 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 occurNature of the reactants – some substances are more reactive than othersTemperature – higher temp= more KE A.E.Catalyst – Lowers activation energy

(provides an alternate path) R P

Two Types of Catalysts: 1. Heterogeneous catalyst-

A 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:

The decomposition of hydrogen peroxide is slow. Iodide ions help speed it up.

H202(l) H20(l) + 02(g)

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 At equilibrioum,Products favored

1% 99%

CaCO3 CaO + CO2 At equilibrioum,Reactants favored

99% 1%Ch18 HW#1 1 - 6

Ch18 HW#1 1 – 61. 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 – 61. Rate of reaction – how fast the reaction converts reactants into products






monthmol

monthsmol 19.03.51




Lower temp, Higher temp = higher KE slows rxn rate = higher rate of rxn

monthmol

monthsmol 19.03.51

4.How does each affect rate of reaction?a. Tempb. Concentrationc. Particle sized. Catalyst

5. Double arrows in equation:

6.How do the amounts of reactants & products change once a reactionhas achieved equilibrium?

4.How does each affect rate of reaction?a. Temp – higher temp = higher KE = faster rateb. Concentration – greater concentration = faster ratec. Particle size – smaller particles =more surface area = faster rated. Catalyst – lowers the activation energy = faster rate





Represent a reversible reaction.

Ex:


N2(g) + 3H2(g) 2NH2(g)



Represent a reversible reaction.

Ex:


The amounts stay constant. The rate of the forward reaction equals the rate of the reverse reaction.

N2(g) + 3H2(g) 2NH2(g)

Ch18.2 – Factors Affecting EquilibriumA 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 PrincipleIf 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 productsAdd N2: productsAdd heat: reactantsIncrease pressure: products

Ex2) CO2(g) + H2(g) + heat CO(g) + H2O(g)

Decrease temp: reactantsAdd heat: productsAdd H2O: reactantsIncrease 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 Worksheet1. CANTC:

2. What do catalysts do to speed up a reaction?

3. Draw catalyst path on energy diagram:

4. 2 types of catalysts:


ConcentrationSurface AreaNature of the reactantsTempCatalyst

2. What do catalysts do to speed up a reaction?

3. Draw catalyst path on energy diagram:

4. 2 types of catalysts:



2. What do catalysts do to speed up a reaction?Provide alternate path/lowers activation energy

3. Draw catalyst path on energy diagram: a.e.

4. 2 types of catalysts: RP



2. What do catalysts do to speed up a reaction?Provide alternate path/lowers activation energy

3. Draw catalyst path on energy diagram: a.e.

4. 2 types of catalysts: RP

Heterogeneous – provides a place for the reaction to occur

Homogeneous – becomes part of the reaction mechanismthen 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: __________


a. CaCO3(s) added: productsb. Heat added: productsc. CaO removed: productsd. Pressure increased: reactantse. 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: __________


a. CaCO3(s) added: productsb. Heat added: productsc. CaO removed: productsd. Pressure increased: reactantse. Temp decreased: reactants

6. 2H2(g) + O2(g) 2H2O(g) + heat

a. H2 is added: productsb. Heat is added: productsc. O2 is removed: reactantsd. Pressure is increased: productse. 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: productsb. Heat is added: reactantsc. HCl is removed: productsd. Pressure is increased: no effecte. Temp decreased: products

8. ____CH2(g) + ____ O2(g) ____CO2(g) + ___H2O(g) + 890kJ


7. H2(g) + Cl2(g) 2HCl(g) + 93.2kJ


8. 2 CH2(g) + 3 O2(g) 2 CO2(g) + 2 H2O(g) + 890kJ


7. H2(g) + Cl2(g) 2HCl(g) + 93.2kJ


8. 2 CH2(g) + 3 O2(g) 2 CO2(g) + 2 H2O(g) + 890kJ

a. CO2 is added: reactantsb. Heat is added: reactantsc. CO2 is removed: productsd. Pressure is increased: productse. 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:


a. Br2(g) is added: reactantsb. Heat is added: productsc. Br2(g) is removed: productsd. Pressure is decreased: productse. Temp decreased: reactants


