5 THERMOCHEMISTRY & EQUILIBRIUM

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THERMO-CHEMISTRY

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THERMO-CHEMISTRY

 WITH THE EXCEPTION OF ENERGY FROM

THE SUN, MOST OF THE ENERGY USED IN OUR DAILY LIVES COMES FROM CHEMICAL REACTIONS.

 THE COMBUSTION OF GASOLINE,

ELECTRICITY FROM COAL, BATTERIES, AND PHOTOSYNTHESIS TO NAME A FEW.

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THERMO-CHEMISTRY

 THERMODYNAMICS: THE STUDY OF

ENERGY AND ITS TRANSFORMATIONS.

 THE RELATIONSHIPS BETWEEN CHEMICAL

REACTIONS AND ENERGY CHANGES THAT INVOLVE HEAT, A PORTION OF

THERMODYNAMICS CALLED THERMOCHEMISTRY.

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THERMO-CHEMISTRY

 ENERGY: THE CAPACITY TO DO WORK OR

TRANSFER HEAT.

 WORK: THE ENERGY USED TO CAUSE AN

OBJECT TO MOVE AGAINST A FORCE.

 HEAT: THE ENERGY USED TO CAUSE THE

TEMPERATURE OF AN OBJECT TO INCREASE.

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THERMO-CHEMISTRY

 KINETIC ENERGY: THE ENERGY OF

MOTION.

 THE MAGNITUDE OF THE KINETIC

ENERGY, E_k, OF AN OBJECT DEPENDS ON ITS MASS, m, AND SPEED, v.

 E

k = ½ mv2

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THERMO-CHEMISTRY

 IN CHEMISTRY, WE ARE INTERESTED IN

THE KINETIC ENERGY OF ATOMS AND

MOLECULES. ALTHOUGH TOO SMALL TO BE SEEN, THESE PARTICLES HAVE MASS AND ARE IN MOTION AND, THEREFORE, POSSESS KINETIC ENERGY.

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THERMO-CHEMISTRY

 ALL OTHER KINDS OF ENERGY, LIKE THE

ENERGY STORED IN A CHEMICAL BOND, HAS POTENTIAL ENERGY . POTENTIAL ENERGY IS STORED ENERGY.

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ENERGY

 AN EXAMPLE IS OCTANE C₈H₁₈

 OCTANE IS ONE OF THE MAIN

COMPONENTS OF GASOLINE.

 THE CHEMICAL POTENTIAL ENERGY OF

OCTANE RESULTS FROM THE

ARRANGEMENT OF THE CARBON AND

THERMOCHEMISTRY

 SO, IF OCTANE HAS A MOLECULAR

FORMULA OF C₈H₁₈ AND IS A SATURATED HYDROCARBON THAT IS COVALENTLY

BONDED; DRAW IT’S STRUCTURAL FORMULA…..

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ENERGY

 WHEN GASOLINE BURNS IN A CARS

ENGINE, SOME OF THE OCTANES STORED ENERGY IS CONVERTED TO WORK IN

MOVING THE PISTONS, WHICH ULTIMATELY MOVES THE CAR.

 MUCH OF THE POTENTIAL ENERGY OF

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ENERGY

 A WELL TUNED CAR CAN ONLY CONVERT

THERMOCHEMISTRY

 A 4:1 RATIO OF H-C IN GASOLINE YIELDS

52 KJ/G ENERGY CONTENT AND RELEASES 1.2 MOLES OF CO₂.

 A 1:1 RATIO OF H-C IN COAL YIELDS

39.3 KJ/G ENERGY CONTENT AND RELEASES 2 MOLES OF CO₂.

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THERMO-CHEMISTRY

 THE SI UNIT FOR ENERGY IS THE JOULE

(J).

 BECAUSE A JOULE IS NOT A LARGE

AMOUNT OF ENERGY, WE OFTEN USE

KILOJOULES (kJ) IN DISCUSSING THE

ENERGIES ASSOCIATED WITH CHEMICAL REACTIONS.

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THERMO-CHEMISTRY

 A calorie (c) SMALL “c”, IS AN INFORMAL

NON-SI UNIT USED FOR HEAT ENERGY.

