thermodynamics: basic concepts · thermodynamics: basic concepts thermodynamics: ... solutions of...

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Lecture 4. THE FIRST LAW OF THERMODYNAMICS

THERMODYNAMICS: Basic Concepts Thermodynamics: (from the Greek therme,

meaning "heat" and, dynamis, meaning "power") is the study of energy conversion between heat and mechanical work, and subsequently the macroscopic variables such as temperature, volume and pressure

Energy: capacity to do work Work: motion against an opposing force System: part of the world which we have interest or

being investigated. Surrounding: is where we make our observations

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Definition of Work: •  Work is motion against opposing force. •  Work is defined as a force acting through a displacement x, the

displacement being in the direction of the force.

xfw Δ=

Consider a Gas Expansion Work: Initial State Final State

P1, V1, T P2, V2, T

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Consider a Gas Expansion Work:

:is pressure, opposing external The ,

)( 12

exP

hmgwhhmgw

Δ−=

−−=

VPwVVPhAPw

AmgP

ex

exex

ex

Δ−=

−−=Δ−=

=

)( 12

or


2

1

1

2

2

1

lnln

,2

1

2

1

PPnRT

VVnRTw

VdVnRTw

VnRTP

PdVPw

PP

dVPw

v

v

in

in

v

v in

ex

v

v ex

−=−=

−=

=

−=

−=

∫

∫

∫

and

gas, the of pressure theis

than greater mallyinfinitesi onlyis instant, every at

volume in increase ssimalinfiniteti against done work a For

in

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Consider a Gas Expansion Work: q  the external pressure is infinitetissimally smaller than the internal pressure at all stages of the expansion - reversible process. A reversible change in thermodynamics is a change that can be reversed by an infinitesimal modification of a variable. q  in reversible processes, there is maximum amount of work done that could possible extracted from a process

Gas Expansion Work: Sample Problem A sample of 4.50 g of methane occupies 12.7 L at 310 K. (A) Calculate the work done when the gas expands isothermally against a constant external

pressure of 200 Torr until its volume has increased by 3.3 L. (B) Calculate the work that would be done if the same expansion occurred reversibly.

J 87.97

atm L 1J101.3

atm L 0.87

atm L 0.87

L) torr760

atm 1 torr

process) ble(irreversi formula the use we constantis pressure opposing since A.

:SOLUTION

:

−=

×−=

−=

×−=

Δ−=

w

w

w

w

VPw ex

3.3)(200(

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J -

KmolK J 1-mol g

g

:expansiongas reversible a for B.:SOLUTION

1-1-

4.1677.12)7.123.3(ln310)3145.8()

165.4(

ln

=

+×××−=

−=

wLLw

VV

nRTwi

f

Definition of HEAT:

Heat is the transfer of energy between two bodies that are at different temperatures. Heat appears at the boundary of the system. 0th Law: heat is transferred from the hotter object to the colder one. Heat is path dependent.

heat specificsetemperatur in change T mass,m

=

=Δ=

Δ= Tmsq

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THE FIRST LAW OF THERMODYNAMICS:

“Energy can neither be created nor destroyed but can be converted from one form to another.” Law of Conservation of Energy. ENERGY OF THE UNIVERSE:

surrsys

surrsysuniv

surrsysuniv

EEEEE

EEE

Δ−=Δ

=Δ+Δ=Δ

+=

therefore

system, given a forenergies inchanges the and0

If a system undergoes an energy change, the surroundings must also undergo a change equal in magnitude but opposite in direction

Total ENERGY of a system; Internal Energy, U

TOTAL ENERGY:

UE

UPEKEE

total

total

=

++=

0 are PE and KE rest, at system a for

Internal Energy , U: q  energy associated with the chemical system q  depends on the thermodynamic parameters such as T, P, V,

composition, etc. q  consists of translational, rotational, vibrational, electronic energies,

as well as intermolecular interactions

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Internal Energy, U:

For a given system, we do not know the exact nature of U, we will be only interested in the changes in U for a particular process that the system undergoes.

12 UUU −=Δ

Mathematical expression of the first law of thermodynamics:

wqU +=Δ

or for an infinitesimal change:

dwdqdU +=

The change in the internal energy of a system in a given process is the sum of the heat exchange, q, between the system and its surroundings and the work done on or by the system.

