Thermodynamics

Heat, disorder, spontaneity

Energy

•The capacity to perform work–often measured as heat

Energy

•A tub is filled with water at 35°C–Dip a cup into the water and fill it.

–What is the temperature of the water in the cup?

Energy

•Which amount of water, that in the tub or in the cup, can melt the greater amount of ice during the same time frame?

Energy

•Two substances may have the same temperature but different amounts of heat energy.

Energy

•Temperature is the measure of average KE of a substance

Energy

•Heat is the measure of the total energy transferred from an object with a higher temperature to an object with a lower temperature.

Energy

•Heat is measured in either Joules (J) or calories (cal)

•A calorie is defined as the amount of heat needed to raise 1 g of water 1°C.

•1 cal = 4.18 J

Energy

•Graph the following data for two experiments on the same hand-drawn graph.

Time for ice to melt…

0°C 0 0

0°C (ice disappears) 25 190

25°C 33 250

50°C 41 310

75°C 49 370

100°C (water begins to boil)

57 429

100°C (water disappears)

226 1701

Temperature 1 cube (s) 8 cubes (s)

Time

Energy

Temperature (°C)

Tim

e (

s)

Specific Heat Capacity

•the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius

•Measured in J/g°C or cal/g°C

Specific Heat Capacity

•When a substance’s SHC (or C) is greater, more heat is required to make that substance equal in temperature to a substance with a lesser SHC

Specific Heat Capacity

•Which has the greater SHC, silicone or iron?

heat = (T)(mass)(SHC)

Heating Curve for H2O

Heat (cal)

Tem

pera

ture

(°C

)

G

H

G

Heating Curve for H2O•BC has value of 80 cal/g

–Known as the heat of fusion (sl) or heat of solidification (ls)

•DE has a value of 540 cal/g–Known as the heat of vaporization (lg) or heat of condensation (gl)

Heating Curve for H2O

•G has a value of 0°C–known as the melting point or the freezing point

•H has a value of 100°C–Known as the boiling point or the condensation point

Calorimetry•Measurement of heat energy

•Two types of calorimeters–Constant pressure (coffee-cup calorimeter)

–Constant volume (bomb calorimeter)

Biological Calorimetry•Nutrients

–Carbohydrates–Proteins–Lipids–Water–Vitamins–minerals

Biological Calorimetry•Carbohydrates

–4 kcal/g or 17 kJ/g

•Proteins–4 kcal/g or 17 kJ/g

•Lipids–9 kcal/g or 38 kJ/g

Heat of ReactionHrxn

•amount of heat absorbed or released in a chemical reaction

•If absorbed, it is a reactant and the process is endothermic

•If released, it is a product and the process is exothermic

Heat of Reaction•Deviations

Hformation is amount of heat absorbed or released during synthesis of one mole of an element or compound at

298 K and 1atm of pressure

Heat of Reaction•Deviations

Hsolution is amount of heat absorbed or released when a substance dissolves in a solvent

Heat of Reaction•Deviations

Hcombustion is amount of heat released when a substance reacts with O2 to form CO2 and H2O

Heat of Reaction•Is part of the stoichiometry of

a reaction…the heat of combustion of methane is 803 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + 803 kJ

• If there were 5 moles of CH4 present, how many kJ would be produced?

Heat Energy Practice Problems

1. How many kJ are released by a reaction that raises the temperature of 1.00 kg of water in a coffee-cup calorimeter from 25.0°C to 27.0°C? Psst…you know the SHC of water

2. A swimming pool measures 6.0 m x 12.0 m and is 3.0 m deep all around. The pool is filled with water at a temperature of 20.0°C. How many kJ must be released by the pool’s heater to raise the water temperature to 25.0°C? Psst…the density of water is 1 g/cm3, you know the SHC of water, and 1 m = 100 cm, so 1 m3 would equal how many cm3?

3. Gaseous butane, C4H10, is burned in lots of lighters. Write the balanced equation for the complete combustion of butane. Butane’s heat of combustion is 2878 kJ. How many kJ of heat energy would be released by the combustion of 10.0 g of butane?

4. Use the table on the next slide to calculate the number of kiloJoules provided by the fat in one serving of each of the following foods:a. french friesb. cheeseburger

4. (continued)

French fries(3.4 oz.)

320

36.3 4.4 17.1

Cheeseburger

(4.1 oz.)

310

31.2 15.0

13.8

Food (amt.) kcal carb(g) prot(g) fat(g)

5. Is more energy released when 428 g of H2 or 428 g of isooctane, C8H18, react with an excess of oxygen? Psst…balance the equations.

2H2 + O2 2H2O + 484 kJ

2C8H18 + 25O2 16CO2 + 18H2O + 4893 kJ

1. 8.36 kJ2. 4.51 x 106 kJ3. 248 kJ4. a. 650 kJ

b. 524 kJ5. 428 g H2 will release 5.13 x

104 kJ while 428 g of C8H18 will release 9.16 x 103 kJ. So, the 428 g H2 will release more energy.