 A calorie (cal) IS DEFINED AS THE

AMOUNT OF ENERGY REQUIRED TO RAISE THE TEMPERATURE OF ONE GRAM OF

WATER BY ONE DEGREE CELSIUS.

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THERMO-CHEMISTRY

 A calorie (cal) IS NOW DEFINED IN TERMS

OF THE JOULE.

 1 cal = 4.184 J

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THERMO-CHEMISTRY

 A RELATED ENERGY UNIT IN NUTRITION IS

THE NUTRITIONAL CALORIE (NOTE THE CAPITOL “C”)

 1 CAL = 1000 cal = 1 kcal

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ENERGY

 WHEN HUMANS BREAK DOWN SUGAR

AND FAT FORMING CO₂ AND H₂O, THESE EXOTHERMIC REACTIONS GENERATE

ENERGY

 THE HUMAN BODY IS PRIMARILY C,H,O,N

WITH TRACE AMOUNTS OF OTHER ELEMENTS.

 O: 10 $, C: 18 CENTS, H: 2 $, K: 104 $,

Na: 30 $, Rb: 8 $, AND F: 6 $

 YOU ARE WORTH ABOUT 160 DOLLARS

ENERGY

 THE INTERNAL COMBUSTION ENGINE

TURNING CHEMICAL ENERGY INTO MECHANICAL ENERGY IS 20-40% EFFICIENT.

 OUR BODIES ARE 20-30 % EFFICIENT. WE

NEED ~2000 CALORIES PER DAY AND GIVE OFF THE EQUIVALENT OF 100

WATTS OF HEAT PER HOUR.

ENERGY

 CALCULATE THE NUMBER OF CALORIES

YOU REQUIRE:

 WOMEN: 655 + (4.35 X WEIGHT) + (4.7 X

HEIGHT) – (4.7 X AGE)

 MEN: 66 + (6.23 X WEIGHT) + (12.7 X

HEIGHT) – (6.8 X AGE)

THERMOCHEMISTRY

REMEMBER, IN CHEMICAL REACTIONS

BONDS ARE BROKEN IN THE REACTANTS AND NEW BONDS ARE MADE IN THE

PRODUCTS.

THE ENERGY FOUND IN A C-H BOND IS

410 KJ PER MOLE. IN A C-C BOND 347, IN A C=C BOND 611.

THERMOCHEMISTRY

 THE COMBUSTION OF CH₄, METHANE.

 CH

4 + 2 O2 YIELDS CO2 + 2 H2O

 ENERGY INPUT IS REQUIRED TO BREAK

THE BONDS (ENDOTHERMIC PROCESS) (+)

 ENERGY IS RELEASED ON FORMING NEW

BONDS (EXOTHERMIC PROCESS) (+)

THERMOCHEMISTRY

 CHEMICAL BONDS STORING ENERGY:

 CHEMICAL BONDS POSSESS POTENTIAL

ENERGY. REMEMBER THAT ELECTRONS HAVE QUANTUMS OF ENERGY.

 ELECTRONS HAVE THE GROUND STATE

AND THE EXCITED STATE (HIGHER POTENTIAL ENERGY)

 WHEN ELECTRONS FALL TO A LOWER

ENERGY LEVEL, THEY GIVE OFF ENERGY (EMISSION)

THERMOCHEMISTRY

 WHEN CH₄ METHANE FORMS, THE

VALENCE ELECTRONS END UP IN A

STABLE LOWER ENERGY C-H BOND THAT ARE RELATIVELY STRONG AND INERT.

 IF YOU ADD ENERGY TO METHANE IN THE

FORM OF A FLAME OR SPARK IN THE

PRESENCE OF OXYGEN, THE C-H BONDS BREAK.

THERMOCHEMISTRY

 ONCE THE BONDS BREAK (REACTANTS),

THEY REARRANGE TO CO₂ AND H₂O

(PRODUCTS); THE EXCESS ENERGY 794 KJ/MOLE IS RELEASED AS HEAT ENERGY AND CAN BE USED TO COOK FOOD.