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:


a. Br2(g) is added: reactantsb. Heat is added: productsc. Br2(g) is removed: productsd. Pressure is decreased: productse. Temp decreased: reactants


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 corkcreates 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 CO(g) + H2(g)

a) lower tempb) Increase pressurec) Remove H2

d) Add H2Oe) Catalystf) Raise temp and decrease pressure

9. Affect equilibrium: 2SO3(g) + CO2(g) + heat CS2(g) + 4O2(g)

a) Add CO2

b) Add heatc) Decrease pressured)Remove O2

e) Add catalyst


HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) + heatNot when there are no gases present, or equal numbers.


a) lower tempb) Increase pressurec) Remove H2

d) Add H2Oe) Catalystf) Raise temp and decrease pressure


a) Add CO2


e) Add catalyst




a) lower temp Reactants b) Increase pressure Reactantsc) Remove H2 Productsd) Add H2Oe) Catalystf) Raise temp and decrease pressure


a) Add CO2


e) Add catalyst




a) lower temp Reactants b) Increase pressure Reactantsc) Remove H2 Productsd) Add H2O Productse) Catalyst No Changef) Raise temp and decrease pressure Both favor products


a) Add CO2


e) Add catalyst






a) Add CO2 Productsb) Add heat Productsc) Decrease pressure Productsd)Remove O2

e) Add catalyst






a) Add CO2 Productsb) Add heat Productsc) Decrease pressure Productsd)Remove O2 Productse) Add catalyst No Change

Ch18.3 – Spontaneous Reactions2 things contribute to deciding if a reaction will proceed on its own:

Free Energy & Entropy

Free Energy – energy released from reactionsGasoline(g) + O2(g) CO2(g) + H2O(g) + heat

can do work

Exothermic reactions are usually spontaneousEndothermic 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 These will oppose each other.

Entropy: Order Disorder One will beat the other

Most Least order orderSolids ↔ Liquids ↔ GasesLeast Most energy energy

Ex) Are the following reactions spontaneous? Predicta) H2O(l) → H2O(g)

b) C(s) + O2 (g) → CO2(g) + 393.5 kJ

Ex) Are the following reactions spontaneous? Predicta) H2O(l) → H2O(g) more order → more disorderless energy ← more energy

b) C(s) + O2 (g) → CO2(g) + 393.5 kJ order disorder → disorder

less E more E ← More E (wins)

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 cardsb) 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?






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 body13) Products in one spontaneous reactions are more ordered then reactants.

Is the entropy change favorable or unfavorable?reactants products If this rxn is spontaneousless 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 Usedb) sugar cube dissolved in water or cube?

Dissolvedc) 1g salt crystal or 1g powdered. Powder








14) Higher entropy: a) new pack playing cards or used cards Usedb) sugar cube dissolved in water or cube?

Dissolvedc) 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) Guessa) 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.


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)



order order disorder (spont) low E low E high E (non-spont)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)




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)




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)




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) + 890kJdisorder disorder disorder (no effect) high E high E high E high E exothermic (spont)e)Br2(l) Br2(g)



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) + 890kJdisorder 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)


H2(g) + I2(g) 2HI(g)

At equilibrium: 2% 98%

Chemists don’t use % to indicate the position of equilibrium, instead use molarity and ratios.


