Thermochemistry

Unit 13Thermochemistry

Dr. Jorge L. Alonso

Miami-Dade College – Kendall Campus

Miami, FL

Textbook Reference: •Chapter # 15 (sec. 1-11)

•Module # 3 (sec. I-VIII)

CHM 1046: General Chemistry and Qualitative Analysis

Thermochemistry

EnergyPotential Energy an object possesses by virtue of its position or chemical composition (bonds).

Kinetic Energy an object possesses by virtue of its motion.

12

KE = mv2

2 C8H18 (l) + 25 O2 (g)

16 CO2(g) + 18 H2O(g)

Energy (-E): work + heat

Chemical bond

energy

Thermochemistry

Energy

Work (w): Energy (force) used to cause an object that has mass to move (a distance) = kinetic energy. W = F x d

2 C8H18 (l) + 25 O2 (g) 16 CO2(g) + 18 H2O(g)

H = - 5.5 x 106 kJ/

Energy produced by chemical reactions = heat or work.

gas=2534= +9

E =

Heat (q): spontaneous transfer (flow) of energy (energy in transit) from one object to another; causes molecular motion & vibrations (kinetic energy); measured as temperature.

9 x 22.4L/ = 202 L

q + wLaw of Conservation of Energy:

Chemical bond

energy

Thermochemistry

Units of Energy: Joule & calorie• The joule (J) = SI unit of energy (work)

• The calorie (cal) = heat required to raise 1 g of H2O 10C

1 cal = 4.184 J

{Nutritional Calorie = 1000 calories (1kcal)=4,184 J}.

= kg m2

s2Joule (J) = N·m =

kg m

s2m

The Newton (N) is the amount of force that is required to accelerate a kilogram of mass at a rate of one meter per second squared.

Energy (Work) = F x d

Thermochemistry

System and Surroundings

• The system includes the molecules we want to study.

• The surroundings are everything else (here, the cylinder and piston).

System

Surroundings

Work

Heat-Eq(heat)

-Ework

+Eq

+Ew

E = q + w

Thermochemistry

• Energy is neither created nor destroyed.

First Law of Thermodynamics

• In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

• Also known as the Law of Conservation of Energy

System

Surroundings

Work

Heat-Eq

-Ew

+Eq

+Ew

Esys + Esurr = 0

Esys = -Esurr

orE sys = (q + w)surr

Thermochemistry

Changes in Internal Energy (E)

E = q + w

E, q, w, and Their Signs

> q

> w

+ q

+ w

- q

- w

Syst gains heat

Syst looses heat

: energy released/absorbed as either work or heat, is looked at from the perspective of the

system (the chemicals)

Work done on Syst

Work done by Syst

+ q + w- q - w- E =

+ E =

± q ± w± E =

Thermochemistry

Why consider Energy Change (ΔE) and not the total Internal Energy (E)?

• But when a change occurs we can measure the change in internal energy:

E = Efinal − Einitial

• Total Internal Energy (E) = the sum of all kinetic and potential energies of all components of the system. How can it be determined?

• Usually we have no way of knowing the total internal energy of a system; finding that value is simply too complex a problem.

Esys + Esurr = 0 Esys + Esurr = ETotal (constant)

Thermochemistry

Energy as Work (w) of Gas Expansion

{WorkGasExpansion}

Work = - (Force x distance)

w = - (F x h)

w = - (P x A x h)

w = - P V(@ constant P)

F = P A

w = - nRT(@ constant V)

Thermochemistry

Exchange of Heat (H) by chemical systems

• When heat is released by the system to the surroundings, the process is exothermic.

• When heat is absorbed by the system from the surroundings, the process is endothermic.

- H

+ H

q

q

Thermochemistry

Enthalpies of Reaction (H)The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:

H = Hproducts − Hreactants

This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

Notice it is the same value, but of opposite sign for the reverse reaction

Reactants:

Products

Thermochemistry

State Functions

A state function depends only on the present state of the system, not on the path by which the system arrived at that state.

• Other state functions are P, V and T.

∆E=q+w

• Are E, q and w all state functions?

• However, q and w are not state functions.