THERMOCHEMISTRY

 IN BIOCHEMISTRY, CELLULAR

RESPIRATION PROVIDES ENERGY TO

BREAK THE BONDS OF C₆H₁₂O₆ TO FORM STRONGER BONDS OF CO₂ AND H₂O.

PRODUCES 2870 KJ/MOLE.

 C₆H₁₂O₆ + 6 O₂ YIELDS 6 CO₂ + 6 H₂O

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THERMO-CHEMISTRY

 WHEN ANALYZING ENERGY CHANGES, WE

FOCUS ON A WELL DEFINED PART OF THE UNIVERSE TO KEEP TRACK OF THE

ENERGY CHANGES THAT OCCUR.

 THE PORTION WE SINGLE OUT FOR STUDY

IS CALLED THE SYSTEM; EVERYTHING ELSE IS CALLED THE SURROUNDINGS.

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THERMO-CHEMISTRY

 WHEN WE STUDY THE ENERGY CHANGE

OF A CHEMICAL REACTION, THE

REACTANTS AND PRODUCTS ARE THE

SYSTEM. THE CONTAINER AND

EVERYTHING ELSE BEYOND IT ARE THE SURROUNDINGS.

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THERMO-CHEMISTRY

 THE SYSTEMS WE CAN MOST READILY

STUDY IN THERMOCHEMISTRY ARE

CLOSED SYSTEMS, SYSTEMS THAT CAN EXCHANGE ENERGY BUT NOT MATTER WITH THEIR SURROUNDINGS

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THERMO-CHEMISTRY

 THE FIRST LAW OF

THERMODYNAMICS, ALSO CALLED THE LAW OF CONSERVATION OF ENERGY,

STATES THAT ENERGY IS CONSERVED. THE ENERGY OF THE UNIVERSE IS

CONSTANT. ENERGY CAN NIETHER BE

CREATED OR DESTROYED, BUT IT CAN BE CHANGED TO OTHER FORMS.

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THERMO-CHEMISTRY

 ENDOTHERMIC: A PROCESS OCCURS IN

WHICH THE SYSTEM ABSORBS HEAT. DURING AN ENDOTHERMIC PROCESS, SUCH AS THE MELTING OF ICE, HEAT FLOWS INTO THE SYSTEM.

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THERMO-CHEMISTRY

 A PROCESS IN WHICH A SYSTEM LOSES

HEAT IS CALLED EXOTHERMIC. DURING AN EXOTHERMIC PROCESS, SUCH AS THE COMBUSTION OF GASOLINE, HEAT EXITS OR FLOWS OUT OF THE SYSTEM INTO THE SURROUNDINGS.

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THERMO-CHEMISTRY

 EVERY CHEMICAL REACTION AND

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THERMO-CHEMISTRY

 WHEN YOU REMOVE THE PLASTIC WRAP

FROM A HEAT PACK/HAND WARMER,

OXYGEN ENTERS THE PACK AND REACTS WITH IRON IN THE PACK IN AN

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THERMO-CHEMISTRY

 THIS IS REPRESENTED BY THE

FOLLOWING EQUATION:

 4Fe + 3O

2 YIELDS 2Fe2O3 + 1625 kJ

 ENERGY IS SHOWN AS A PRODUCT OF

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THERMO-CHEMISTRY

 IF YOU DISSOLVE AMMONIUM NITRATE

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THERMO-CHEMISTRY

 THE EQUATION FOR THIS REACTION IS:

 27 kJ + NH₄NO₃ YIELDS NH₄+ + NO₃

- ENERGY IS ON THE REACTANT SIDE

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THERMO-CHEMISTRY

 IN ORDER TO MEASURE AND STUDY

ENERGY CHANGES THAT ACCOMPANY CHEMICAL REACTIONS, WE NEED TO

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THERMO-CHEMISTRY

 ENTHALPY (H): THE HEAT CONTENT OF

A SYSTEM AT CONSTANT PRESSURE.

 WE ARE ACTUALLY MEASURING THE

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THERMO-CHEMISTRY

 THE CHANGE IN ENTHALPY OF REACTION

IS CALLED ENTHALPY (HEAT) OF REACTION AND IS DESIGNATED BY:



^H

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THERMO-CHEMISTRY

rxn is the difference between the enthalpy

of the substance that exist at the end of the reaction and the enthalpy of the substance at the start.



