thermodynamics: the first law yongsik lee 2004. 3

Thermodynamics: the first law

Yongsik Lee2004. 3.

Branches of Thermodynamics Thermodynamics

Classical thermodynamics Phenomenal, bulk properties

Statistical thermodynamics Connection between atomic and bulk thermodynamic pr

operties Branches

Thermochemistry Electrochemistry bioenergetics

The conservation of energy

Matter & Energy

물질 (matter) 공간을 차지하고 질량 (mass) 을 가진다

에너지 (energy) 일을 할 수 있는 능력 열 (heat) 과 일 (work) 의 형태를 가진다

E=mc2

물질과 에너지의 연관성

System and surroundings Consider matter & energy tr

ansfer Open system Closed system Isolated system

Wall Diathermic adiabatic

Work Gas push back the

surrounding atmosphere

Do wok on its surroundings

Energy transfer as work/heat

The measurement of work Transfer of energy that can cause motion against a

n opposing force Work = force x displacement

Physical work No force means no work

Work = pressure x volume change Expansion work Pressure = force / area Volume = area x displacement Work = F dx = P dV

에너지의 단위 에너지의 크기는 줄 (J) 이라는 단위로 표시 1 J 은 어떤 물체를 1 뉴턴 (N) 의 힘으로 1 m 만큼 움직이는 데

필요한 일의 양 1 J = 1 N-m = 1 kg m2 s-2

1 J 은 1 kg 의 책을 중력에 대해서 10 cm 만큼 들어올리는데 필요한 일의 양

인간의 심장이 한번 박동하는데 필요한 에너지 1840 년대 줄의 실험에 의해 물 1g 의 온도를 1℃ 올리는 데

필요한 열은 4.18J 의 일 (1cal) 에 해당한다는 사실이 밝혀졌다 . 1 cal = 4.184 J

따라서 J 은 일과 열 , 그리고 에너지를 나타내는 단위

다른 에너지 단위들 calorie

식품 분야에서 많이 쓰이는 에너지 단위 1 cal = 4.184 J 1 Cal = 1 kcal

Erg 1 J = 107 erg

BTU (British thermal unit) 일 파운드의 물을 화씨 일도 올리는데 필요한 에너지

Heat

열 (q) 높은 온도에서 낮은 온도의 물체로 흐른다 분자운동이 활발하면 열이 많다

열과 온도 계에 열을 가하면 보통 온도가 올라간다

Heat capacity C = q/ΔT

Work and Heat

계가 열을 받으면 부호는 +, 계가 열을 잃으면 - 가 된다 .

Expansion work W=-PexΔV W=- ∫PexdV Free expansion

When external pressure is zero

Means no work

Indicator Diagram

Reversible expansion Reversible

A change that can be reversed By an infinitesimal change in a variable Maximum work is possible

At Constant volume 일정부피에서 반응이

일어나면 ,  △V = 0, 그러므로 w = 0

Isothermal expansion

Perfect gas Isothermal (ΔT=0) process

Kinetic energy is constant

The measurement of heat Heat capacity

C = q/ΔT [C]=[J/K] Large C means small ΔT

heat capacity at constant pressure (Cp) at constant volume (Cv)

Heat capacity Molar heat capacity Specific heat capacity

일과 열은 결국 같은 종류가 아닐까 ?

일이나 열은 같은 효과를 얻을 수 있다 .

Joule 은 일과 열을 모두 ' 에너지 ' 라는 새로운 개념의 한 형태로 정의

에너지는 계의 상태가 변할 때 일이나 열의 형태로 나타낼 수 있는 추상적인 양

James P. Joule (1818-1889) born in Salford, Lancashir

e worked in a brewery Hero in Thermodynamics

질문 : 무게 25℃ 물이 1 g 있다 . 이 물의 온도를 올릴 수 있는 방법은 무엇인가 ?

두 가지 해결 방법일 ( 실험 2) 과 열 ( 실험 1) 은 같은 효과를 얻을 수 있다 .

Internal energy and enthalpy

Thermodynamic variables Intensive property Extensive property

dependent of the amount of the system Path function State function

Depends only on the current state of the system (independent of the path)

Molar property – per number Specific property – per mass

State function

Internal energy Measure of the energy

reserves of the system

ΔU=w+q Isothermal change

ΔT=0 ΔU=w+q=C ΔT=0 U is constant

Expansion at constant volume △U = q + w △U = q + 0 = qv

제 1 법칙 : 에너지 보존법칙 Crate energy from nothing?

