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# Thermodynamics: Energy Relationships in Chemistry - PowerPoint PPT Presentation

Thermodynamics: Energy Relationships in Chemistry. The Nature of Energy. What is force:. A push or pull exerted on an object. What is work:. An act or series of acts which overcome a force. Thermodynamics: Energy Relationships in Chemistry. Mechanical work.

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The Nature of Energy

• What is force:

A push or pull exerted on an object

• What is work:

An act or series of acts which overcome a force

• Mechanical work

The amount of energy required to move an

object over a certain distance

w = F * d

• What is energy:

The capacity to do work

potential energy

Kinetic energy

Ek= 1/2 mv 2

E= joule = 1kg-m2/s2

4.184 J = 1 cal

(4.184 J)

Thermodynamics: Energy Relationships in Chemistry

Sample problem: A 252 g baseball is thrown with a speed

of 39.3 m/s. Calculate the kinetic energy of the ball in joules

and calories

Ek= 1/2 mv2

=1/2 (0.145 kg)(25m/s)2 = 45 kg m2/s2 = 45J

(45 J)

= 11 cal

System and Surroundings

surroundings

system

First law of thermodynamics:

• Energy can neither be created nor destroyed

• The energy lost by a system equals the energy gained

• by its surroundings

• Everything wants to go to a lower energy state

E = E final - E initial

endothermic

exothermic

E final < E initial

E final > E initial

q

Thermodynamics: Energy Relationships in Chemistry

Sample problem: During the course of a reaction a system loses 550 J

of heat to its surroundings. As the gases in the system expand, the piston

moves up. The work on the piston by the gas is determined to be 240 J.

What is the change in the internal energy of the system,

E = q + w

E = (-550 J) + (-240 J

E = -790 J

• A State Function is independent of

• pathway and is capitalized.

• E, energy is an extensive property

• and is a State function.

• heat (q) and work (w) are not state

• functions.

Let work w = -P  V

If E = q + w, then E = q + -P  V

When a reaction is carried out in a constant-volume

container ( V = 0) then, E = q v

When a reaction is carried out at constant pressure

container then, E = q p- P  V, or

q p = E+ P  V

• Chemical reactions usually occur under conditions where

• the pressure is held constant, therefore:

change in enthalpy: H = E+ P  V

H = q p

• Since chemical reactions usually occur under conditions

• where the volume of the system undergoes little change:

H =E

• Since H= H final + H initial, then for any type of chemical reaction, H= H products - H reactants

Some things you may never have wished to know about enthalpy

• Enthalpy is an extensive property

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

H = -802 kJ

-75 kJ 0kJ -393.5kJ -242kJ

[(-393.5) + 2(-242)] – [(-75) + 2(0)] = -802.5kJ

when 4.50 g of methane gas (CH4) is burned in

a constant pressure environment

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

(1mol CH4)

(-802 kJ)

(4.50 g CH4)

= -226 kJ

(1 mole CH4)

(16.0 g)

Assume:

CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)

H = -890 kJ

• The following process would also produce the same

• result

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

H = -802 kJ

H = -88kJ

2H2O(g)  2H2O(l)

CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)

H = -890 kJ

CH4 (g) + 2O2(g)  CO(g) + 2H2O + 1/2 O2

CO(g) + 2H2O + 1/2 O2CO2(g) + 2H2O

CH4 (g) + 2O2(g) CO2(g) + 2H2O

Calorimetry: Things are heating up

Calorimetry: Measurement of heat flow

Molar Heat capacity: The energy required to raise the

temperature of 1 mole of a substance by 1C (C = q/T, J/mol-C )

q = n (molar heat capacity)T

Specific Heat: The energy required to raise the temperature

of 1 gram of a substance by 1C (C = q/T, J/g-C )

q = m sT

Sample exercise: The specific heat of Fe2O3 is 0.75 J/g-C. A.) What

is the heat capacity of a 2.00 kg brick of Fe2O3. B.) What quantity of heat is required to increase the temperature of 1.75 g of Fe2O3 from 25 C to 380 C .

A.) (2.00 kg)

(1000g)

(0.75 J)

= 1.50 x 103 J/ C

(1 g - C)

(1kg)

B.) q = mST = 1.75 g (0.75 J/g-C ) (355 C) =465 J

Constant Pressure Calorimetry

reacted together in a ‘coffee cup’ calorimeter.* The temperature of the

resulting solution increased from 21.0 C to 27.5 C . Calculate the

enthalpy change of the reaction (the specific heat of water = 4.18 J/g-C).

q = mST

q = (100g)(4.18 J/ g-C )(6.5 C )

q = 2717 J = 2.7 kJ

=54kJ/Mol

1mol .050L

L

* assume the calorimeter absorbs negligible heat and that the density of the solution is 1.0 g/ml.

