6-1

Thermochemistry: Energy Flow and Chemical Change

Chapter 6

6-2

화학반응의 에너지 관계

• 에너지의 본성과 에너지의 형태• 화학반응에서의 에너지 변화 , ΔE• 열역학 . 계와 상태함수 열역학 제 1 법칙 , 일 (w) 과 열 (q)• 화학반응에서의 엔탈피 , H• 열계량법• 표준생성 (ΔH0

f) 및 반응엔탈피 (ΔH0rxn)

6-3

Thermochemistry: Energy Flow and Chemical Change

6.1 Forms of Energy and Their Interconversion

6.2 Enthalpy: Heats of Reaction and Chemical Change

6.3 Calorimetry: Laboratory Measurement of Heats of Reaction

6.4 Stoichiometry of Thermochemical Equations

6.5 Hess’s Law of Heat Summation

6.6 Standard Heats of Reaction (DH0rxn)

6-4

Thermochemistry is a branch of thermodynamics that deals withthe heat involved with chemical and physical changes.

Thermodynamics is the study of heat and its transformations.

Fundamental premise

When energy is transferred from one object to another, it appears as work and/or as heat.

For our work we must define a system to study; everything elsethen becomes the surroundings.

The system is composed of particles with their own internal energies (E or U).Therefore the system has an internal energy. When a change occurs, theinternal energy changes.

6-5

isolated

Various Systems

system matter energy

open( 열린 계 ) allowed allowed

closed( 닫힌 계 ) forbidden allowed

isolated( 고립 계 )

forbidden forbidden

Some other thermodynamics terms

state, state functions,path, process,extensive properties,intensive properties.

closed

open

6-6

계 물질 에너지

열린계 ○ ○

닫힌계 × ○

고립계 × ×

우주 = 계 + 주위Universe = System + Surroundings

열역학의 용어들

상태 1(state)State functionsP, V, T, S, H, E, m, n, …

상태 2

경로 (path)q, w과정 (process)단열 , 정압 , 정적 ,가역 , 비가역 ,발열 , 흡열 , …

6-7

Euniv = Esys + Esurr

Units of Energy

Joule (J)

Calorie (cal)

British Thermal Unit

1 cal ≡ 4.184J

1 J = 1N∙m = 1 kg∙m2/s2

1 Btu = 1055 J

6-8

-- Heat and work are forms of energy transfer and energy is conserved.

The First Law of Thermodynamics

E = Q + W

work doneon the system

change intotal internal energy

heat added

to systemState Function Process Functions

6-9

Calculating the change in internal energy

We see that the person’s internal energy falls by 704 kJ. Later, that energy will be restored by eating.

Suppose someone does 622 kJ of work on an exercise bicycle and loses 82 kJ of energy as heat. What is the change in internal energy of the person? Disregard any matter loss by perspiration.

Solution w = −622 kJ (622 kJ is lost by doing work on the bicycle),q = −82 kJ (82 kJ is lost by heating the surroundings).

Then the first law of thermodynamics gives us

E = q + w = (-82 kJ )+ (-622 kJ) = -704 kJ

6-10

Work, and the expansion(p-V) work

F F

dW

pex=F/A

exdW Fdy p Ady Increase in volume, dV

exdW p dV (p-V work)

Work = force • distance

force acting on the piston = pex × Areadistance when expand = dy

+y

6-11

Total Work Done

exdW p dV

f

i

V

exVW p dV

To evaluate the integral, we must know how the pressure depends (functionally) on the volume.

We will consider the following cases0. Constant volume work1. Free expansion2. Expansion against constant pressure

6-12

0. Work for the constant volume process

0dV

0f

i

V

exVW p dV

For the constant volume process, there is no p-V work

6-13

Heat and Internal Energy

E = q + w = qV

For the constant volume process, there is no p-V work, so

If we add heat q to the system, the temperature of the system increases by T , and the internal energy increases by E.

V

dEC

dT

Constant volume heat capacity

6-14

A constant-volume bomb calorimeter.

The constant-volume heat capacity is the slope of a curve showing how the internal energy varies with temperature. The slope, and therefore the heat capacity, may be different at different temperatures.

6-15

Determining the Change in Internal Energy of a System

When gasoline burns in a car engine, the heat released causes the products CO2 and H2O to expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (DE) in J, kJ, and kcal.