HEAT and WORK sign conventions:

Process Sign

Work done by the system on the surroundings

-

Work done on the system by the surroundings

+

Heat absorbed by the system from the surroundings

+

Heat absorbed by the surroundings from the system

-

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Constant volume adiabatic bomb calorimeter

Adiabatic means no heat exchange with the surroundings

Sample inside an oxygen filled container is ignited by an electrical discharge. Heat released is measured by the increase in the temperature of the water.

v

v

v

qU

VPqwqU

V

=Δ

Δ−=

+=Δ

=Δ 0 VP constant at ,

,

ENTHALPHY In processes carried out under constant P,

PVUHEnthalpyPVUPVUq

VVPqUUorVPwqU

p

p

v

+=

+−+=

−−=−

Δ=+=Δ

or

,

)()(

),(,

1122

1212

Total enthalpy of a system can not be measured directly, so the change in enthalpy, ΔH, is a more useful value than H itself.

For a constant P process;

VPdUdH

VPUH

Δ+=

Δ+Δ=Δ

change, simalinfiniteti for

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Entalphy, ΔH vs. Internal Energy, ΔU

Consider the following reaction:

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

Heat evolved by the reaction is 367.5 kJ, reaction takes place at constant pressure,

qp= ΔH = -367.5KJ.

Internal Energy ΔU:

The volume of the H2(1 mole) generated by the reaction occupies 24.5 L, therefore –PΔV = -24.5 L at or -2.5 kJ Finally,

ΔU = -367.5 kJ – 2.5 kJ

ΔU = -370 kJ.

- Difference is due to expansion work.

VPHU Δ−Δ=Δ

SAMPLE PROBLEM 4.02

In which of the following processes is ΔH = ΔE? A.  2.0 moles of NH3 are colled from 325oC to 300oC at 1.2 atm B.  1.0 g of water is vaporized at 100oC and 1.0 atm C.  2.0 moles of hydrogen iodide gas react to form hydrogen gas and iodine gas

in a 40 L container D.  Calcium carbonate is heated to form calcium oxide and carbon dioxide in a

container with variable volume E.  1.0 mole of carbon dioxide sublimes to gas phase.

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SAMPLE PROBLEM 4.03

For which of the following reactions will ΔH be equal to ΔE?

A. H2(g) + Br2(g) 2HBr(g)

B. H2O(l) H2O(g)

C. CaCO3 CaO(s) + CO2(g)

D. 2H2(g) + O2(g) 2H2O(l)

E. CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

Constant Pressure Calorimeter:

A constant-pressure calorimeter made of two plastic cups. The outer cup helps insulate the reacting mixture from the surroundings. Two solutions of known volume containing the reactants at the same temperature are carefully mixed in the calorimeter. The heat produced or absorbed by the reaction can be determined by the temperature change, the quantities and specific heats of the solutions used, and the heat capacity of the calorimeter.

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HEAT CAPACITIES: Calorimetry: measurement of heat changes in chemical and physical processes.

When heat is added to the system, the corresponding temperature rise will depend on a) The amount of heat delivered, b) The amount of the substance, c) The chemical nature and the physical state of the substance, d) The conditions at which the energy is added to the system.

HEAT CAPACITIES:

Thus for a given amount of a substance, change in temperature is related to the heat added:

capacity heat molaris C mol K J

capacity heatis C K J

1-1-

1-

,

,

,

TnqC

nC

TqC

orTCq

Δ==

Δ=

Δ=

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HEAT CAPACITY at constant volume

TCnTTCdTCU

dTCdUTU

TqC

vv

T

T v

v

vv

Δ=−==Δ

=Δ

Δ=

Δ=

=Δ

∫ )( 122

1

or

q U volume, constant at v

HEAT CAPACITY at constant pressure processes

TCnTTCdTCH

dtCdHTH

Tq

C

pp

T

T p

p

pp

Δ=−==Δ

=Δ

Δ=

Δ=

=Δ

∫ )( 122

1

or

q H process, pressure constant at p

for gases, Cp > Cv because of the work to be done to the surroundings in constant pressure processes for condensed phases, Cp and Cv are identical for most purposes for ideal gases, RCC vp =−

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Ch 45.5 First Long Exam 8:30 – 9:25 July 14 Multiple Choice July 16 Problem Solving

Coverage: Properties of Gases; Kinetic Molecular Theory of Gases, First Law of Thermodynamics and Thermochemistry.