Thermochemistry

Calorimetry• The experimental methods of measuring of

heats (H) involved in chemical reactions

• Methods utilized:1. Constant Pressure

Calorimetry (open container in contact with atmosphere)

2. Constant Volume Calorimetry: closed container (bomb)

1 atm

Thermochemistry

Thermochemical Equations

2 C8H18 (l) + 25 O2 (g) 16 CO2(g) + 18 H2O(g)

H = - 5.5 x 106 kJ/ C8H18

1. H is per mole of reactant, at the indicated states (s, l, g).

2. Direct quantitative relationship between coefficients of balanced equation and H.

3. Reverse equation has H of opposite sign.

Problem: Calculate H when 25g of C8H18 (l) (MM= 114) are burned in excess oxygen?

HC 25g kJ? 188

g114

1 kJ 10x 1.2- 6

1

kJ 10 x 5.5

6

Thermochemistry

E = q + w

E + PV = qp

E = qp - PV

qp = Heat of Reaction

=

or Enthalpy of Reaction

H = E + PV

SOLVE FOR

1 atm

• Most chemical reactions are carried out in beakers or flasks that are open to the atmosphere (i.e., at constant pressure, isobaric).

(1) Calorimetry at Constant Pressure

H = Epv and when the volume is constant

q = m c T

E + PV = H Big Problem:

calorimeter also

absorbs heat

Thermochemistry

(a) Determining the Heat Capacity of a Calorimeter

50 mL of H2O @ 650C are added…….

…..to 50 mL of H2O

@ 250C inside the calorimeter

The final temperature of the 100 mL of H2O inside the calorimeter is 420C

Heat lost by = heat gained by + heat gained by hot water cold water calorimeter

qhot = qcold + qcalorimeter

50 mL H2O

50 mL H2O

Thermochemistry

qhot = qcold + qcalorimeter

Determining the Heat Capacity of a Calorimeter (Ccal)

(mcT)H2O = (mcT)H2O + (Ccal T)

(mcT)H2O - (mcT)H2O

(T)cal

Ccal =

2542

2542/184.450 - 6542/184.450g

0

0000

C

CCgJgCCgJCcal

C/J492

C)17(

J)4.3556 - (J6.4811-C 0

0cal

Includes both the mass and heat capacity

Thermochemistry

(b) Calculating the Heats of Reactions (H)

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

qrxn = (mc*T)NaClsoln + (CcalT)

qrxn = - Hrxn

= heat of neutralization rxn* How do you determine the cNaCl soln?

50 mL@25°C + 50 mL@25°C 100 mL of NaCl(aq) @ 30°C

H(qrxn) = qNaCl soln + qcal

Heat of = heat gained by + heat gained by reaction NaCl solution calorimeter

Thermochemistry

How do you determine the cNaCl soln?

50 mL of H2O @ 650C are added…….

…..to 50 mL of NaCl solution @ 250C inside the calorimeter

The final temperature of the 100 mL of solution inside the calorimeter is 410C

50 mL H2O

50 mL NaCl soln

Heat lost by = heat gained by + heat gained by hot water salt water soln calorimeter

qh = qNaClsoln + qcal

(mcT)H2O = (mc*T)NaClsoln + (CcalT)

Thermochemistry

(2) Calorimetry at Constant Volume (Bomb Cal.)

E = q + w

Bomb calorimeter

w = P V V =0 w = 0

E = qv

q = m c T

Thermochemistry

• Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H.

• For most reactions, the difference is very small.

Bomb Calorimetry

E = qv

H = E + nRT

Thermochemistry

Review of Thermochemical Equations

E = q + w

E = qv

E = qp - nRT since ∆H= qp

1 atm

E = qp - PV

E = H - PV

H = E + PV

E = q + w

Thermochemistry

Comparing H and E

H = E + PV

2 C8H18 (l) + 25 O2 (g) 16 CO2(g) + 18 H2O(g)

E = H - PV (useful at const P)

E = H - nRT (useful at const V)

Problem: calculate E @ 25°C for above equation @ const. V

R= 8.314J/K. gas=34-25 = +9 = 4.5 C8H18

/kJ10 x 5.5/kJ11/kJ10 x 5.5E

)K298)(J10/kJ1)(K/J314.8)(5.4(/kJ10 x 5.5E66

36

H = - 5.5 x 106 kJ/

Thermochemistry

Constant-Volume CalorimetryProblem: When one mole of CH4 (g) was combusted at 250C in a bomb

calorimeter, 886 kJ of energy were released. What is the enthalpy change for this reaction?