Activation Energy

Rxn progress (s)

En

erg

y (

kJ)

Ea

H

Reactants

Products

Activation Energy

Rxn progress (s)

En

erg

y (

kJ)

Ea

Uncatalyzed

Catalyzed

H

Enthalpy

• Enthalpy can be equated with heat energy

• represented by H H is also known as change in

enthalpy

Hess’s Law

• states that in going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

Finding H using Hess’s Law

• If a reaction is reversed, the sign of H is also reversed.

• If the coefficients in a balanced equation are multiplied by an integer, then the value of H is multiplied by that same integer.

Hess’s Law

• Consider the following reaction:

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

It does not necessarily occur as we see it. It can, in fact, occur in a few additive steps, known as elementary steps.

Hess’s Law

Plausible elementary steps:

a. N2(g) + 2O2(g) 2NO(g) H = 180 kJ

b. 2NO(g) + O2(g) 2NO2(g) H = -112 kJ

Hess’s Law

a. N2(g) + O2(g) 2NO(g) H = 180 kJ

b. 2NO(g) + O2(g) 2NO2(g) H = -112 kJ

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

N2(g) + 2O2(g) + 2NO2(g) H = 68 kJ

So, the reaction is endothermic.

Hess’s Law

Two forms of carbon are graphite anddiamond. Using the enthalpies ofcombustion for graphite and diamond asyour elementary steps, calculate the H

forthe conversion of graphite to diamond

andstate whether it is an endo- or exothermicprocess.

C(graphite)(s) C(diamond)(s) H = ?

Hess’s Law

The elementary steps are:

a. C(graphite)(s) + O2(g) CO2(g) H = -394kJ

b. C(diamond)(s) + O2(g) CO2(g) H = -396kJ

Hess’s Law

a. C(graphite)(s) + O2(g) CO2(g) H = -394kJ

b. CO2(g) C(diamond)(s) + O2(g) H = +396kJ

C(graphite)(s) + O2(g) + CO2(g) CO2(g) + C(diamond)(s) + O2(g)

C(graphite)(s) C(diamond)(s) H = 2kJ

So, it is endothermic.

Finding H using standard heats of formation

H = ∑Hf°products − ∑Hf

°reactants

• Use pages Zumdahl Chemistry II textbook

• All elements in their natural states will have Hf

° equal to zero.

Finding H using standard heats of formation

Find the H of the followingreaction using Hf

° values:

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

Finding H using standard heats of formation

(2mol NO2 x 34kJ) mol

(1mol N2 x 0kJ) + (2mol O2 x 0kJ) mol mol

68kJ; the reaction is endothermic

Finding H using standard heats of formation

Find the H of the reaction which

converts graphite to diamond using

Hf° values.

Entropy

• Is the measure of disorder or chaos present in a substance.

• Chemical reactions may result in increasing disorder or decreasing disorder.

• Represented by S…thus, change in entropy is S

Entropy

• When there are more moles of products than reactants, entropy usually increases.

• When phase changes from more organized to less organized, entropy increases.

• If S is positive, entropy increases; if negative, entropy decreases.

Finding S using standard entropy values

S = ∑S°products − ∑S°

reactants

• Use Zumdahl Chemistry II textbook

Finding S using standard entropy values

Find the S of the followingreaction using S°

values:

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

Finding S using standard entropy values

(2mol NO2 x 240J) K·mol

(1mol N2 x 192J) + (2mol O2 x 205J) K·mol K·mol

-122 J/K; entropy is decreasing

Finding S using standard entropy values

Find the S of the reaction whichconverts graphite to diamond

usingS°

values.

Spontaneity

refers to whether a reaction will happen

without outside intervention or not. It says

nothing about how quickly the reaction will

happen only that it will or will not occur.

Free Energy

• is symbolized by G and is used to determine the spontaneity of a reaction

G = H TS

Free Energy

• If G is positive, it is a nonspontaneous process and is known as an endergonic reaction.

• If G is negative, it is a spontaneous process and is known as an exergonic reaction.

Free Energy

Find the G of the followingreaction at 25°C:

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

Free Energy

G = 68 kJ (298 K)(-0.122 kJ) K

G = 104 kJ; the reaction is nonspontaneous or endergonic

Free Energy

Find the G of the reaction which

converts graphite to diamond at100°C and state whether it isspontaneous or not.

More Practice Problems…

6. Acetylene gas, C2H2, is used in some welding applications and can be made via the following reaction:

2C(s) + H2(g) C2H2(g)

Determine its H using the elementarysteps on the following slide.

a. C2H2(g) + 2½O2(g) 2CO2(g) + H2O(l)

H = -1300kJ

b. C(s) + O2(g) CO2(g)

H = -394kJ

c. H2(g) + ½O2(g) H2O(l)

H = -286kJ

7. Think about photosynthesis… recall that carbon dioxide gas reacts with water to produce solid glucose (C6H12O6) and oxygen gas. Write a balanced equation, and determine the H, S, and G at 25°C. State whether the reaction is endo- or exothermic, whether entropy increases or decreases, and whether it spontaneous or not.

6. H = +226 kJ

7. H = +2802 kJ; endothermicS = -262 J/K; entropy decreasesG = +2880 kJ; nonspontaneous