Calculate the values of ΔE and ΔH for heating of 55.40 g of Xenon from 300 K to 400 K in a closed vessel. Assume ideal gas behavior and the heat capacities are independent of temperature. Xenon is a monoatomic gas. Its Cv is equal to 12.47 J K-1 mol-1.

SAMPLE PROBLEM 4.04

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When 229 J of energy is supplied as heat at constant pressure to 3.00 mole CO2(g), the temperature of the sample increases by 2.06 K. Calculate the molar heat capacities at constant volume (Cv) and constant pressure (Cp) of the gas.

SAMPLE PROBLEM 4.05

Use the Cp and Cv values from the previous problem to calculate the change in a) molar enthalpy b) molar internal energy when carbon dioxide is heated from 15oC (temperature when air is inhaled) to 37oC (blood temperature, temperature in our lungs).

SAMPLE PROBLEM 4.06

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A 50.0 mL sample of 0.400 M Copper (II) sulfate solution at 23.35oC is mixed with 50 mL of 0.600M sodium Hydroxide solution, also at 23.35oC in a coffee cup calorimeter. After the reaction occurs the temperature of the resulting mixture is measured to be 25.23oC. The density of the final solution is 1.02 g/mL. Calculate the amount of heat evolved. Specific heat of solution is 4.184 J/g oC

SAMPLE PROBLEM 4.07

GAS EXPANSION: ISOTHERMAL EXPANSION

q- w thatfollows it therefore 0U ,then

T, on dependentis energy internal the gases, ideal For.maintainedis mequilibriu thermal the that rate slow a

such at happen must typically system the of out or into transfer heat The constant. heldis etemperatur :process Isothermal

==Δ

0 H then T, constant at constantis PV and 0, U ncesi

PVUH

=Δ

=Δ

Δ+Δ=Δ

For isothermal expansion by an ideal gas, the heat absorbed is equal to the work done by the ideal gas on its surroundings.

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GAS EXPANSION: ADIABATIC EXPANSION

VPVP or

CC where

PP

VV

,0qgssurroundin and system between exchange heat no :process diabaticA

2211

v

p

1

2

2

1

γγ

γ

=

=γ=⎟⎟⎠

⎞⎜⎜⎝

⎛

=

57 that so

R57C and R2

5C gases, diatomic for3

5 that so

,R25C and R2

3C gases, monoatomic for

pv

pv

=γ

==

=γ

==

Calculate the values of ΔU, and ΔH for the reversible adiabatic expansion of 1.0 mole of a monoatomic ideal gas from 2 L to 6 L. The temperature of the gas is initially 298 K.

SAMPLE PROBLEM 4.08

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THERMOCHEMISTRY:

Heat of reaction is the heat change in the transformation of reactants at a given temperature and pressure to products at the same conditions.

For constant pressure processes, the heat of reaction qp is equal to enthalpy change of the reaction, ΔH.

positive.is H gs.surroundin the from heatabsorbs system :reactions cEndothermi

negative.is H gs.surroundin to heat offgives system :reactions rmicExothe

r

r

Δ

Δ

THERMOCHEMISTRY:

2H2(g) + O2(g) 2HO2(g)

241.8 kJ of heat is given off. The enthalpy change for this process is called standard enthalphy of reaction, ΔHo

In general, standard enthalpy change of a chemical reaction is the total enthalpy of the products minus the total enthalpy of the reactants

tcoefficien tricstoichiomeis v enthalpy; molar standard theis H

)reactants(Hv)products(HvHo

ooor ∑∑ −=Δ

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THERMOCHEMICAL EQUATIONS:

C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) + 1367 kJ

The coefficient in the balanced thermochemical equations refer to the number of moles of reactants and products. The numerical value of ΔH refers to the number of moles of substances specified by the equation The physical states of the species are important and must be specified. The value of ΔH does not change significantly with moderate changes in temperature.