^H

rxn

= H

– H

 _OR



^H

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THERMO-CHEMISTRY

 Remember the heat pack reaction:

 4Fe(s) + 3O

2(g) yields 2Fe2O3(s) + 1625 kJ

 According to this equation, as the

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THERMO-CHEMISTRY

 So, in this instance,

H

products

is <

H

.

 Enthalpy changes for exothermic

reactions are always negative.

 In our heat pack reaction:



^H

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THERMO-CHEMISTRY

 The cold pack reaction:

 27kJ + NH₄NO₃(s) yields NH₄+ (aq) + NO₃

-(aq)



^H

products

> ^H



A positive ^H

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THERMO-CHEMISTRY

 Enthalpy changes for endothermic

reactions is always positive.



^H

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THERMO-CHEMISTRY

 The heat pack and ice pack reactions are

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THERMO-CHEMISTRY



4Fe(s) + 3O

2

(g) YIELDS

2Fe

O

(s) ^H = -1625kJ



NH

4

NO

YIELDS NH

+

+ NO

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THERMO-CHEMISTRY

 THERMOCHEMICAL EQUATION: A

BALANCED CHEMICAL EQUATION THAT INCLUDES THE PHYSICAL STATES OF THE REACTANTS AND PRODUCTS AND THE

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THERMO-CHEMISTRY

 C

6H12O6(s) + 6O2(g) YIELDS 6CO2(g) +

6H₂O(l) ^H_COMB = -2808 kJ

 Is this an endothermic or exothermic

reaction?

 Exothermic

 How did we know?

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THERMO-CHEMISTRY

 IS IT A PROPERLY WRITTEN

THERMOCHEMICAL EQUATION?

 YES. IT IS BALANCED, THE PHYSICAL

STATES ARE GIVEN, AND ^H IS RECORDED.

 THE comb STANDS FOR COMBUSTION.

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THERMO-CHEMISTRY

 The energy released -2808 kJ is the

enthalpy of combustion.

 The enthalpy (heat) of combustion

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THERMO-CHEMISTRY

 Look at the following:

 C₆H₁₂O₆(s) + 6O₂(g) yields 6CO₂(g) +

6H₂O(l) ^H_comb= -2808 kJ

 Question:

 How much heat is evolved when 54

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THERMO-CHEMISTRY

 HOW MANY GRAMS OF C₆H₁₂O₆ ARE IN

ONE MOLE?

 180 GRAMS

 CHANGE 54 GRAMS OF C

6H12O6 TO

MOLES:

 54g / 180 g

 = .300 MOLES C

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THERMO-CHEMISTRY

 .300 MOL C₆H₁₂O₆ X 2808 kJ / 1 mole

C₆H₁₂O₆

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THERMO-CHEMISTRY

 The compound methanol has a molecular

formula of CH₃OH and ^H_fus 3.22 kJ.

 Calculate the heat required to melt 25.7 g

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THERMO-CHEMISTRY

 Change 25.7 g to moles:

 25.7 g CH

3OH / 32 g CH3OH

 = .803 moles CH

3OH

 .803 moles x 3.22 kJ / 1 moles CH

3OH

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THERMO-CHEMISTRY

 The molecular formula for ammonia is

NH₃. ^H = 23.3 kJ

 How much heat evolved when 275 g of

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THERMO-CHEMISTRY

 275 g NH₃ / 17g NH₃

 = 16 moles NH

3

 16 mol NH

3 x 23.3 kJ / 1 mol NH3

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THERMO-CHEMISTRY

 The molecular formula for methane is

CH₄. its ^H = - 891 kJ.

 What mass of methane must be burned

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THERMO-CHEMISTRY

 12,888 kJ / 891 kJ = 14.46 kJ

 One mole CH

4 = 16 g

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THERMO-CHEMISTRY

 SPECIFIC HEAT: THE SPECIFIC HEAT OF

ANY SUBSTANCE IS THE AMOUNT OF HEAT REQUIRED TO RAISE THE

TEMPERATURE OF ONE GRAM OF THAT SUBSTANCE BY ONE DEGREE CELSIUS.

