Thermochemistry: Energy Flow and Chemical Reactions · PDF fileEnthalpy of Formation ∆H =...

Thermochemistry:

Energy Flow and

Chemical Reactions

Enthalpy of Formation

∆H formation = ∆H f = enthalpy associated with the

formation of a compound from its constituent elements

Examples of formation equations:

C (graphite) + O2 (gas) → CO2 (gas)2C (graphite) + H2 (gas) → C2H2 (gas)C (graphite) + 2H2 (gas) + 1/2 O2 (gas) + N2 (gas) → CH4N2O(urea)

The reactants should be elements in their natural form

Write balanced formation equations at standard conditions for each of the following substances:

a) CaCl2b) NaHCO3

c) CCl4d) HNO3

Answer:a) Ca(s) + Cl2(g) → CaCl2(s)

b) Na(s) + ½ H2(g) + Cgraph + 3/2 O2(g) → NaHCO3(s)

c) Cgraph + 2Cl2(g) → CCl4(l)

d) ½ H2(g) + ½ N2(g) + 3/2 O2(g) → HNO3(l)



∆H = depends of various conditions such as T, P, state of matter, etc., of the reactants and products

To compare enthalpies, it is convenient to define a set of conditions called a standard state at which the enthalpies are tabulated.

Standard state = pure form of the substance at 1.0 atmospheric pressure

and at 25 C (298 K)

∆Ho = standard enthalpy = enthalpy at standard conditions of all species

∆Hof = standard enthalpy of formation

= change in enthalpy of reaction that forms 1.0 mol of compound from its elements with all the substances in their standard states

If the element exists in more than one form under standard conditions, the most stable form of that element is used for ∆Ho

f


2C (graphite) + 3H2 (gas) + ½ O2 (gas) → C2H5OH ∆Hof = -277.7 kJ

C (graphite) + O2 (gas) → CO2 (gas) ∆Hof = - 393.5 kJ

2C (graphite) + 3H2 (gas) → C2H6 (gas) ∆Hof = - 84.68 kJ

The standard enthalpy of formation, ∆Hof values are tabulated in books

and can be consulted every time it is needed.

Table 6.5 Selected Standard Heats of Formation at 250C(298K)

Formula DH0f(kJ/mol)

calcium

Ca(s)

CaO(s)

CaCO3(s)

carbon

C(graphite)

C(diamond)

CO(g)

CO2(g)

CH4(g)

CH3OH(l)

HCN(g)

CSs(l)

chlorine

Cl(g)

0

-635.1-1206.9

0

1.9

-110.5

-393.5

-74.9

-238.6

135

87.9

121.0

hydrogen

nitrogen

oxygen


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

-92.3

0

218

-45.9

90.3

143

-241.8

-285.8

107.8


silver

Ag(s)

AgCl(s)

sodium

Na(s)

Na(g)

NaCl(s)

sulfur

S8(rhombic)

S8(monoclinic)

SO2(g)

SO3(g)

0

0

0

-127.0

-411.1

2

-296.8

-396.0


DH0rxn = S mDH0

f(products) - S nDH0f(reactants)

where m = number of moles of each product

n = number of moles of each reactant

Using the tabulated values, we can calculate the enthalpy of reaction

Sample Problem

Calculate the enthalpy of reaction of the equation below using standard enthalpies of formation.

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

Solution:∆Ho reaction = 3∆Ho

f [CO2(g)] + 4Hof [H2O (l)] - ∆Ho

f [C3H8 (g)] -5∆Ho

f [O2 (g)]= 3 (-393.5 kJ) + 4 (-285.8 kJ) – (-103.85 kJ) – 5( 0 kJ)= - 2220 kJ

Sample Problem

Calculate the enthalpy of reaction of the equation below using standard enthalpies of formation.

C6H6 (g) + 15/2 O2 → 6 CO2 (g) + 3 H2O (l)

Solution:∆Ho reaction = 6∆Ho

f [CO2(g)] + 3Hof [H2O (l)] - ∆Ho

f [C6H6 (g)] -15/2 ∆Ho

f [O2 (g)]= 6 (-393.5 kJ) + 3 (-285.8 kJ) – (49.0 kJ) – 15/2(0 kJ)= - 3267.4 kJ

Acetylene burns in air according to the following equation:

C2H2 + 5/2 O2 → 2CO2 + H2O ΔH = -1255.8 kJ

Given that ΔHfo of CO2 = - 393.5 kJ/mol and ΔHf

o of H2O = -241.8 kJ/mol, what is the ΔHf

o of acetylene?