What is the heat capacity of a 5.05g chunk

of an unknown metal. The metal was heated

in boiling water and then placed in 50 mL

of water in a coffee cup calorimeter at a

temperature of 24.5ºC. The highest

temperature achieved was 28.9ºC.

What is the heat capacity of the metal.

50g x 4.184j/gºC x 4.5ºC = 5.05g x X x 71.1ºC

X = 2.62J/gºC

Bomb Calorimetry

Sample exercise:When 1.00 g of the rocket fuel, hydrazine (N2H2) is burned

in a bomb calorimeter, the temperature of the system increases by 3.51 C.

If the calorimeter has a heat capacity of 5.510 kJ/ C what is the quantity of

heat evolved.What is the heat evolved upon combustion of one mole of

N2H4.

qevolved = -Ccalorimeter x T

-(5.510 kJ)

19.3 kJ =

(3.51 C)

(C)

(32 g)

(19.3 kJ)

(1.00mol)

=618 kJ

(1.00 mol)

(1.00 g)

in a series of steps,H for the reaction

will be equal to the sum of the enthalpy

changes for the individual steps.

CH are4 (g) + 2O2(g)  CO(g) + 2H2O + 1/2 O2 H = -607 kJ

CO(g) + 2H2O + 1/2 O2CO2(g) + 2H2O H = -283 kJ

CH4 (g) + 2O2(g) CO2(g) + 2H2O H = -890 kJ

Sample exercise: Calculate the areH for the reaction:

2C(s) + H2(g) C2H2(g)

given the following reactions and their respective enthalpy changes

C2H2(g) + 5/2O2  2CO2(g) + H2O(l) H = -1299.6 kJ

C(s) + O2(g)  CO2(g) H = -393.5 kJ

H2(g) + 1/2O2  H2O(l) H = -285.9 kJ

2CO2(g) + H2O(l)  C2H2(g) + 5/2O2 H = 1299.6 kJ

2C(s) + 2O2(g)  2CO2(g) H = -787.0 kJ

H2(g) + 1/2O2  H2O(l) H = -285.9 kJ

2C(s) + H2(g)  C2H2(g) H = 226.7 kJ

Practice Exercise : Calculate the areH for the reaction:

NO(g) + O(g) NO2(g)

given the following reactions and their respective enthalpy changes

NO(g) + O3  NO2(g) + O2(g) H = -198.9 kJ

O3(g)  3/2O2(g) H = -142.3 kJ

O2(g) 2O (g) H = 495.0 kJ

NO(g) + O3  NO2(g) + O2(g) H = -198.9 kJ

3/2O2(g)  O3(g) H = 142.3 kJ

O (g)  1/2O2(g) ) H = -247.5 kJ

NO(g) + O(g) NO2(g) H = -304.1 kJ

Heats of formation, areHºf

• A thermodynamic description of the formation of

• compounds from their constituent elements.

Heat of vaporization: H for converting liquids to gases

Heat of fusion: H for melting solids

Heat of combustion: H for combusting a substance in

oxygen

• A thermodynamic description of the formation of

• compounds under standard conditions (1 atm, 298 K

• (25 C)) is called the standard heat of formation, Hºf

(products) - are

 m

(reactants)

 n

H rxn =

H f

H f

Thermodynamics: Energy Relationships in Chemistry

The standard heat of formation for one mole of ethanol

is the enthalpy change for the following reaction

 C2H5OHH f = -277.7 kJ

2C(graphite) + 3H2(g) + ½ O2(g)

note: the standard heat of formation of the most stable form of any element is 0.

15 are

15

2

2

(products) -

 m

(reactants)

H rxn =

 n

H f

[1H f

3H f

H f

H f

[6H f

Thermodynamics: Energy Relationships in Chemistry

Sample exercise: The quantity of heat produced from one gram

of propane (C3H8) is -50.5 kJ/gram. How does this compare with the

heat produced from one gram of benzene (C6H6)?

C6H6(l) + O2  6CO2(g) + 3H2O(l)

(-285.8 kJ)]

(-393.5 kJ) +

H rxn =

(0)]

-

(-49.04 kJ) +

-

H rxn =

[6(-393.5 kJ) +

3(-285.8 kJ)]

(- 49.04 kJ)= -3267 kJ

note: the standard heat of formation of the most stable form of any element is 0.