SOLUTION: q = - 325 J w = - 451 J

DE = q + w = -325 J + (-451 J) = -776 J

-776J103J

kJ= -0.776kJ -0.776kJ

4.18kJ

kcal= -0.185 kcal

6-16

molar constant volume heat capacity of monatomic gases

(at 1 atm, 25 °C)

Monatomic gas CV, m (J/(mol·K)) CV, m/R

He 12.5 1.50

Ne 12.5 1.50

Ar 12.5 1.50

Kr 12.5 1.50

Xe 12.5 1.50

6-17

1. Work for free expansion case

0exp

0f

i

V

exVW p dV

For the free expansion process, there is no p-V work

E = q + w = q

6-18

2. Work for expansion against constant pressure

exp p f

i

V

exVW p dV p V

E = q + w = q - p V

qp = E + p V = (E + p V )

If we define Enthalpy as, H ≡ E + pV

qp = HEnthalpy change is the heat given to the system at constant pressure.

6-19

Heat and Enthalpy

H = qP

For the constant pressure process,

If we add heat q to the system, the temperature of the system increases by T , and the internal energy increases by H.

P

dHC

dT

Constant pressure heat capacity

6-20

Enthalpy and Internal Energy

qp = ΔE + P Δ V = Δ H

Δ H ≈ ΔE in

1. Reactions that do not involve gases.

2. Reactions in which the number of moles of gas does not change.

3. Reactions in which the number of moles of gas does change but q is >>> PDV.

H = E + PV

qV = ΔE

ΔE = q + w

6-21

An exothermic process

발열반응

Fe2O3 (s) + 2Al(s) → Al2O3(s) + 2Fe (s)

if the reaction carried out at constant pressure, and the heat out -qp = - H > 0,

it is called an exothermic reaction.

Fe2O3 (s) + 2Al(s)

heat outhe

at o

utheat out

Al2O3(s) + 2Fe (s)

Ent

halp

y, H

6-22

An endothermic process

흡열반응

NH4NO3 (s) + H2O (l) → NH4+(aq) + NO3

−(aq)

if the reaction carried out at constant pressure, and the heat in qp = H > 0,

it is called an endothermic reaction.E

ntha

lpy,

H

heat in

heat in

heat in

6-23

Specific Heat Capacities of Some Elements, Compounds, and Materials

Specific Heat Capacity (J/g*K)

SubstanceSpecific Heat Capacity (J/g*K)

Substance

Compounds

water, H2O(l)

ethyl alcohol, C2H5OH(l)

ethylene glycol, (CH2OH)2(l)

carbon tetrachloride, CCl4(l)

4.184

2.46

2.42

0.864

Elements

aluminum, Al

graphite,C

iron, Fe

copper, Cu

gold, Au

0.900

0.711

0.450

0.387

0.129

wood

cement

glass

granite

steel

Materials

1.76

0.88

0.84

0.79

0.45

6-24

Finding the Quantity of Heat from Specific Heat Capacity

A layer of copper welded to the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 250C to 300.0C? The specific heat capacity (c) of Cu is 0.387 J/g*K.

q = 0.389 J/gK×125 g×(300-25)0C= 1.33×104 J

6-25

Coffee-cup calorimeter.

온도계

막대젓개

이중 보온 컵

뚜껑

6-26

Determining the Heat of a Reaction

You place 50.0 mL of 0.500 M NaOH in a calorimeter at 25.000C and carefully add 25.0 mL of 0.500 M HCl, also at 25.000C. After stirring, the final temperature is 27.210C. Calculate qsoln (in J) and DHrxn (in kJ/mol). (Assume the total volume is the sum of the individual volumes and that the final solution has the same density and specfic heat capacity as water: d = 1.00 g/mL and c = 4.18 J/g*K)

50.0 mL of 0.500 M NaOH

25.0 mL of 0.500 M HCl

75.0 mL

27.210C25.000C

6-27

total volume after mixing = 0.0750 L

0.0750 L ×103 mL/L ×1.00 g/mL= 75.0 g of water

q = mass × specific heat ×DT

= 75.0 g × 4.18 J/g*0C x (27.21-25.00)0C

= 693 J

(693 J/0.0125 mol H2O)(kJ/103 J) = 55.4 kJ/ mol H2O formed

For NaOH 0.500 M × 0.0500 L = 0.0250 mol OH-

For HCl 0.500 M ×0.0250 L = 0.0125 mol H+

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

H+(aq) + OH-(aq) H2O(l)

HCl is the limiting reactant.