Solution:

∆Ho reaction = 2∆Hof [CO2(g)] + Ho

f [H2O (l)] - ∆Hof [C2H2 (g)] -

5/2 ∆Hof [O2 (g)]

∆Hof [C2H2 (g)] = 227 kJ/mol

Sample Problem

The general process for determining DH0rxn from DH0

f values.

Enth

alp

y, H

Elements

Reactants

Products

DH0rxn = S mDH0

f(products) - S nDH0f(reactants)

de

com

po

siti

on

-DH0f DH0

f

form

atio

nDH0

rxn

Hinitial

Hfinal


Worksheet # 9 - 5

1. Calculate ∆Hreaction for each of the following:

a) 2H2S (g) + 3O2 (g) → 2 SO2 (g) + 2 H2O (g)b) CH4 (g) + Cl2 (g) → CCl4 (l) + HCl (g) (unbalanced)

c) C2H6 (g) + O2 (g) → CO2 (g) + H2O (l) (unbalanced)

2. The standard enthalpy change for the reaction:

CaCO3 (s) → CaO (s) + CO2 (g) ∆H = 178.1 kJ

Given that ∆Hof of CO2 (g) = - 393.5 kJ/mol and that ∆Ho

f of CaO (s) = - 635.5 kJ/mol, what is the ∆Ho

f of CaCO3(s)?

Worksheet # 9 – 5: Answers

1. Calculate ∆Hreaction for each of the following:

a) 2H2S (g) + 3O2 (g) → 2 SO2 (g) + 2 H2O (g) ∆H = - 1036.8 kJb) CH4 (g) + 4Cl2 (g) → CCl4 (l) + 4HCl (g) ∆H = - 433 kJc) C2H6 (g) + 7/2 O2 (g) → 2CO2 (g) + 3H2O (l) ∆H = - 1560 kJ

2. The standard enthalpy change for the reaction:

CaCO3 (s) → CaO (s) + CO2 (g) ∆H = 178.1 kJ

Given that ∆Hof of CO2 (g) = - 393.5 kJ/mol and that ∆Ho

f of CaO (s) = - 635.5 kJ/mol, what is the ∆Ho

f of CaCO3(s)?Answer = -1207.1 kJ/mol

Bond Energy

Examine closely the following reaction:

Bond Energy

Bond Energy – the energy required to break a bond or to form a bond

Bond Energy

When is bond energy negative?

When is bond energy positive?

Bond Energy

It takes energy to break a bond. Thus, bond breakage is positive (endothermic).

Energy is released when a bond is formed. Thus, bond formation is negative (exothermic).

Bond energy for a particular bond has the same numerical value. The sign changes depending on the whether bond is broken or formed.

Bond Energy

Bond Energy

What is the relationship of bond energy and bond strength?

The stronger is the bond, the greater the energy required to break the bond and larger is the bond energy.

Bond Energy

What is the relationship of bond energy and bond order (number of bonds)?

The higher the number of bonds, the greater the energy required to break them, the larger is the bond energy!

Bond Energy

What is the relationship of bond energy and bond length?

The shorter the bond length, the greater the energy required to break the bond and the larger is the bond energy!

Enthalpy of Reaction and Bond Energies


Using bond energies to calculate DH0rxn.

DH0rxn = ΣDH0

reactant bonds broken +

Σ DH0product bonds formed


DH01 = + sum of BE DH0

2 = - sum of BE

DH0rxn

Enth

alp

y, H

BOND BREAKING

BOND FORMATION

2CO + O2 → 2CO2 ∆H = ?


2CO + O2 → 2CO2 ∆H = ?

∆H reaction = sum of bond energies reactants + sum of bond energies products

= 2[BE(CΞO)] + [BE(O2)] + 4 [-BE(C = O)]= 2(1070) + 498 + 4(-799) kJ= - 588 kJ

O

O

O O

The structures of each of the substances have to be known!


