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Ch18.1 – Reaction Rates Reaction Rates explained by collision theory. -must collide with the correct o rientation -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. - PowerPoint PPT Presentation
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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
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
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
monthmol
monthsmol 19.03.51
Ch18 HW#1 1 – 61. 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
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
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
5. Double arrows in equation:
Represent a reversible reaction.
Ex:
6.How do the amounts of reactants & products change once a reactionhas achieved equilibrium?
N2(g) + 3H2(g) 2NH2(g)
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
5. Double arrows in equation:
Represent a reversible reaction.
Ex:
6.How do the amounts of reactants & products change once a reactionhas achieved equilibrium?
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:
Ch18.1,18.2 Review Worksheet1. CANTC:
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:
Ch18.1,18.2 Review Worksheet1. CANTC:
ConcentrationSurface AreaNature of the reactantsTempCatalyst
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
Ch18.1,18.2 Review Worksheet1. CANTC:
ConcentrationSurface AreaNature of the reactantsTempCatalyst
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
Ch18.1,18.2 Review Worksheet1. CANTC:
ConcentrationSurface AreaNature of the reactantsTempCatalyst
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: __________
5. CaCO3(s) + heat CaO(s) + CO2(g)
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: __________
5. CaCO3(s) + heat CaO(s) + CO2(g)
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
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. 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: productsb. Heat is added: reactantsc. HCl is removed: productsd. Pressure is increased: no effecte. Temp decreased: products
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:
9. Br2(l) + heat Br2(g)
a. Br2(g) is added: reactantsb. Heat is added: productsc. Br2(g) is removed: productsd. Pressure is decreased: productse. 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: reactantsb. Heat is added: productsc. Br2(g) is removed: productsd. Pressure is decreased: productse. 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 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
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) + heatNot when there are no gases present, or equal numbers.
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
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) + heatNot 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 Reactantsc) Remove H2 Productsd) 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
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) + heatNot 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 Reactantsc) Remove H2 Productsd) Add H2O Productse) Catalyst No Changef) Raise temp and decrease pressure Both favor products
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
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) + heatNot 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 Reactantsc) Remove H2 Productsd) Add H2O Productse) Catalyst No Changef) Raise temp and decrease pressure Both favor products
9. Affect equilibrium: 2SO3(g) + CO2(g) + heat CS2(g) + 4O2(g)
a) Add CO2 Productsb) Add heat Productsc) Decrease pressure Productsd)Remove O2
e) Add catalyst
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) + heatNot 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 Reactantsc) Remove H2 Productsd) Add H2O Productse) Catalyst No Changef) Raise temp and decrease pressure Both favor products
9. Affect equilibrium: 2SO3(g) + CO2(g) + heat CS2(g) + 4O2(g)
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?
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?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 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?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!
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?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 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?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
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 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
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.
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) + 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) Guessa) 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)
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) Guessa) 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) Guessa) 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) Guessa) 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) + 890kJdisorder disorder disorder (no effect) high E high E high E high E exothermic (spont)e)Br2(l) Br2(g)
17) Guessa) 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) + 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)
Ch18.4 – Equilibrium Constants
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.
Ch18.4 – Equilibrium Constants
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)
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)
][][][
22
2
22
OHOH
Keq
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)
][][][
22
2
22
OHOH
Keq
][][][
22
2
BrNONOBrKeq
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)
][][][
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)
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
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
32
OHCHHCO
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
32
OHCHHCO
Keq
22
3
]][[][
HCOOHCH
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
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?)
9. Write the equilibrium expression for the following reaction, and solve for Keq
C(s) + CO2(g) 2CO(g)
(1.3) (2.25) (.5)Who’s favored?
10. Write the equilibrium expression for the following reaction, and solve for Keq
FeO(s) + CO(g) Fe(s) + CO2(g)
(1) (2) (.5) (.5) Who’s favored?
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?)
products favored9. Write the equilibrium expression for the following reaction, and solve for Keq
C(s) + CO2(g) 2CO(g)
(1.3) (2.25) (.5)Who’s favored?
10. Write the equilibrium expression for the following reaction, and solve for Keq
FeO(s) + CO(g) Fe(s) + CO2(g)
(1) (2) (.5) (.5) Who’s favored?
108]10][.25[.]60][.75[.
]][[]][[
2
2
22
22
eq
eq
K
COSnOCOSn
K
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?)
products favored9. Write the equilibrium expression for the following reaction, and solve for Keq
C(s) + CO2(g) 2CO(g)
(1.3) (2.25) (.5)Who’s favored?
reactants favored10. Write the equilibrium expression for the following reaction, and solve for Keq
FeO(s) + CO(g) Fe(s) + CO2(g)
(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
9. Write the equilibrium expression for the following reaction, and solve for Keq
C(s) + CO2(g) 2CO(g)
(1.3) (2.25) (.5)Who’s favored?
reactants favored
10. Write the equilibrium expression for the following reaction, and solve for Keq
FeO(s) + CO(g) Fe(s) + CO2(g)
(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)
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] [.1]2
Keq = = 11.9 [.25] [.15]3
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] [.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
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
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 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
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 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
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 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 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 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 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) (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