0.0125 mol of H2O will form during the rxn.

6-28

Calculating the Heat of Combustion

A manufacturer claims that its new dietetic dessert has “fewer than 10 Calories per serving.” To test the claim, a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in O2(the heat capacity of the calorimeter = 8.15 kJ/K). The temperature increases 4.9370C. Is the manufacturer’s claim correct?

qcalorimeter = heat capacity × DT

= 8.151 kJ/K × 4.937 K

= 40.24 kJ

40.24 kJ ×(kcal/4.184 kJ) = 9.63 kcal or Calories

The manufacturer’s claim is true.

6-29

Using the Heat of Reaction (DHrxn) to Find Amounts

The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by

If aluminum is produced this way, how many grams of aluminum can form when 1.000 × 103 kJ of heat is transferred?

Al2O3(s) → 2Al(s) + 3/2O2(g) DHrxn = 1676 kJ

1.000x103 kJ ×(2 mol Al/1676 kJ)×(26.98 g Al/mol) = 32.20 g Al

6-30

Using Hess’s Law to Calculate an Unknown DH

Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation:

CO(g) + NO(g) → CO2(g) + 1/2N2(g) DH = ?

Given the following information, calculate the unknown DH:

(1) CO(g) + 1/2O2(g) → CO2(g) DHA = -283.0 kJ

(2) N2(g) + O2(g) → 2NO(g) DHB = 180.6 kJ

Multiply Equation (2) by 1/2 and reverse it.

DHB = -90.3 kJ

CO(g) + 1/2O2(g) CO2(g) DHA = -283.0 kJ

NO(g) 1/2N2(g) + 1/2O2(g)

DHrxn = -373.3 kJCO(g) + NO(g) CO2(g) + 1/2N2(g)

6-31

Table 6.3 Selected Standard Heats of Formation at 250C(298K)

Formula DH0f(kJ/mol)

calciumCa(s)CaO(s)CaCO3(s)

carbonC(graphite)C(diamond)CO(g)CO2(g)CH4(g)CH3OH(l)HCN(g)CSs(l)

chlorineCl(g)

0-635.1

-1206.9

01.9

-110.5-393.5-74.9

-238.6135

87.9

121.0

hydrogen

nitrogen

oxygen

Formula DH0f(kJ/mol)

H(g)H2(g)

N2(g)NH3(g)NO(g)

O2(g)O3(g)H2O(g)

H2O(l)

Cl2(g)

HCl(g)

0

0

0

-92.30

218

-45.990.3

143-241.8

-285.8

107.8

Formula DH0f(kJ/mol)

silverAg(s)AgCl(s)

sodium

Na(s)Na(g)NaCl(s)

sulfurS8(rhombic)S8(monoclinic)SO2(g)

SO3(g)

0

0

0

-127.0

-411.1

2-296.8

-396.0

6-32

Writing Formation Equations

Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include DH0

f.

(a) Silver chloride, AgCl, a solid at standard conditions.

(b) Calcium carbonate, CaCO3, a solid at standard conditions.

(c) Hydrogen cyanide, HCN, a gas at standard conditions.

DH0f = -127.0

kJ(a) Ag(s) + 1/2Cl2(g) → AgCl(s)

DH0f = -1206.9 kJ(b) Ca(s) + C(graphite) + 3/2O2(g) → CaCO3(s)

DH0f = 135 kJ(c) 1/2H2(g) + C(graphite) + 1/2N2(g) → HCN(g)

6-33

The general process for determining DH0rxn from DH0

f values.

aA + bB → cC + dD

ΔH0rxn = {c Δ H0

f(C) + d Δ H0f(D) } – {a Δ H0

f(A) + b Δ H0

f(B) }products reactants

6-34

Calculating the Heat of Reaction from Heats of Formation

Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

Calculate DH0rxn from DH0

f.

DHrxn = [4DH0f(NO(g) + 6DH0

f (H2O(g)] - [4DH0f(NH3(g) + 5DH0

f(O2(g)]

= (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) -

[(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)]

DHrxn = -906 kJ

6-35

Figure 6.11

The trapping of heat by the atmosphere.