All failed 한 계가 받아들인 열에너지는 그 계의 내부

에너지를 올리거나 , 계가 한 일로 간다 . dq = dU - dW The internal energy of an isolated system is

constant Not a single violation!

Fuel

Combustion of propane

C3H8(g) + 5 O2(g) → 3CO2(g) + 4H2O(l) at 298 K 1 atm

(1 atm = 101325 Pa), -2220 kJ = q

What is the work done by the system?For an ideal gas, propane;

pV = nRT (p = pex)n – no. of molesR – gas constantT = temperatureV – volumep = pressure

V= nRT/p or Vi = niRT/pex

6 moles of gas:Vi = (6 × 8.314 × 298)/ 101325 = 0.1467 m3

3 moles of gas:Vf = (3 × 8.314 × 298)/ 101325 = 0.0734 m3

work done = -pex × (Vf – Vi)

= -101325 (0.0734 – 0.1467) = +7432 Jat constant pressure Pex

work done = - pex (nfRT/pex – niRT/pex)

= (nf – ni) RT

Work done = - n∆ gasRT

i.e. work done = - (3 – 6) × 8.314 × 298 = + 7432.7 J

Can also calculate U∆∆U = q +w q = - 2220 kJ

w = 7432.7 J = 7.43 kJ

∆U = - 2220 + 7.43 = - 2212.6 kJ

NB:qp U ∆ why?

Only equal if no work is done i.e. V = 0∆

i.e. qv = U∆

Example : energy diagram

C3H8 + 5 O2 (Ui)

3CO2 + 4H2O(l) (Uf)

U

U

progress of reaction

reaction path

Reversible work For an infinitesimal change in volume, dV Work done on system = PdV

For ideal gas, PV = nRT

P = nRT/ V work = P(V) dV = nRT dV/ V

= nRT ln (Vf/Vi) because dx/x = ln x Work done by system= -nRT ln(Vf/Vi)

Vf

Vi

Vf

Vi

From heat to work = Engine Compare the work

C-D-A C-A

enthalpy

Definition H=U+PV

Properties State function ΔH=qp

Relation between Heat capacities For an ideal gas: H = U + p V For 1 mol: dH/dT = dU/dT + R Cp = Cv + R

Cp / Cv = γ (Greek gamma) At constant pressure

Some of the energy supplied as heat escapes as work when the system expands

The temperature does not rise as much

Variation of H with temperature

Suppose do reaction at 400 K, need to know H

f at 298 K for comparison with literature value. How?

As temperature ↑, Hm ↑

Hm ∝ T

Hm = Cp,m T

where Cp,m is the molar heat capacity at constant pressure.

H vs. Temperature qv = Cv ( T2 – T1) or Cv

T = U Cp = H / T Cv = U /T

For small changes Cp = dH / dT Cv = dU / dT

Chapter 3Thermochemistry

Yongsik Lee2004. 3

반응열 반응열

표준상태에서 이루어진 반응에서 발생한 열 STP (standard temperature and pressure)

반응열 측정 직접법 : 통열량계를 이용하여 직접 측정 , 또는

표에서 직접 계산 간접법 : 원소로부터 직접 합성이 안 되는 경우

Hess 의 법칙 이용

Bomb calorimeter

난 통 열량계는 잘 몰라요

통아저씨 이양승

Parr 통열량계

Using Bomb calorimeter submerges the reaction inside an insulated

container of water. An electrical heating device starts the

reaction inside a sealed reaction vessel and the temperature rise of the water which surrounds it is measured.

heat loss from a calorimeter (calibration) to keep the temperature of the water

surrounding the reaction vessel constant by heating or cooling it

Use well-known standard

Determine the heat of reaction

Diphenyl + Oxygen reaction combustion reaction

Using benzoic acid for reference Benzoic acid 0.9150 g (0.0217-0.0111 g) Fe wire

Bomb calorimeter with Fe wire Diphenyl 0.8214 g + 0.0031 g Fe wire

Heat capacity calculation

Heat of combustion calculation

enthalpy

Definition H=U+PV For a perfect gas

H=U + RTFor a solid/liquidH=U

Properties State function ΔH=qp

Physical change

Phase

Definition A specific state of matter Uniform throughout in composition and

physical state More specific than the state

Two solid phases of sulfur solid state Rhombic or monoclinic sulfur

Sulfur allotropes Sulfur has three allotropes:

Rhombic, Monoclinic, Plastic At room temperature, monoclinic sulfur is the most stable form. When he

ated, monoclinic sulfur melts to form a viscous liquid at 119 ℃ at the atmosphere pressure.

