Download Welcome to Thermochemistry!

Welcome to
Thermochemistry!
Energy in Chemistry
(11:23)

Energy is the ability to do work or produce heat. The
sum of the potential energy and kinetic energy is equal
to the internal energy of the system.

Potential energy is the energy stored in chemical bonds.
To break a bond, energy must be added to the system
and so the P.E. goes up. (ie: changing liquid water to
water vapor, you must add heat or lower pressure). The
attraction between the nucleus of an atom and the
shared electrons is called the coulombic attraction
which is the electrostatic attraction between two
charged particles.
Coulombic attractions

Electrostatic attraction (coulombic forces) are the strongest of all
intermolecular forces. They cause opposite charges to attract and
like charges to repel. When they act between separate charged
particles (ion-ion interactions) they are especially strong.

The energy it takes to pull a Na+ and a Cl- apart in a sodium
chloride crystal is much greater than that needed to pull apart a
hydrogen molecule.

When atoms bond, the coulombic attraction results in energy being
released and a subsequent lower potential energy.

Kinetic energy is the energy of motion. KE = ½ mv2

First Law of Thermodynamics is the Law of Conservation of Energy.

Heat vs. Temperature: Heat is the total amount of kinetic energy in
a system and is represented by the letter q. Temperature is the
average kinetic energy in a system and is measured in oC or K.
Enthalpy (H): The flow of energy (heat exchange) at
constant pressure when two systems are in contact.

Enthalpy of reaction: ΔHrxn is the amount of heat released in an
exothermic reaction (negative values) or amount of heat absorbed
in an endothermic reaction (positive values). kJ/molrxn

Enthalpy of combustion: ΔHcomb is the heat absorbed or released by
burning something usually with oxygen. kJ/molrxn

Enthalpy of formation: ΔHf is the heat released or absorbed when
ONE mole of a compound is formed from elements in their standard
state. kJ/molrxn

Enthalpy of fusion: ΔHfus is the heat absorbed to melt (overcome
IMFs) of 1 mole of a solid to a liquid at its melting point. kJ/molrxn

Enthalpy of vaporization: Δhvap is the heat absorbed to vaporize 1
mole of a liquid to vapor at its boiling point. kJ/molrxn
Other definitions

Endothermic: Energy absorbed by the system. + ΔH

Exothermic: Energy released by the system. - ΔH

Entropy: The measure of the dispersal of matter and energy; increase
dispersal +ΔS; decrease dispersal –ΔS

Gibbs Free Energy (G): criteria for determining thermodynamic
favorability and calculating the theoretical amount of energy to do
work.

Standard Conditions: 1 atm, 25oC, and 1.0 M solutions. Standard
conditions are indicated by adding the symbol o to G, S, or H. So if you
see ΔHo , ΔSo, ΔGo then you know you’re dealing with STP.

ΔE = q(heat) + w (work)

+ w if work is done on the system (compression of gases)

-w if work is done by the system (expansion of gases)

Work = -PΔV
Gibbs Free Energy
(13:00)
It is the energy associated with a chemical reaction that can be used
to do work and is the sum of its enthalpy (H)
plus the product of the temperature and the entropy (S) of the system.
This quantity can be defined as:
G=H−TS
or more completely as
G=U+PV−TS
where
•U = internal energy (SI unit: joule)
•P = pressure (SI unit: pascal)
•V = volume (SI unit: m 3 )
•T = temperature (SI unit: kelvin)
•S = entropy (SI unit: joule/kelvin)
•H = enthalpy (SI unit: joule)
Enthalpy ΔH

Measure only the change in enthalpy (the difference between the
potential energies of the products and reactants.)

Enthalpy can be calculated from several sources including:

Stoichiometry

Calorimetry

Tables of standard values

Hess’s Law

Bond energies
Potential energy diagrams
ΔH = PEproducts – PEreactants
EA is always +
Potential Energy Diagram with
Catalyst
Calorimetry
(12:00)
q = m· C · ΔT
or
J = g · ΔT · SH
q = heat energy
m = mass of water
C = specific heat capacity
ΔT = the change in temperature
SH of water = 4.184 J/goC
Calculate the amount of energy it takes to heat
235.0 g of water from 0oC to boiling at sea level.
Hess’s Law

2C(s) + H2(g) ---> C2H2(g)ΔH° = ??? kJ

Given the following thermochemical equations:

C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(l)ΔH° = -1299.5 kJ

C(s) + O2(g) ---> CO2(g)ΔH° = -393.5 kJ

H2(g) + (1/2)O2(g) ---> H2O(l)ΔH° = -285.8 kJ

a) first eq: flip it so as to put C2H2 on the product side

b) second eq: multiply it by two to get 2C

c) third eq: do nothing. We need one H2 on the reactant side and
that's what we have.