)K ( J

kJK) J/η.η) ( (- kJ H 27325

1000

13182886

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

E = qv = - 886 kJ

R = 8.31 J/ .K

n = 1 – (1+2) = -2

H = E + PV

= E + nRT

kJ891 kJ 95.4 kJ - 886 - ΔH

Thermochemistry

Energy in FoodsMost of the fuel in the food we eat comes from carbohydrates & fats.

Thermochemistry

Methods of determining H1. Calorimetry (experimental)

2. Hess’s Law: using Standard Enthalpy of Reaction (Hrxn) of a series of reaction steps (indirect method).

3. Standard Enthalpy of Formation (Hf ) used with Hess’s Law (direct method)

4. Bond Energies used with Hess’s Law

Experimental data combined with theoretical concepts

Thermochemistry

(2) Determination of H using Hess’s Law

Hrxn is well known for many reactions, but it is inconvenient to measure Hrxn for every reaction.

The Standard Enthalpy of Reaction (Hrxn) of a series of reaction steps are added to lead to reaction of interest (indirect method).

However, we can estimate Hrxn for a reaction of interest by using Hrxn values that are published for other more common reactions.

Standard conditions (25°C and 1.00 atm pressure).

(STP for gases T= 0°C)

Thermochemistry

Hess’s Law

“If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

- 1840, Germain Henri Hess (1802–50), Swiss

Thermochemistry

Calculation of H by Hess’s Law

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

3 C(graphite) + 4 H2 (g) C3H8 (g) H= -104

3 C(graphite) + 3 O2 (g) 3 CO2 (g) H=-1181

4 H2 (g) + 2 O2 (g) 4 H2O (l) H=-1143

C3H8 (g) 3 C(graphite) + 4 H2 (g) H= +104

• Appropriate set of Equations with their H values are obtained (or given), which containing chemicals in common with equation whose H is desired.

• These Equations are all added to give you the desired equation.

• These Equations may be reversed to give you the desired results (changing the sign of H).

• You may have to multiply the equations by a factor that makes them balanced in relation to each other.

• Elimination of common terms that appear on both sides of the equation .

Thermochemistry

Calculation of H by Hess’s Law

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

3 C(graphite) + 3 O2 (g) 3 CO2 (g) H=-1181

4 H2 (g) + 2 O2 (g) 4 H2O (l) H=-1143 Hrxn =

+ 104 kJ

-1181 kJ

- 1143 kJ

- 2220 kJ

C3H8 (g) 3 C(graphite) + 4 H2 (g) H= +104

Thermochemistry

Calculate heat of reaction

W + C (graphite) WC (s) ΔH = ?

Given data:

2 W(s) + 3 O2 (g) 2 WO3 (s) ΔH = -1680.6 kJ

C (graphite) + O2 (g) CO2 (g) ΔH = -393.5 kJ

2 WC (s) + 5 O2 (g) 2 WO3 (s) + CO2 (g) ΔH = -2391.6 kJ

Calculation of H by Hess’s Law

C (graphite) + O2 (g) CO2 (g) ΔH = -393.5 kJ

½(2 W(s) + 3 O2 (g) 2 WO3 (s) ) ½(ΔH = -1680.6 kJ)

½(2 WO3 (s) + CO2 (g) 2 WC (s) + 5 O2 (g)) ½ (ΔH = +2391.6 kJ)

W + C (graphite) WC (s) ΔH = - 38.0

W(s) + 3/2 O2 (g) WO3 (s) ) ΔH = -840.3 kJ

WO3 (s) + CO2 (g) WC (s) + 5/2 O2 (g) ΔH = + 1195.8 kJ)

Thermochemistry

Problem: Chloroform, CHCl3, is formed by the following reaction: Desired ΔHrxn equation: CH4 (g) + 3 Cl2 (g) → 3 HCl (g) + CHCl3 (g)

Determine the enthalpy change for this reaction (ΔH°rxn), using the following:

