Chapter 12

12-1: Chemical Reactions That Involve Heat

Suggested Reading:

Pages 381 - 387

Chemical Reactions Involve ENERGY

Changes in ENERGY result from bonds being broken and new bonds being formed.

Requires ENERGY

Breaking Bonds

Releases ENERGY

Bond Formation

Both absorption and release of energy occurs.

In a chemical reaction

We detect the net result.

Measure the temperature of the surroundings.

System: Reactants and Products

System & Surroundings

Surroundings: Solvent, container, atmosphere above the reaction, etc.

The study of the changes in heat in chemical reactions.

Thermochemistry

Types of Reactions

RELEASE HEAT!

Exothermic Reactions

CC33HH88(g)(g) + 5O + 5O22(g)(g)

3CO3CO22(g)(g) + 4H + 4H22OO(g)(g) + 2043 kJ + 2043 kJ

Combustion reactions are exothermic!

HEAT is listed as a product in the reaction! HEAT ENERGY

Exothermic Reactions Energy

needed to break bonds must be LESS THAN the energy released when new bonds are formed.

Surroundings will have a higher

temperature after the reaction!

NaOH(s) Na+(aq) + OH-(aq)

Beginning Temp of Surroundings: 25.4°C

Ending Temp of Surroundings: 29.5°C

Change in Temp of Surroundings: +4.1°C

+4.1°C

Means heat was GIVEN OFF

EXOTHERMIC

NaOH(s) Na+(aq) + OH-(aq)

+4.1°CCalculate HEAT per MOLE if you have 5.0 grams of NaOH to start.

5.0 g NaOH

40 g NaOH

1 mol NaOH 0.125 mol NaOH=

+4.1°C

0.125 mol NaOH= + 33°C per mole NaOH

Means HEAT is given off.

ABSORB HEAT!

Endothermic Reactions

CC (s)(s) + H + H22O O (g)(g) + 113 kJ + 113 kJ

COCO (g)(g) + H + H22 (g)(g)

HEAT is listed as a reactant in the reaction!

HEAT ENERGY

Endothermic Reactions The energy

needed to break bonds is GREATER THAN the energy released when new bonds are formed.

Surroundings will have a lower

temperature after the reaction!

Exo & Endo Demos

Fireworks

More about fireworks

Section 2Section 2

Heat & Enthalpy Heat & Enthalpy ChangesChanges

ENTHALPYENTHALPYENTHALPYENTHALPY

Heat content of a system at constant pressure.

Heat absorbed or released in a reaction depends on the difference in enthalpy.

ENTHALPYENTHALPYENTHALPYENTHALPY

Represented by capital H.

(delta) means a change or difference.

H = change in enthalpy.

ENTHALPY CHANGE ENTHALPY CHANGE HH ENTHALPY CHANGE ENTHALPY CHANGE HH

HRXN = HPRODUCTS - HREACTANTS

Heat of ReactionΔH

Exothermic ReactionsExothermic ReactionsExothermic ReactionsExothermic Reactions

ENTHALPY CHANGE IS NEGATIVE

Heat content of productsIs LOWER!

Heat content of reactantsIs HIGHER

HEAT IS RELEASED!

Endothermic ReactionsEndothermic ReactionsEndothermic ReactionsEndothermic Reactions

ENTHALPY CHANGE IS POSITIVE

Heat content of reactantsIs LOWER!

Heat content of productsIs HIGHER

HEAT IS ABSORBED!

ENTHALPY PROBLEMSENTHALPY PROBLEMSENTHALPY PROBLEMSENTHALPY PROBLEMS

Similar to “Stoich” problems.

The change in energy depends on the number of moles of the reactants.

Calculate the amount of heat Calculate the amount of heat released when 5.50 g of released when 5.50 g of methane reacts with excess methane reacts with excess oxygen.oxygen.

Calculate the amount of heat Calculate the amount of heat released when 5.50 g of released when 5.50 g of methane reacts with excess methane reacts with excess oxygen.oxygen.

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

ΔH°= -890kJ5.50 g CH4 1 mol CH4

16 g CH4

890 kJ

1 mol CH4

= 306 kJ

How much heat will be How much heat will be released when 6.44 g of released when 6.44 g of sulfur reacts with excess sulfur reacts with excess oxygen?oxygen?

