Thermodynamics: Energy Relationships in Chemistry The Nature of Energy What is force: What is work:...

Thermodynamics: Energy Relationships in Chemistry

The Nature of Energy

•What is force:

•What is work:

A push or pull exerted on an object

An act or series of acts which overcome a force


•Mechanical work

The amount of energy required to move anobject 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


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) (1 cal)

(4.184 J) = 11 cal


System and Surroundings

system

surroundings


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


E final < E initial E final > E initial

endothermicexothermic


E = q + w

+

+


+ -


-

+

+

-

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

q

w

•A State Function is independent of pathway and is capitalized.E, energy is an extensive propertyand is a State function. heat (q) and work (w) are not statefunctions.



P-V work


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

Sample problem: How much heat is produced when 4.50 g of methane gas (CH4) is burned in a constant pressure environment

(4.50 g CH4) (1mol CH4)

(16.0 g) = -226 kJ

(-802 kJ)

(1 mole CH4)

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

•The enthalpy change for a reaction is equal in magnitude but opposite in sign to H for the reverse reaction

•The enthalpy change for a reaction depends on the state of the reactants and productsAssume:CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJ

•The following process would also produce the same resultCH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

2H2O(g) 2H2O(l) H = -88kJ

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

•Here is a second reaction pathway which produces the same results

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 1C (C = q/T, J/mol-C )

q = n (molar heat capacity)T

Specific Heat: The energy required to raise the temperatureof 1 gram of a substance by 1C (C = q/T, J/g-C )

q = m sT


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

(1kg) = 1.50 x 103 J/ C(0.75 J)

(1 g - C)

B.) q = m ST = 1.75 g (0.75 J/g-C ) (355 C) =465 J


Constant Pressure Calorimetry

Sample exercise: 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH arereacted together in a ‘coffee cup’ calorimeter.* The temperature of theresulting solution increased from 21.0 C to 27.5 C . Calculate theenthalpy change of the reaction (the specific heat of water = 4.18 J/g-C).

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

q = mS T

q = (100g)(4.18 J/ g-C )(6.5 C )

q = 2717 J = 2.7 kJ1mol .050L L

=54kJ/Mol

Heat capacity of a metalWhat 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 burnedin 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 ofheat evolved. What is the heat evolved upon combustion of one mole ofN2H4.

qevolved = -Ccalorimeter x T

(3.51 C)19.3 kJ = -(5.510 kJ) (C)

(1.00mol) =618 kJ(32 g)

(1.00 mol)

(19.3 kJ)

(1.00 g)

HESS’S Law: If a reaction is carried out in a series of steps, H for the reaction will be equal to the sum of the enthalpy changes for the individual steps.

CH4 (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 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 kJC(s) + O2(g) CO2(g) H = -393.5 kJH2(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 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 kJO3(g) 3/2O2(g) H = -142.3 kJO2(g) 2O (g) H = 495.0 kJ

NO(g) + O3 NO2(g) + O2(g) H = -198.9 kJ3/2O2(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, Hºf

• A thermodynamic description of the formation of •compounds from their constituent elements.

• 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

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


The standard heat of formation for one mole of ethanol is the enthalpy change for the following reaction

C2H5OH H f = -277.7 kJ

H fH rxn = n (products) - H f m (reactants)

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

2C(graphite) + 3H2(g) + ½ O2(g)



Sample exercise: The quantity of heat produced from one gramof propane (C3H8) is -50.5 kJ/gram. How does this compare with theheat produced from one gram of benzene (C6H6)?

C6H6(l) + O2 6CO2(g) + 3H2O(l)15

2

H fH rxn = n (products) - H f m (reactants)

[6H fH rxn = (-393.5 kJ) + 3H f(-285.8 kJ)]

[1H f (-49.04 kJ) + H f(0)] -

15

2

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

H rxn = [6(-393.5 kJ) + 3(-285.8 kJ)] (- 49.04 kJ)= -3267 kJ -

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Thermodynamics: Energy Relationships in Chemistry The Nature of Energy What is force: What is work:...

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