 DIFFERENT SUBSTANCES HAVE

DIFFERENT COMPOSITIONS, EACH

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THERMO-CHEMISTRY

 WATER HAS A SPECIFIC HEAT OF 4.184 J,

ETHANOL 2.44, IRON 0.449, AND GOLD 0.129.

 IF THE TEMPERATURE OF WATER IS TO

RISE BY ONE DEGREE, 4.184 J MUST BE ABSORBED BY EACH GRAM OF WATER.

 ONLY 0.129 J IS REQUIRED TO RAISE THE

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THERMO-CHEMISTRY

 BECAUSE OF ITS HIGH SPECIFIC HEAT,

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THERMO-CHEMISTRY

 THE AMOUNT OF ENERGY ABSORBED OR

RELEASED CAN BE CALCULATED BY THE FORMULA:

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THERMO-CHEMISTRY



q = heat absorbed or

released



c = specific heat of

substance



m = mass of sample in

grams



^T = final temp.



T

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THERMO-CHEMISTRY

 GIVEN THE FOLLOWING INFO:

 A 5 g PIECE OF LEAD AT 850 CELSIUS AND

A SPECIFIC HEAT OF 0.129 J.

 A 5 g PIECE OF ALUMINUM AT 650

CELSIUS AND A SPECIFIC HEAT OF 0.897 J.

 THERMAL EQUILIBRIUM OF THE TWO

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THERMO-CHEMISTRY



A) CALCULATE “q” FOR EACH

METAL.



B) WHICH METAL LOSES MORE

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THERMO-CHEMISTRY

 GIVEN THE FOLLOWING:



A pond made of 14,500 kg of

granite rock contains 22,500 kg of

water. The specific heat of water

is 4.184 J . The specific heat of

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THERMO-CHEMISTRY



22,500 kg of water is 2.25 x 10

g



14,500 kg of granite is 1.45 x 10

g.

 Because each substance has its own

specific heat, the amount of heat

absorbed and released are calculated separately.

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THERMO-CHEMISTRY



The specific heat of ethanol is

2.44 J. If the temperature of

34.4 g of ethanol increases from

25

C to 78.8

C, how much heat

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THERMO-CHEMISTRY



In the construction of bridges,

gaps must be left between

adjoining steel beams to allow

for expansion and contraction of

the metal due to heating and

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THERMO-CHEMISTRY



The temperature of a sample of

iron with a mass of 10g changed

from 50.4

C to 25.0

C with the

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THERMO-CHEMISTRY



A 155g sample of an unknown

substance was heated from

25

C to 40.0

C. In the process,

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THERMO-CHEMISTRY

 HEAT CHANGES THAT OCCUR DURING

CHEMICAL AND PHYSICAL PROCESSES

CAN BE MEASURED ACCURATELY USING A CALORIMETER.

 CALORIMETER: INSULATED DEVICE

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THERMO-CHEMISTRY



Using a calorimeter to

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THERMO-CHEMISTRY



Example: You put 125 g of water in

a calorimeter with an initial temp of

25.6

C. You heat 50g of unknown

metal sample to a temp of 115

C

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THERMO-CHEMISTRY



Both the water and metal have

attained a final temp of 29.3

C.

Assuming no heat is lost to the

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THERMO-CHEMISTRY

 FIRST, CALCULATE THE QUANTITY OF

HEAT LOST BY THE METAL OR HEAT GAINED BY THE WATER USING THE:



q = c x m x ^t



(4.184)(125)(3.7)

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THERMO-CHEMISTRY

 THE HEAT LOST BY THE METAL IS 1935 J.

 HOW DID WE SOLVE FOR THE SPECIFIC

HEAT OF THE METAL EARLIER?



C

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THERMO-CHEMISTRY

 WHAT IS q?

 1935 J

 WHAT IS m?

 50g

 WHAT IS ^t

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THERMO-CHEMISTRY

 SOLVE FOR THE SPECIFIC HEAT OF THE

UNKNOWN METAL:



(1935 J) / (50)(85.7)



C

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THERMO-CHEMISTRY

 CALCIUM HAS A SPECIFIC HEAT OF .647 J,

STRONTIUM .301 J, AND BARIUM .204 J.