At higher pressure, the monoclinic sulfur → the rhombic sulfur.

Both crystalline forms have the S8 crown shaped molecule and the plastic sulfur has a chain structure of unspecified number of atoms Sn

Enthalpy of phase transition Phase transition

Conversion of one phase to another phase

State function ΔsubH = ΔfusH + ΔvapH ΔfusH = - ΔfreezeH Path independent

Enthalpy of vaporization (ΔvapH)

Atomic and molecular change Succession of ionization

First ionization enthalpy Second ionization enthalpy

More energy is needed to separate an electron from a positively charged ion

Electron gain enthalpy Reverse of ionization Cl(g) + e-(g) → Cl-(g)

ΔH = -349 kJ

Born-Haber cycle : 1 mol of NaCl

Bond Enthalpy

C-H bond enthalpy in CH4

CH4 (g) → C (g) + 4 H (g) , at 298K.

Need: Hf of CH 4 (g) =- 75 kJ mol-1

Hf of H (g) = 218 kJ mol-1

Hf of C (g) = 713 kJ mol-1

Hdiss = nHf (products) - nHf

( reactants)

= 713 + ( 4x 218) – (- 75) = 1660 kJ mol-1 Since have 4 bonds : C-H = 1660/4 = 415 kJ mol-1

Mean bond enthalpies Definition

Average of bond enthalpies Table 3.4 ΔH(AB) selected bond enthalpies Table 3.5 ΔHB mean bond enthalpies

Example H-O-H H2O(g) → 2H(g) + O(g) ΔH=+927 kJ H2O(g) → HO(g) + H(g) ΔH=+479 kJ HO(g) → H(g) + O(g) ΔH=+428 kJ Mean bond enthalpy ΔHB (O-H) = 463 kJ/mol

Chemical change

Standard enthalpy changes

ΔrH° Standard reaction enthalpy

Reactants in their standard states Products in their standard states

Standard state Pure substances at exactly 1 bar Temperature is not defined

Hess’ Law

the standard enthalpy of a reaction is the sum of the standard enthalpies of the reaction into which the overall reaction may be divided.

ΔrH°(overall reaction) = ΣΔrH°(each step) C (g) + ½ O2(g) → CO (g)

Hcomb =? at 298K

C (g) + ½ O2(g) → CO(g) Hcomb=?

From thermochemical data: C(g) +O2(g) → CO 2(g)

Hcomb =-393.5 kJmol-1

CO (g) +1/2 O2 (g) →CO 2(g) H

comb = -283.0 kJ mol-1

C (g) + O2 (g) – CO (g) – 1/2 O2 (g) → CO2 (g) – CO2 (g) C (g) + ½ O 2 (g) → CO (g) , H

comb= -393.5 – (-283.0) = - 110.5 kJ mol-1

standard enthalpies of formation

ΔfH° The standard enthalpy (per mole of the substan

ce) for its formation from its elements in their reference states

Reference state The most stable form under the prevailing condi

tions At specified temperature and 1 bar

화학반응 = 결합의 재배치 화학결합

결합을 끊는데 에너지가 필요하다 . 분자에 에너지를 가해주어야 한다 . 흡열반응

화학반응 화학결합을 끊고 새로운 화학결합을 만든다 약한 결합을 끊고 강한 결합을 만들면 에너지가

남는다 . ( 반응열의 발생 )

Standard enthalpies of formation Standard reaction enthalpy, ΔrHº ΔrHº =Σ ν Hº(products)- Σ νHº(reactants)

화학결합과 에너지 (kJ/mol)

Hydrazine synthesis reaction

N2 + 2H2 → N2H4

Energy absorbed for bond breaking +946 + 2 x (+436) = +1818 kJ

Energy released by bond making (-389) x 4 + (-163) = -1719 kJ

Energy change during the reaction (+1818) + (-1719) = +99 kJ

Water Generation reaction

Kirchoff’s equation : H(T)

Cp,m = Hm/ T (= [J mol-1/ K])

(= [J K-1 mol-1]) HT2°= HT1°+ Cp ( T2 - T1)

Cp = n Cp(products)- nCp(reactants)

For a wide temperature range: Cp ∫ dT between T1 and T2. Hence : qp = Cp( T2- T1) or H = Cp T

exercises

2-4, 2-5, 2-9, 2-17 3-5, 3-15

References

http://www.whfreeman.com/ECHEM/

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