H2(g) + I2(g) 2HI(g)

At equilibrium: 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

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)

]reactants[]products[

eqK

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: N2O4 2NO2

b) Write Keq:

c) Calc Keq:

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

][]N[

42

22

ONO

Keq 2.0]0045.0[]03.0[ 2

eqK

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)


2H2(g) + O2(g) 2H2O(g)




NO(g) + O3(g) O2(g) + NO2(g)

][][][

22

2

22

OHOH

Keq


2H2(g) + O2(g) 2H2O(g)




NO(g) + O3(g) O2(g) + NO2(g)

][][][

22

2

22

OHOH

Keq

][][][

22

2

BrNONOBrKeq


2H2(g) + O2(g) 2H2O(g)




NO(g) + O3(g) O2(g) + NO2(g)

][][][

22

2

22

OHOH

Keq

][][][

22

2

BrNONOBrKeq

]][[]][[

3

22

ONOONO

Keq

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)





]][[]][[

24

3

ClCHHClClCH

Keq





]][[]][[

24

3

ClCHHClClCH

Keq

]][[]][[

24

32

OHCHHCO

Keq





]][[]][[

24

3

ClCHHClClCH

Keq

]][[]][[

24

32

OHCHHCO

Keq

22

3

]][[][

HCOOHCH

Keq





]][[]][[

24

3

ClCHHClClCH

Keq

]][[]][[

24

32

OHCHHCO

Keq

22

3

]][[][

HCOOHCH

Keq

72

262

62

42

][][][][

OHCOHCO

Keq

8. Write the equilibrium expression for the following reaction, and solve for Keq

SnO2(s) + 2CO(g) Sn(s) + 2CO2(g)

(.25) (.10) (.75) (.60)(Which side of this reaction is favored?)


C(s) + CO2(g) 2CO(g)

(1.3) (2.25) (.5)Who’s favored?


FeO(s) + CO(g) Fe(s) + CO2(g)

(1) (2) (.5) (.5) Who’s favored?




products favored9. Write the equilibrium expression for the following reaction, and solve for Keq


(1.3) (2.25) (.5)Who’s favored?



(1) (2) (.5) (.5) Who’s favored?

108]10][.25[.]60][.75[.

]][[]][[

2

2

22

22

eq

eq

K

COSnOCOSn

K




products favored9. Write the equilibrium expression for the following reaction, and solve for Keq


(1.3) (2.25) (.5)Who’s favored?

reactants favored10. Write the equilibrium expression for the following reaction, and solve for Keq


(1) (2) (.5) (.5) Who’s favored?

108]10][.25[.]60][.75[.

]][[]][[

2

2

22

22

eq

eq

K

COSnOCOSn

K

085.0]25.2][3.1[

]5[.

]][[][

2

2

22

2

eq

eq

K

COCCOK



(1.3) (2.25) (.5)Who’s favored?

reactants favored



(1) (2) (.5) (.5) Who’s favored?

reactants favored

085.0]25.2][3.1[

]5[.

]][[][

2

2

22

2

eq

eq

K

COCCOK

125.0]2][1[]5][.5[.

]][[]][[ 2

eq

eq

K

COFeOCOFe

K

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?

Keq is too high, so reaction will proceed backwards towards the reactants, until it lowers to 0.11.

384.0]5.2[]2][2.1[][]][[

2

222

eq

eq

K

IClClI

K

Ch18.5 Solving Keq‘sEx1) 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 arethe 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 18-2218. 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 18-2218. 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)


Keq = [N2] [H2]3

19. Cale Keq N2(g) + 3H2(g) ↔ 2NH3(g)

[.25] [.15] [.10] [.1]2

Keq = = 11.9 [.25] [.15]3

20. Keq for 2HI(g) ↔ H2(g) + I2(g)


Keq = [N2] [H2]3

19. Cale Keq N2(g) + 3H2(g) ↔ 2NH3(g)

[.25] [.15] [.10] [.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 Pb) Keq = 0.003 Rc) Keq = 3.5 Pd) Keq = 6 x 10-4 R


Start: [X] [0] [0] Equilibrium: [x – 1.0] [.5] [.5]

21.Products or reactants?a) Keq = 1 x 102 Pb) Keq = 0.003 Rc) Keq = 3.5 Pd) Keq = 6 x 10-4 R


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 CalculationsEntropy (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 when1 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 CalculationsEntropy (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 when1 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/KΔSo = So

products – 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 – react480 – 626.2 = -145.2 J/K

Dec in disorder

Ch18 HW#6 23,24

Ch18 HW#6 23,2423. How does the magnitude of the standard entropy of an element

in gas state compare to that of the liquid state?