2 C (graphite) + H2 (g) + 3Cl2 (g) → 2CHCl3 (g)ΔH°f = – 103.1 kJ/mol CH4 (g) + 2 O2 (g) → 2 H2O (l) + CO2 (g) ΔH°rxn = – 890.4 kJ/mol2 HCl (g) → H2 (g) + Cl2 (g) ΔH°rxn = + 184.6 kJ/molC (graphite) + O2 (g) → CO2(g) ΔH°rxn = – 393.5 kJ/molH2 (g) + ½ O2 (g) → H2O (l) ΔH°rxn = – 285.8 kJ/mol

answers: a) –103.1 kJ b) + 145.4 kJ c) – 145.4 kJ d) + 305.2 kJ e) – 305.2 kJ f) +103.1 kJ

This is a hard question. To make is easer give:C (graphite) + ½ H2(g) + 3/2 Cl2(g) → CHCl3(g) ΔH°f = – 103.1 kJ/mol

Hess’s Law

Thermochemistry

Methods of determining H1. Calorimetry (experimental)

2. Hess’s Law: using Standard Enthalpy of Reaction (Hrxn) of a series of reaction steps (indirect method).

3. Standard Enthalpy of Formation (Hf ) used with Hess’s Law (direct method)

4. Bond Energies used with Hess’s Law

Experimental data combined with theoretical concepts

Thermochemistry

(3) Determination of H using Standard Enthalpies of Formation (Hf )

Standard Enthalpy of formation Hf are measured under standard conditions (25°C and 1.00 atm pressure).

Enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

C + O2 CO2 ∆Hf = -393.5 kJ/

Thermochemistry

Calculation of H

We can use Hess’s law in this way:

H = nHf(products) - mHf(reactants) where n and m are the stoichiometric coefficients.

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

C + 2H2(g) CH4(g) ΔHf = -74.8 kJ/ŋC(g) + O2(g) CO2(g) ΔHf = -393.5 kJ/ŋ

2H2(g) + O2(g) 2H2O(g) ΔHf = -241.8 kJ/ŋ

H = [1(-393.5 kJ) + 1(-241.8 kJ)] - [1(-74.8 kJ) + 1(-0 kJ)]

= - 560.5 kJ

n CO2(g) + n H2O(g)n CH4(g) + n O2(g) -

Thermochemistry

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

Calculation of H

H = [3(-393.5 kJ) + 4(-285.8 kJ)] - [1(-103.85 kJ) + 5(0 kJ)]

= [(-1180.5 kJ) + (-1143.2 kJ)] - [(-103.85 kJ) + (0 kJ)]

= (-2323.7 kJ) - (-103.85 kJ)

= -2219.9 kJ

H = nHf(products) - mHf(reactants)

Table of Standard Enthalpy of formation, Hf

Thermochemistry

• Most simply, the strength of a bond is measured by determining how much energy is required to break the bond.

• This is the bond enthalpy.• The bond enthalpy for a Cl—Cl bond,

D(Cl—Cl), is measured to be 242 kJ/mol.

(4) Determination of H using Bond Energies

Thermochemistry

Average Bond Enthalpies (H)

NOTE: These are average bond enthalpies, not absolute bond enthalpies; the C—H bonds in methane, CH4, will be a bit different than theC—H bond in chloroform, CHCl3.

• Average bond enthalpies are positive, because bond breaking is an endothermic process.

Thermochemistry

Enthalpies of Reaction (H )

• Yet another way to estimate H for a reaction is to compare the bond enthalpies of bonds broken to the bond enthalpies of the new bonds formed.

• In other words, Hrxn = (bond enthalpies of bonds broken)

(bond enthalpies of bonds formed)

Thermochemistry

Hess’s Law: Hrxn = (bonds broken) (bonds formed)

Hrxn = [D(C—H) + D(Cl—Cl) [D(C—Cl) + D(H—Cl)

= [(413 kJ) + (242 kJ)] [(328 kJ) + (431 kJ)]

= (655 kJ) (759 kJ)

= 104 kJ

CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)

Thermochemistry

Bond Enthalpy and Bond Length

• We can also measure an average bond length for different bond types.

• As the number of bonds between two atoms increases, the bond length decreases.

Thermochemistry

2003 B Q3

Thermochemistry

Thermochemistry

2005 B

Thermochemistry

Thermochemistry

2002

Thermochemistry

Thermochemistry

Thermochemistry

2002 #8

Thermochemistry

Thermochemistry

2003 A