When 2.61 g of dimethyl ether, CH3OCH3, is burned at constant pressure, 82.5 kJ of heat is given off and producing only CO2 and water. Find the ΔH for the reaction.

SAMPLE PROBLEM 4.09

SAMPLE PROBLEM 4.10

Write the thermochemical equation for for the reaction in sample problem 4.07.

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When aluminum metal is exposed to atmospheric oxygen, (as in aluminum doors and windows), it is oxidized to form aluminum oxide. How much heat is released by the complete oxidation of 24.2 grams of aluminum at 25oC and 1.0 atm. (ΔH = - 3352 kJ)

SAMPLE PROBLEM 4.11

STANDARD STATES AND STANDARD ENTHALPY CHANGES

The thermodynamic standard state of a substance is the most stable pure form under standard pressure (1.0 atm) and at standard temperature (25oC) Thermochemical Standard States:

For pure substances in the liquid or solid states, the standard state is the pure liquid or solid

For gases, the standard state is the gas at P = 1.0 atm and T at 25oC. For substances in solution, standard state refers to 1.0 M

concentration. The standard enthalpy change, refers to the ΔH when a specified number of moles of reactants, all at standard states are converted completely to a specified number of moles of products.

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STANDARD MOLAR ENTHALPIES OF FORMATION

The standard molar enthalpy of formation of a substance is the enthalpy change for the reaction in which one mole of the substance in a specified state is formed from its elements in their standard states. By convention the ΔH of formation value for any element in its standard state is zero.

SAMPLE PROBLEM 4.11

Write the standard formation reaction for ethanol a) C2H5OH(l) B) ammonium perchlorate

THERMOCHEMISTRY:

∑∑ Δ−Δ=Δ

+−+=Δ

+→+

)tstanreac(Hv)products(HvH

general; in or))B(Hb)A(Ha()D(Hd)C(HcH

:is reaction the of enthalpy standard thedDcCbB Aa

of

of

or

ooooor

formation. of enthalpy molar standard theis H ofΔ

- is the enthalpy change when 1 mole of a compound is formed from its constituent elements at 1 bar and 298 K.

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THERMOCHEMISTRY:

0))g(O(H ;0))g(H(H

0H :etemperatur particular a atforms allotropic stable most their inelements for

:H

2o

f2o

f

of

of

=Δ=Δ

=Δ

Δ

2H2(g) + O2(g) H2O(g)

kJ/mole 8.241H of −=Δ

MEASUREMENTS OF ΔH of reaction

DIRECT METHOD: measure ΔfHo of direct synthesis from their elements:

2H2(g) + O2(g) H2O(g)

[ ]8.241H

))g(O(H))g(H(H2))g(OH(H2H

)tstanreac(Hv)products(HvH

or

2o

f2o

f2o

fo

r

of

of

or

−=Δ

Δ+Δ−Δ=Δ

Δ−Δ=Δ ∑∑

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Calculate the reaction enthalpy of the following reaction:

3HN3(l) + 2NO(g) H2O2(l) + 4N2(g) Using their standard enthalpies of formation. ΔfHo values: kJ/mol HN3(l) 264.0 NO(l) 90.25 H2O2(l) -187.78

SAMPLE PROBLEM 4.12

SAMPLE PROBLEM 4.13

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SAMPLE PROBLEM 4.14

SAMPLE PROBLEM 4.15

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MEASUREMENTS OF ΔH of reaction

INDIRECT METHOD: HESS’S LAW: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

ofHΔ for carbon monoxide:

C(graphite) + 1/2O2(g) CO(g)

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

CO + 1/2O2(g) CO2(g)

kJ/mole 5.393H or −=Δ


THERMOCHEMICAL EQUATIONS

Enthalpy changes are additive. •  the ΔH of the reverse reaction will have the opposite sign •  multiplying a reaction by a factor, you also multiply the ΔH by the same factor.

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MEASUREMENTS OF ofHΔ

ofHΔ for carbon monoxide:


CO2(g) 1/2O2(g) + CO(g)


kJ/mole 0.283H or =Δ

C(graphite) + 1/2O2(g) CO(g) kJ/mole 5.110H or −=Δ

MEASUREMENTS OF ofHΔ

C(graphite) + O2(g) CO2(g) + 393.5kJ

kJ/mole 5.393−=Δ orHC(graphite) + O2(g) CO2(g)

Hess’s Law can be rephrased as: ‘ The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which the reaction might be divided.’