How much heat will be How much heat will be released when 6.44 g of released when 6.44 g of sulfur reacts with excess sulfur reacts with excess oxygen?oxygen?

2S + 3O2 2SO3

ΔH°= -791.4 kJ

6.44 g S 1 mol S

32 g S

791.4 kJ

2 mol S= 79.6 kJ

Suggested Reading: Online

Pages 539-542

Suggested Reading: Online

Pages 539-542

Hess’s LawHess’s Law

HESS’S LAWHESS’S LAWHESS’S LAWHESS’S LAW

If a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the enthalpy changes for the individual steps.

Hess’s LawHess’s LawHess’s LawHess’s Law

Start Finish

Enthalpy is Path independent.

Both lines accomplished the same result, they went from start to finish. Net result = same.

Hess’s law can be used to determine the

enthalpy change for a reaction

that cannot be measured directly!

HNET = H1 + H2

N2(g) + O2(g) 2NO(g) ΔH1=

+181kJ2NO(g) + O2(g) 2NO2(g) ΔH2= -

113kJADD THEM UP

ALEGBRAICALLY

N2(g) + O2(g) 2NO(g) ΔH1=

+181kJ2NO(g) + O2(g) 2NO2(g) ΔH2= -

113kJN2(g) + 2O2(g)

First, add up the chemical

equations.

+ 2NO(g) 2NO(g) + 2NO2(g)

Notice that 2NO(g) is on both the reactants and

products side and can be cancelled

out.N2(g) + O2(g) 2NO(g) ΔH1=

+181kJ2NO(g) + O2(g) 2NO2(g) ΔH2= -

113kJN2(g) + 2O2(g) + 2NO(g) 2NO(g) +

2NO2(g)

Write the net equation:

N2(g) + 2O2(g) + 2NO(g) 2NO(g) + 2NO2(g)

N2(g) + 2O2(g) 2NO2(g)

HNET = H1 + H2

ΔH1= +181kJ

ΔH2= -113kJ

Apply Hess’s Law to calculate the enthalpy for the

reaction.

HNET = H1 + H2

ΔHNET = (+181kJ) + (-113kJ)

ΔHNET = +68kJOverall, the formation of NO2 from N2 and O2 is an

endothermic process, although one of the steps is exothermic.

ΔH

Reaction Progress

N2(g) + 2O2(g)

2NO(g) + O2(g)

2NO2(g)

ΔHNET = +68kJ

ΔH1 = +181kJ

ΔH2 = -113kJ

RULES for Hess’s Law Problems

1.If the coefficients are multiplied by a factor, then the enthalpy value MUST also be multiplied by the same factor.

2.If an equation is reversed, the sign of ΔH MUST also be reversed.

C(s) + ½O2(g) CO(g) ΔH1= -110.5kJCO(g) + ½O2(g) CO2(g) ΔH2= -283.0kJ

C(s) + O2(g) + CO(g) CO(g) + CO2(g)

Practice Problem: #1

C(s) + O2(g) CO2(g) HNET = H1 + H2

HNET = (-110.5kJ) + (-283.0kJ)

HNET = -393.5kJ

Net Equation

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g)

Practice Problem: #3

CH3OCH3(l) + 3O2(g) 2CO2(g) + 3H2O(g)

ΔH1= -1234.7kJ

ΔH2= -1328.3kJ

You have to REVERSE equation 2to get the NET equation.

DON’T forget to change the sign Of ΔH2

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g)

Practice Problem: #3

2CO2(g) + 3H2O(g) CH3OCH3(l) + 3O2(g)

ΔH1= -1234.7kJ

ΔH2= +1328.3kJ

C2H5OH(l) + 3O2(g) + 2CO2(g) + 3H2O(g) 2CO2(g) + 3H2O(g) + CH3OCH3(l) +

3O2(g)

Net EquationC2H5OH(l) CH3OCH3(l)

Net EquationC2H5OH(l) CH3OCH3(l)

HNET = H1 + H2

HNET = (-1234.7kJ) + (+1328.3kJ)

HNET = +93.6kJ

H2(g) + F2(g) 2HF(g) ΔH1= -542.2kJ

2H2(g) + O2(g) 2H2O(g) ΔH2= -571.6kJ

Practice Problem: #5

You have to REVERSE equation 2to get the NET equation.