Ch18 HW#6 23,2423. 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 2424a) 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) (2mol)(69.94J/K·mol)466.2J/K 139.88J/K

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


= 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)






= 253.35 – 88.7 = +164.65 J/K (Inc)

b) 2H2(g)+ O2(g)→ 2H2O(l)


ΔS = 139.88 – 466.2 = -326.3 J/K (Dec)





= 253.35 – 88.7 = +164.65 J/K (Inc)

b) 2H2(g)+ O2(g)→ 2H2O(l)


ΔS = 139.88 – 466.2 = -326.3 J/K (Dec)



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) (1)(213.6)+(2)(188.7)

596J/K 591J/K

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) (1)(213.6)+(2)(188.7)

596J/K 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) (1)(213.6)+(2)(188.7)

596J/K 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) CO2(g)

C(s) O2(g) CO2(g)

Standard Heats of Formation, Hf0 0 kJ/mol 0 kJ/mol -393.5kJ/mol

Standard Entropies, S0 5.694J/mol.K 205J/mol.K 213.6J/mol.K

Ex4) Use ΔH & S to determine if rxn is spontaneous

2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6H2O(g)

Ch18 HW#7

C2H6(g) O2(g) CO2(g) H2O(g)

Hf0 -84.7kJ/mol 0 kJ/mol -393 kJ/mol -286kJ/mol

S0 68.2 J/mol.K 64.1 J/mol.K 63.8 J/mol.K 80.7 J/mol.K

Ch18 HW#725. Calculate the standard free energy change ΔG for the reaction between iron (III) oxide and carbon (graphite). Refer to the table of free energies.

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?

Fe2O3(s) C(s) (graphite) Fe(s) CO2(g)Standard Gibbs Free Energies of Formation ( Gf)

-741.0 0.0 0.0 -394.4

28. Calculate the standard free energy change ΔG for the reaction between nitrogen and hydrogen, given the ΔH and ΔS values.

N2(g) + 3H2(g) 2NH3(g)

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

ΔHfº (kJ/mol) 0 0 – 46.19

ΔSº (J/K.mol) 191.5 130.6 192.5

30. In each of the following pairs choose the system with the higher entropy.a. ___ A heap of loose stamps or ___ in an albumb. ___ Ice cubes in their tray or ___ ice cubes in a bucketc. ___ 10 mL of water at 100˚C or ___ 10 mL of steam at 100˚Cd. ___ 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. Add Cl2 __________b. Remove O2 __________c. Increase pressure __________d. 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.

a. 2H2O2(l) 2H2O(l) + O2(g)

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

H2O(g) Cl2(g) HCl(g) O2(g)Standard entropy,S0 (J/K.mol) 188.7 223.0 186.7 205.0Standard enthalpy,ΔH0 (kJ/mol) -241.8 0 -92.3 0

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)

Mg3N2(c) H2O(l) Mg(OH)2(c) 2NH3(g)

ΔHf0 -461 -286 -925 -46.1

ΔG -422 -237 -834 -16.5

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) ___ Mg(OH)2(c) + ___ NH3(g)

ΔH | ↓ |

Ch18 Review1. The industrial production of ammonia is described by this reversible reaction.

N2(g) + 3H2(g) 2NH3(g) + 92 kJWhat effect do the following changes have on the equilibrium position?a. addition of heatb. increase in pressurec. addition of catalystd. removal of heate. 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 750C 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 25C. 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 spontaneousor nonspontaneous:

H2S(g) H2(g) + S(s)

Standard Gibbs Free Energies H2S(g) H2(g) S(s)

of Formation (Gf) (in kJ/mol) -33.02 0.0 0.0

ch18.1 – reaction rates reaction rates explained by collision theory

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