A piece of metal with a mass of 4.68 g

absorbs 256 J of heat when its temp

increases by 182

C. Which of the metals

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THERMO-CHEMISTRY



(4.68g) (182

C) = 851



256 J / 851



= .301 SPECIFIC HEAT



THE UNKNOWN METAL IS

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THERMO-CHEMISTRY

 HESS’S LAW: IF A REACTION IS CARRIED

OUT IN A SERIES OF STEPS, ^H FOR THE OVERALL REACTION WILL EQUAL THE

HESS’S LAW

 HESS’S LAW PROVIDES A USEFUL MEANS

OF CALCULATING ENERGY CHANGES THAT ARE DIFFICULT TO MEASURE DIRECTLY.

 REGARDLESS OF THE # OF STEPS OF A

REACTION, TOTAL ENTHALPY CHANGE FOR THE REACTION IS THE SUM OF ALL CHANGES.

HESS’S LAW

 EXAMPLE #1

 HYDROGEN GAS CAN BE GENERATED BY

THE REACTION OF CARBON AND WATER.

 C

(s) + 2 H2O(g) YIELDS CO2(g) + 2 H2(g)

HESS’S LAW

 THIS REACTION REQUIRES THE INPUT OF

90.1 kj OF HEAT FOR EVERY MOLE OF CARBON CONSUMED.

HESS’S LAW

 WHEN HEAT IS ABSORBED DURING A

REACTION, THE QUANTITY OF HEAT IS A POSITIVE NUMBER….

 q > 0 FOR AN ENDOTHERMIC REACTION

HESS’S LAW

 WHEN HEAT IS EVOLVED (RELEASED),

THE REACTION IS EXOTHERMIC…

 q < 0 FOR AN EXOTHERMIC REACTION

HESS’S LAW

 WHERE DOES THE INPUT ENERGY OF

90.1 kj GO WHEN THE REACTION OCCURS?

HESS’S LAW

 TO COMPARE THE ENERGY AVAILABLE IN

EACH FUEL, WE CAN MEASURE THE HEAT EVOLVED IN THE COMBUSTION OF EACH FUEL WITH ONE MOLE OF OXYGEN GAS.

HESS’S LAW

 C(s) + O

2(g) YIELDS CO2(g)

 q = - 393.5 kj

HESS’S LAW

 2 H

2(g) + O2(g) YIELDS 2 H2O(g)

 q = - 483.6 kj

HESS’S LAW

 MORE ENERGY IS AVAILABLE FROM THE

COMBUSTION OF THE HYDROGEN THAN FROM COMBUSTION OF THE CARBON.

 REMEMBER THAT THE CONVERSION OF

THE CARBON FUEL TO HYDROGEN FUEL REQUIRES THE INPUT OF ENERGY, 90.1 kj

HESS’S LAW

 THE HEAT INPUT IN THE EQUATION

(90.1kj), IS THE DIFFERENCE BETWEEN THE HEAT EVOLVED -393.5 kj IN THE

COMBUSTION OF CARBON AND THE HEAT EVOLVED -483.6 kj IN THE COMBUSTION OF HYDROGEN.

HESS’S LAW

 TAKE THE COMBUSTION OF CARBON AND

ADD THE REVERSE OF THE COMBUSTION OF HYDROGEN:

 C(s) + O

2(g) YIELDS CO2(g)

 2 H₂O(g) YIELDS 2 H₂(g) + O₂(g)

 NOW, COMBINE THESE REACTIONS….

HESS’S LAW

 C(s) + O₂(g) + 2H₂O(g)  CO₂(g) + 2 H₂(g) +

O₂(g)

 CANCEL THE O2(g) FROM BOTH SIDES.

 YOU END UP WITH THE ORIGINAL REACTION:

C(s) + 2 H₂O(g)  CO₂(g) + 2 H₂(g)

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THERMO-CHEMISTRY

 AS AN EXAMPLE: THE COMBUSTION OF

METHANE GAS CH₄ (g), TO FORM CO₂ (g) AND LIQUID WATER CAN BE THOUGHT OF AS OCCURRING IN TWO STEPS.