HH

HH

HO

H

O O+

2

+

∆H reaction = 4 BE(C-H) + 2 BE(O2) + 4[-BE(O-H)] + 2[-BE(C = O)]= 4 (413) + 2 (498) - 4(467) – 2(799) kJ

2

DH0rxn= - 818kJ

O

O

Enth

alp

y,H

BOND BREAKING

4BE(C-H)= +1652kJ

2BE(O2)= + 996kJ

DH0(bond breaking) = +2648kJBOND FORMATION

4[-BE(O-H)]= -1868kJ

DH0(bond forming) = -3466kJ

DH0rxn= -818kJ

2[-BE(C O)]= -1598kJ


Sample Problem

Use bond energies to calculate DH0rxn for the following reaction:

C2H4 + Cl2 C2H4Cl2

Answer: -168 kJ

H Cl

H

ClH

H

Cl Cl+

H

HH

H

Sample Problem

Use bond energies to calculate DH0rxn for the following reaction:

Answer: - 22 kJ

OH

HH

H

H

HH

O

OH

+ C Ξ O

Worksheet # 9 - 6

HH

HH

ClCl

ClCl

ClH

HCl

+ 3 + 3 →

1. Calculate the ∆H reaction (using bond energies) for the chlorination of methane to form chloroform?

2. Calculate the ∆H reaction (using bond energies) for the following reaction.

H

HH

H

H H

HH

H

Br

+ H Br →


HH

HH

ClCl

ClCl

ClH

HCl

+ 3 + 3 →

1. Calculate the ∆H reaction (using bond energies) for the chlorination of methane to form chloroform?

2. Calculate the ∆H reaction (using bond energies) for the following reaction.

H

HH

H

H H

HH

H

Br

+ H Br →

Answer: -330 kJ

Answer: -59 kJ

Table 9.5 Heats of Combustion(DHcomb) of Some Carbon Compounds

Two-Carbon Compounds One-Carbon Compounds

Ethane (C2H6) Ethanol (C2H6O)

Sum of C-C and C-H Bonds

Structural Formulas

DHcomb(kJ/mol)

Sum of C-O and O-H Bonds

Methane (CH4) Methanol (CH4O)

DHcomb(kJ/g)

C C

H

H

H

H

H

H C C

H

H

H

H

H

O H C

H

H

O HHC

H

H

HH

7 6 4 3

0 2 0 2

-1560 -1367 -890 -727

-51.88 -29.67 -55.5 -22.7

Points to Ponder…..

Why are fossil fuels (not other compounds) used as gasoline in vehicles?

What is the difference if one uses bioethanol as a fuel?

Points to Ponder…..

Why are carbohydrates the major source of energy of humans?

Homework

What is the difference between the enthalpy calculated using enthalpy of formation and bond energies?

Calculate and compare the enthalpies of reaction (using enthalpy of formation) in the reactions used in worksheet #9-6.

Worksheet # 9 – 7

O

O

N H

H

H

NHH

NH

H

O

O H

H→

+

+

1. Calculate the ∆H reaction for the following reaction using bond energies and standard enthalpy of formation. The equation is not balanced.

H

H

H

H

2. Calculate the ∆H reaction for the following reaction using bond energies and standard enthalpy of formation.

→

+

Br H

H

HH

Br

Br Br

Worksheet # 9 – 7

NHH

NH

H

OUrea - ∆H, f = - 333.3 kJ/mol

Br H

H

HH

Br

1,2 - dibromoethane - ∆H, f = - 37.8 kJ/mol


O

O

N H

H

H

NHH

NH

H

O

O H

H→

+

+


H

H

H

H


→

+ 2

Ans: - 133.3 kJ (from ∆H, formation)

Ans: - 92 kJ (from bond energy) and – 90.1 kJ(from ∆H, formation)

Br Br

Br H

H

HH

Br

Enthalpies of Reaction

Which is a more accurate calculation?

• the one using the enthalpies of formation (state of substances indicated, position of atoms indicated, values are not averaged, etc..)

• bond energies are used to only estimate the enthalpy of reaction and if enthalpy of formation of some substances are unknown

Hess’s Law of Heat Summation

= the enthalpy change of an overall process is the sumof the enthalpy changes of its individual steps

Steps in calculating an unknown ΔH1. Identify the target equation, the step whose ΔH is unknown and note

the number of moles of reactants and products.2. Manipulate the equations with known ΔH values so that the target

numbers of moles of reactants and products are on the correct side.Remember: - change the sign of ΔH when you reverse an equation- multiply numbers of moles and ΔH by the same factor

3. Add the manipulated equations to obtain the target equation. Add their ΔH values to obtain the unknown ΔH .

Hess Law

Calculate ΔHrxn for:Ca + ½ O2 + CO2 → CaCO3

Given the following set of reactions:Ca + ½ O2 → CaO ΔH = - 635.1 kJCaCO3 → CaO + CO2 ΔH = 178.3 kJ