DON’T forget to change the sign Of ΔH2

H2(g) + F2(g) 2HF(g) ΔH1= -542.2kJ

2H2O(g) 2H2(g) + O2(g) ΔH2= +571.6kJ

Practice Problem: #5

You will need to multiply the first equation by 2.

DON’T forget to multiply the ΔH by 2 also.

2H2(g) + 2F2(g) 4HF(g) ΔH1= -1084.4kJ2H2O(g) 2H2(g) + O2(g) ΔH2= +571.6kJ

Practice Problem: #5

Net Equation

2H2(g) + 2F2(g) + 2H2O(g) 4HF(g) + 2H2(g) + O2(g)2F2(g) + 2H2O(g) 4HF(g) +

O2(g)HNET = H1 + H2

HNET = (-1084.4kJ) + (+571.6kJ)

HNET = -512.8kJ

Suggested Online

Review Pages 530-

545

Suggested Online

Review Pages 530-

545

Section 4Section 4

CalorimetryCalorimetry

CalorimetryCalorimetryCalorimetryCalorimetryThe study of heat flow and heat measurement.

Heat CapacityHeat CapacityHeat CapacityHeat Capacity

The amount of heat needed to raise the temperature of an object by 1 Celsius degree.

Specific HeatSpecific HeatSpecific HeatSpecific Heat

The heat capacity of 1 gram of a substance

Specific Heat of liquid water is 4.184 J/gºC

4.184 J = 1 Calorie

1 Calorie = 1000 calories = 1 kilocalorie

1 Food Calorie = 1000 calories

Calorimetry ExperimentsCalorimetry ExperimentsCalorimetry ExperimentsCalorimetry Experiments

Determine the heats of reaction (ENTHALPY CHANGES) by making accurate measurements of temperature changes using a calorimeter.

Scientists use q to denote measurements made in a calorimeter.

Heat transferred in a reaction is EQUAL, but OPPOSITE in sign to heat absorbed by the surroundings.

qrxn = - qsur

qsur = m x Cp x (Tf –Ti)

Mass of WaterSpecific heat

of Water

Temperature change

Heat Gain/Lost

Practice Problem: #1When a 12.8g sample of KCl

dissolves in 75.0g of water in a calorimeter, the temperature drops from 31.0ºC to 21.6ºC. Calculate H for the process.

KCl(s) K+(aq) + Cl-(aq)

First, calculate qsur and then calculate H.

Practice Problem: #1When a 12.8g sample of KCl

dissolves in 75.0g of water in a calorimeter, the temperature drops from 31.0ºC to 21.6ºC. Calculate H for the process.

KCl(s) K+(aq) + Cl-(aq) qsur = m x Cp x (Tf –Ti)

qsur = 75.0g x 4.184 J/gºC x (21.6 ºC –31.0 ºC )

qsur = -2949.72 J

qsur is negative (as expected)

based on the temperature drop of the water.

KCl(s) K+(aq) + Cl-(aq)

qrxn = - qsur = +2949.72 J

qrxn represents the heat absorbed

due to the reaction of 12.8g KCl.

Now you must convert the KCl to moles.

Convert grams KCl to moles.

12.8 g KCl

74 g KCl

1 mol KCl= 0.173 mol KCl

Calculate H for the reaction

=0.173 mol KCl

KCl(s) K+(aq) + Cl-(aq)

H+2949.72 J

x 1 mol KCl

Coefficient from balanced

equationH = +17050 J

H = +17.1 kJ H Must be positive because it was an endothermic reaction!

Practice Problem: #2What is the specific heat of

aluminum if the temperature of a 28.4 g sample of aluminum is

increased by 8.1ºC when 207 J of heat is added?

qsur = m x Cp x (Tf –Ti)

207J = 28.4g x Cp x 8.1ºC

Cp = 28.4 g x 8.1 ºC 207J

= 0.90 J/g ºC