 1) THE COMBUSTION OF CH

4 (g) TO FORM

CO₂ (g) AND GASEOUS WATER, H₂O (g).

 2) THE CONDENSATION OF GASEOUS

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THERMO-CHEMISTRY

 THE ENTHALPY CHANGE FOR THE

OVERALL PROCESS IS SIMPLY THE SUM OF THE ENTHALPY CHANGES FOR THESE

TWO STEPS:

 CH

4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)

^H = -802 kj

 2 H₂O (g)  2 H₂O (l)

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THERMO-CHEMISTRY

 CH

4 (g) + 2 O2 (g) + 2 H2O (g)  CO2 (g)

+ 2 H₂O (l) + 2 H₂O (g) ^H = - 890 kj

 WATER OCCURS ON BOTH SIDES AND

CAN BE CANCELED:

 THE NET EQUATION:

 CH

4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)

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THERMO-CHEMISTRY

 HESS’S LAW PRACTICE PROBLEM:

HESS’S LAW

 STUDENT PROBLEM: BONUS POINTS

FOR CORRECT ANSWER

 METHYL PROPAN CAN BE CONVERTED TO

PRODUCE DIESEL AND JET FUEL. THE FIRST STEP IN THE PROCESS IS THE PRODUCTION OF 2-METHYL PROPENE.

 C

4H10O(l)  C4H8(g) + H2O(g)

HESS’S LAW

 USE THE FOLLOWING DATA TO

CALCULATE THE ENTHALPY CHANGE:

 1) 4 C(s) + 5 H

2(g) + ½ O2(g)  C4H10O(l)

^ H -335 kj

 2) 4 C(s) + 4 H₂(g)  C₄H₈(g) ^ H -17 kj

 3) H₂(g) + ½ O₂(g)  H₂O(g) ^H -242

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CHEMICAL EQUILIBRIUM

 CHEMICAL EQUILIBRIUM OCCURS WHEN

OPPOSING REACTIONS ARE PROCEEDING AT EQUAL RATES.

 THE RATE AT WHICH PRODUCTS ARE

FORMED FROM REACTANTS EQUALS THE RATE AT WHICH REACTANTS ARE FORMED FROM PRODUCTS.

 CONCENTRATIONS CEASE TO CHANGE,

106 106

CHEMICAL EQUILIBRIUM

 EQUILIBRIUM STATE: A MIXTURE OF

REACTANTS AND PRODUCTS WHOSE

CHEMICAL EQUILIBRIUM

 IF YOU ALLOW A REACTION TO REACH

EQUILIBRIUM AND THEN MEASURE THE EQUILIBRIUM CONCENTRATIONS OF

EVERYTHING, YOU CAN COMBINE THESE CONCENTRATIONS INTO AN EXPRESSION KNOWN AS AN EQUILIBRIUM

CONSTANT.

CHEMICAL EQUILIBRIUM

 LOOK AT A GENERAL EQUATION:

 Aa + Bb YIELDS _{(both directions)} Cc + Dd

 (C)c(D)d

 Kc = ________

 (A)a(B)b

CHEMICAL EQUILIBRIUM

 N₂(g) + 3 H₂(g) YIELDS 2 NH₃(g)

 (NH

3)2

 Kc =

--- (N₂) (H₂)3

110 110

CHEMICAL EQUILIBRIUM

 HABER PROCESS: COMBINE N₂ AND H₂ AT A

PRESSURE OF SEVERAL HUNDRED

ATMOSPHERES AND A TEMPERATURE OF SEVERAL HUNDRED DEGREES CELSIUS.

 THE GASES REACT TO FORM AMMONIA UNDER

THESE CONDITIONS, BUT IN A CLOSED SYSTEM THE REACTION DOES NOT LEAD TO COMPLETE CONSUMPTION OF THE N₂ AND H₂.

 AT SOME POINT THE REACTION APPEARS TO

CHEMICAL EQUILIBRIUM

 PROBLEM: AT EQUILIBRIUM, A MIXTURE

OF N₂, H₂, AND NH₃ GAS AT 500 DEGREES CELSIUS IS DETERMINED TO CONSIST

OF .602 mol/L of N₂, .420 mol/L of H₂, AND .113 mol/L of NH₃. WHAT IS THE EQUILIBRIUM CONSTANT FOR THE

REACTION….