Answer: - 813.4 kJ

Sample Problem

Answer: - 2490 kJ

Calculate ΔHrxn for:C2H4 + 6F2 → 2CF4 + 4HF

Given the following set of reactions:H2 + F2 → 2HF ΔH = - 537 kJC + 2F2 → CF4 ΔH = - 680 kJ2C + 2H2 → C2H4 ΔH = + 52.3 kJ

Sample Problem

Worksheet # 9 - 8

1. Given the following data2ClF (g) + O2 (g) → Cl2O (g) + F2O (g) ∆H = 167.4 kJ2ClF3 (g) + 2O2 (g) → Cl2O (g) + 3F2O (g) ∆H = 341.4 kJ2F2 (g) + O2 (g) → 2F2O (g) ∆H = -43.4 kJCalculate the ∆Ho

reaction for the reactionClF (g) + F2 (g) → ClF3 (g)

2. Calculate ∆Horeaction for

2NOCl (g) → N2 (g) + O2 (g) + Cl2(g)Given the following set of reactions:

½ N2 (g) + ½ O2 (g) → NO (g) ∆H = 90.3 kJNO (g) + ½ Cl2 (g) → NOCl (g) ∆H = - 38.6 kJ


1. Given the following data2ClF (g) + O2 (g) → Cl2O (g) + F2O (g) ∆H = 167.4 kJ2ClF3 (g) + 2O2 (g) → Cl2O (g) + 3F2O (g) ∆H = 341.4 kJ2F2 (g) + O2 (g) → 2F2O (g) ∆H = -43.4 kJCalculate the ∆Ho

reaction for the reactionClF (g) + F2 (g) + ClF3 (g) Ans: -108.7 kJ

2. Calculate ∆Horeaction for

2NOCl (g) → N2 (g) + O2 (g) + Cl2(g)Given the following set of reactions:

½ N2 (g) + ½ O2 (g) → NO (g) ∆H = 90.3 kJNO (g) + ½ Cl2 (g) → NOCl (g) ∆H = - 38.6 kJ

Ans: - 103.4 kJ

The common lead-acid car battery produces a large burst of currrent, even at low temperatures, and is rechargeable. The reaction that occurs while recharging a dead battery is:

2PbSO4(s) + 2H2O(l) → Pb(s) + PbO2(s) + 2H2SO4(l)

a. Use ΔHfo values to calculate ΔHrxn.

b. Use the following equations to check your answer in part (a).

Pb(s) + PbO2(s) + 2SO3(g) → 2PbSO4(s) ΔH = -768 kJSO3(g) + H2O(l) → H2SO4(l) ΔH = -132 kJ

Answer: = 2(-813.99) + (-276.6) + 0 - 2(- 918.39) – 2(-285.84) = 503.88 kJ

Heats of Formation and Hess’s Law

Energy from Foods

Heats of Combustion of Some Fats and Carbohydrates

Fats

Carbohydrates

vegetable oil -37.0

margarine -30.1

butter -30.0

table sugar (sucrose) -16.2

brown rice -14.9

maple syrup -10.4

Substance DHcomb(kJ/g)

Energy from Foods

Which is a better source of energy for our body?

Proteins

Carbohydrates

Fats

Energy from Foods

Calorie determination of food labels:

carbohydrate: (sucrose, a simple sugar)

C12 H22 O11 DH formation = -2226.1 kJ/mol

C12H22O11 + 12O2 → 12 CO2 + 11 H2O= 12 (-393.5 kJ) + 11 (-286 kJ) + 2226.1 kJ= - 5641.9 kJ

Energy from Foods


protein: (glycine, a simple amino acid)

C2H5O2N DH formation = - 537. 25 kJ/mol

4C2H5O2N(s) + 9O2(g) -> 8CO2(g) + 10H2O(l) + 2N2(g)

= 8(-393.5 kJ) + 10(-285.5 kJ) – (-537.25)4 kJ= - 3859 kJ

Energy from Foods


Fats: (lauric acid)

C12 H24 O2 DH formation = - 764 kJ/mol

C12H24O2 + 12O2 → 12 CO2 + 12 H2O= 12 (-393.5 kJ) + 11 (-286 kJ) + 764 kJ= - 7 104 kJ/mol

Points to ponder…….

How do cold packs and hot packs work?

Thermochemistry: Energy Flow and Chemical Reactions · PDF fileEnthalpy of Formation ∆H =...

Documents

Transcript of Thermochemistry: Energy Flow and Chemical Reactions · PDF fileEnthalpy of Formation ∆H =...