2CO2(g) + H2O(l) ---> C2H2(g) + (5/2)O2(g)ΔH° = +1299.5 kJ

2C(s) + 2O2(g) ---> 2CO2(g)ΔH° = -787 kJ

H2(g) + (1/2)O2(g) ---> H2O(l)ΔH° = -285.8 kJ

Examine what cancels:

2CO2 ⇒ first & second equation

H2O ⇒ first & third equation

(5/2)O2 ⇒ first & sum of second and third equation

4) Add up ΔH values for our answer:

+1299.5 kJ + (-787 kJ) + (-285.8 kJ) = +226.7 kJ

Problem #2: The standard molar enthalpy of formation, ΔHf° , of
diborane cannot be determined directly because the compound
cannot be prepared by reaction of boron and hydrogen. However,
the value can be calculated. Calculate the standard enthalpy of
formation of gaseous diborane (B2H6) using the following
thermochemical information:

a) 4 B(s) + 3 O2(g) ---> 2 B2O3(s)ΔH° = -2509.1 kJ

b) 2 H2(g) + O2(g) ---> 2 H2O(l)ΔH° = -571.7 kJ

c) B2H6(g) + 3 O2(g) ----> B2O3(s) + 3 H2O(l)ΔH° = -2147.5 kJ

1) An important key is to know what equation we are aiming for. The
answer is in the word 'formation:‘


2B + 3H2 ---> B2H6
Remember that formation means forming one mole of our target
substance. This means that a one MUST be in front of the B2H6

2) In order to get to our formation reaction, the following must
happen to equations (a), (b) and (c):

equation (a) - divide through by two
equation (b) - multiply through by 3/2
equations (c) – flip

a) 2 B(s) + (3/2) O2(g) ---> B2O3(s)ΔH° = -1254.55 kJ

b) 3 H2(g) + (3/2) O2(g) ---> 3 H2O(l)ΔH° = -857.55 kJ

c) B2O3(s) + 3 H2O(l) ---> B2H6(g) + 3 O2(g) ΔH° = +2147.5 kJ

2B + 3H2 ---> B2H6ΔH° = +35.4 kJ
Using bond enthalpies and Hess’s
Law: ΔH = Σ E
minus Σ E
reactant bonds broken

product bonds broken
Problem #1: Hydrogenation of double and triple bonds is an important
industrial process. Calculate (in kJ) the standard enthalpy change ΔH
for the hydrogenation of ethyne (acetylene) to ethane:

H−C≡C−H(g) + 2H2(g) ---> H3C−CH3(g)

Bond enthalpies (in kJ/mol): C−C (347); C≡C (839); C−H (413); H−H (432)

1) You have to put energy into a bond (any bond) to break it. Bond
breaking is endothermic. Let's break all the bonds of the reactants:

one C≡C ⇒ +839 kJ

two C−H ⇒ 413 x 2 = +826 kJ

two H−H ⇒ 432 x 2 = +864 kJ

The sum is +2529 kJ

Note there are two C−H bonds in one molecule of C2H2 and there is
one H−H bond in each of two H2 molecules. Two different types of
reasons for multiplying by two.

2) You get energy out when a bond (any bond) forms. Bond making
is exothermic. Let's make all the bonds of the one product:

one C−C ⇒ −347 kJ

six C−H ⇒ −413 x 6 = −2478

The sum is −2826 kJ

3) ΔH = the energies required to break bonds (positive sign) plus the
energies required to make bonds (negative sign):

+2529 + (−2825) = −296 kJ/mol
Phase Changes

For water:

Hvap = 2330 J/g

Hfus = 335 J/g

SHsolid = 2.06 J/goC

SHliquid = 4.184 J/goC

SHvap = 1.87 J/goC