 N₂ + 3 H₂ YIELDS _{(BOTH DIRECTIONS)} 2 NH₃

CHEMICAL EQUILIBRIUM

 Kc =((.113)2) / ((.602)(.420)3)

 Kc = .286 mol/L

CHEMICAL EQUILIBRIUM

 PROBLEM: EQUILIBRIUM

CONCENTRATIONS ARE NO .062, H₂ .012, N₂ .019, AND H₂0 .138 FOR THE

FOLLOWING REACTION:

 2 NO + 2 H

2 YIELDS N2 + 2 H2O

 FIND Kc

CHEMICAL EQUILIBRIUM

 Kc = ((.019)(.138)2) / ((.062)2(.012)2)

 Kc = 653

CHEMICAL EQUILIBRIUM

 PROBLEM: A REACTION BETWEEN

GASEOUS SULFUR DIOXIDE AND OXYGEN GAS PRODUCES GASEOUS SULFUR

TRIOXIDE. THE CONCENTRATION OF SO₂ IS 1.5 mol/L, THE CONCENTRATION OF O₂ IS 1.25 mol/L, AND SO₃ IS 3.5 mol/L.

USING A BALANCED EQUATION,

CALCULATE THE EQUILIBRIUM CONSTANT FOR THIS SYSTEM.

CHEMICAL EQUILIBRIUM

 Kc = ((3.5)2 / ((1.5)2(1.25))

 Kc = 4.36

CHEMICAL EQUILIBRIUM

 PROBLEM: AT EQUILIBRIUM AT 2500 K,

HCl = .0625 mol/L , AND H₂ = Cl₂

= .00450 mol/L FOR THE REACTION H₂ + Cl₂ YIELDS 2 HCl.

 FIND THE VALUE FOR K.

CHEMICAL EQUILIBRIUM

 Kc = ((.0625)2 / ((.00450)(.00450))

 Kc = 193

CHEMICAL EQUILIBRIUM

PROBLEM: AN EQUILIBRIUM MIXTURE AT

425 C IS FOUND TO CONSIST OF 1.83 x 10-3

mol/L H₂, 3.13 x 10-3 mol/L OF I

2, AND

1.77 x 10-2 mol/L OF HI. CALCULATE THE

EQUILIBRIUM CONSTANT FOR THE REACTION H₂ + I₂ YIELDS 2 HI.

CHEMICAL EQUILIBRIUM

 Kc = ((1.77 x 10-2))2 /((1.83 x 10-3)(3.13 x

10-3))

 Kc = 54.7

CHEMICAL EQUILIBRIUM

 PROBLEM: NITRYL CHLORIDE, NO₂Cl IS IN

EQUILIBRIUM WITH NO₂ AND Cl₂.

 2 NO₂Cl YIELDS 2 NO₂ + Cl₂

 AT EQUILIBRIUM THE CONCENTRATION OF

THE SUBSTANCES ARE NOCl₂ = .00106 m, NO₂ = .0108 m, AND Cl₂ = .00538 m.

FROM THIS DATA CALCULATE THE EQUILIBRIUM CONSTANT K.

CHEMICAL EQUILIBRIUM

 Kc= ((.0108)2 (.00538)) / ((.00106))2

 Kc = .558

CHEMICAL EQUILIBRIUM

PROBLEM: A MIXTURE OF HYDROGEN AND NITROGEN IN A REACTION VESSEL IS

ALLOWED TO ATTAIN EQUILIBRIUM AT 472C THE EQULIBRIUM MIXTURE OF GASES WAS ANALYZED AND FOUND TO CONTAIN

0.1207m H₂, 0.0402m N₂, AND 0.00272m NH₃. FROM THESE DATA CALCULATE THE EQUILIBRIUM CONSTANT Kc .

N₂ + 3 H₂ YIELDS 2 NH₃

CHEMICAL EQUILIBRIUM

 Kc = ((.00272))2 / ((.0402)(.1207))3

 Kc = .105