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Calculate the standard molar enthalpy of formation of acetylene (C2H2) from its elements.



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

The equations for combustion and the corresponding enthalpy changes are:

(1)  C(graphite) + 1/2O2(g) CO2(g)

(2)  H2(g) + 1/2O2(g) H2O(l)

(3)  2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)


SAMPLE PROBLEM 4.16

SAMPLE PROBLEM: o

fHΔ for carbon monoxide:


CO2(g) 1/2O2(g) + CO(g)


kJ/mole 0.283H or =Δ

C(graphite) + 1/2O2(g) CO(g) kJ/mole 5.110H or −=Δ

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SAMPLE PROBLEM:

Use the thermochemical equation shown below to determine the enthalpy of the reaction:

H2(g) + 1/2O2(g) H2O(g)

CO2(g) C(s) + O2(g) ΔfHo = +590.2 kJ/mol

C2H6(g) 2C(s) + 3H2(g) = 127.0 kJ/mol

C2H6(g) + 7/2O2(g) 2CO2(g) + 3H2O (l) =-2340.7 kJ/mol

SAMPLE PROBLEM 4.17

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SAMPLE PROBLEM 4.18

SAMPLE PROBLEM 4.18

From the following enthalpies of reaction, CaCO3(s) CaO(s) + CO2(g) ΔH = -178.1 kJ CaO(s) + H2O(l) Ca(OH)2(s) ΔH = -65.3 kJ Ca(OH)2(s) Ca+2(aq) + 2OH-(aq) ΔH = -16.2 kJ Calculate ΔHrxn for: Ca+2(aq) + 2OH-(aq) + CO2(g) CaCO3(s) + H2O(l)

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TEMPERATURE DEPENDENCE OF ENTHALPY CHANGE

KIRCHHOFFS’s LAW: The difference in enthalpies of a reaction at two different temperatures, T1 and T2, is just the difference in the enthalpies of heating the products and reactants from T1 to T2.

)( 1212 TTCHH pr −Δ=Δ−Δr

Where ΔCp is the difference in molar heat capacities between the products and reactants

The standard enthalpy change for the reaction: 3O2(g) 2O3(g) Is given by ΔrHo = 285.4 kJ/mol at 298 K and 1 bar. Calculate the value of ΔrHo at 380 K. Assume that the Cp values are all independent of temperature.

SAMPLE PROBLEM 4.18

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Bond Dissociation Enthalpy:

Bond Enthalpy: change in enthalpy when bonds form or break in diatomic molecule Bond Dissociation Enthalpy: average change in enthalpy in polyatomic molecules when individual bonds form or break.

N2(g) 2N(g)

H2O(g) H(g) + OH(g) ΔrHo = 502 kJ/mol OH(g) H(g) + O(g) ΔrHo = 427 kJ/mol

SAMPLE PROBLEM:

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ENTHALPY CHANGE and BOND ENERGIES:

The Enthalpy of a Reaction can be estimated by the enthalpies of the total number of bonds broken and formed in the reaction.

released energy total- input energy total (products))(reactantsr

=

−=Δ ∑ ∑BEBEH o

Bond Enthalpy Changes

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Estimate the enthalpy of combustion of methane:

CH4(g) + O2(g) CO2(g) + H2O(g) Using bond enthalpies in Table 4.4. Compare your result with that calculated from the enthalpies of formation of products and reactants. C-H 410 kJ/mole O=O 494 kJ/mole C=O 563 kJ/mole H-O 460 kJ/mole

SAMPLE PROBLEM 4.19

SAMPLE PROBLEM 4.20

Ethylamine undergoes an endothermic gas phase dissociation to produce ethylene and ammonia. CH3CH2NH2 H2C=CH2 + NH3 ΔH = +53.6 kJ/mol The following average bond energies are given:

C-H = 413 kJ C-C = 346 kJ C=C = 602 kJ N-H = 391 kJ

Calculate the C-N bond energy in ethylamine.

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End of Chapter

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