Thursday, April 10, 2008

Hm

Chapter 5
Reactions in Aqueous Solutions
Chapter goals
Understand the nature of ionic substances dissolved in water.
Recognize common acids and bases and understand their behavior in aqueous solution.
Recognize and write equations for the common types of reactions in aqueous solution.
Recognize common oxidizing and reducing agents and identify common oxidation-reduction reactions.
Define and use the molarity in solution stoichiometry.
Solution
homogeneous mixture
can be gas, liquid, or solid
solvent: component present in highest proportion
exception - water
solute: component(s) in solution other than solvent

We will mostly study aqueous solutions: human body is 2/3 water.
Examples
mixture of 35% naphthalene
and 65% benzene
solvent - benzene
solute – naphthalene

mixture of 10% ethanol, 40% methanol, and 50% propanol
solvent - propanol
solute - ethanol and methanol
Examples
mixture of 40% ethanol, 40% methanol, and 20% butanol
solvent - ethanol/methanol
mixed solvent
solute – butanol

mixture of 40% ethanol, 50% propanol, and 10% water
solvent - water
solute - ethanol and propanol
Next…
We will focus on compounds that
produce ions in aqueous solutions.
They are salts, acids, and bases.
Salts
Salts: ionic compounds made of cations other than H+ and anions other than OH− or O2−

NaCl: Na+ & Cl−

K2SO4: K+ & SO42−

FeBr3: Fe3+ & Br−

Zn3(PO4)2: Zn2+ & PO43−

Ca(HCO3): Ca2+ & HCO3−
Electrolyte
substance that dissolves to produce an electrically conducting medium
form ions in solution (dissociates/ionizes)
examples
soluble ionic compounds
H2O
KBr(s)  K+(aq) + Br–(aq)

H2O
Acids, HCl(g)  H+(aq) + Cl–(aq)

bases, NH3 + H2O NH4+ + OH–
Nonelectrolytes
do not form ions in solution
do not form electrically conducting media upon dissolution
Examples: molecular compounds (alcohols, sugars & acetone)
H2O
CH3OH(l)  CH3OH(aq) N.R.
Glucose C6H12O6(s)  C6H12O6(aq) N.R.
Sucrose C12H22O11(s)  C12H22O11(aq) N.R.
N.R. = no reaction;
no dissociation/ionization
Types of Electrolytes
Strong: dissociate ~100%
most ionic compounds (soluble salts), strong acids, and strong bases
H2O
KBr(s)  K+(aq) + Br–(aq)
HCl(g)  H+(aq) + Cl–(aq)

Weak: insoluble salts, weak acids and bases, water, and certain gases (e.g. CO2)
dissociate only slightly in water
H2O
HF(g) H+(aq) + F–(aq)
Solubility of Ionic compounds in Water: Solubility Rules
Soluble Compounds
1. alkali metal salts (Li+, Na+, K+, Rb+…, ) except potassium perchlorate
2. ammonium (NH4+) salts
3. all nitrates(NO3−), chlorates (ClO3−), perchlorates (ClO4−), and acetates (C2H3O2−), except silver acetate and potassium perchlorate
4. all Cl−, Br−, and l− are soluble except for Ag+, Pb2+, and Hg22+ salts
5. all SO42− are soluble except for Pb2+, Sr2+, and Ba2+ salts

Solubility of Ionic compounds in Water: Rules
Insoluble or slightly soluble Compounds
6. metal oxides (O2−) except those of the alkali metals, Ca2+, Sr2+, and Ba2+
7. hydroxides (OH−) except those of the alkali metals, Ba2+, and Sr2+. calcium hydroxide is slightly soluble
8. carbonates, phosphates, sulfides, and sulfites except those of the alkali metals and the ammonium ion (NH4+)
9. for salts of Cr2O72−, P3−, CrO42−, C2O42−,
assume they are insoluble except for IA metals and NH4+ salts
Precipitation Reactions:
A Driving Force in Chemical Reactions
formation of insoluble solid (precipitate, ppt) is a common reaction in aqueous solutions:
reactants are generally water-soluble ionic compounds
once substances dissolve in water they dissociate to give the appropriate cations and anions
if the cation of one compound forms an insoluble compound with the anion of another, precipitation will occur
Precipitation Reaction:
A Double Replacement (Metathesis) Reaction
Both ionic compounds trade partner ions
__________
| |
AB(aq) + CD(aq)  AD(s) + CB(aq)
|_______|
AD is an insoluble or slightly soluble salt
A+, B−, C+, and D− are ions

AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
weak electrolyte
(unionized precipitate)
Precipitation Reaction:
A Double Replacement (Metathesis) Reaction
A (solid) precipitate is formed.
Example: complete and balance the equation
(NH4)3PO4(aq) + MgSO4(aq)  MgPO4(s) + NH4SO4(aq)
we will write the right subscripts later
Using the solubility rules, predict if at least one
product is going to be insoluble in water.
According to rule 8, MgPO4 (subscripts not
right) is not soluble in water.
Ions are Mg2+, PO43−, NH4+, and SO42−;
subscripts
(NH4)3PO4(aq) + MgSO4(aq)  Mg3(PO4)2(s) + (NH4)2SO4(aq)

balancing
2(NH4)3PO4(aq) + 3MgSO4(aq)  Mg3(PO4)2(s) + 3(NH4)2SO4(aq)
Example: complete and balance the equation
Na2SO4(aq) + BaBr2(aq) 
Na2SO4(aq) + BaBr2(aq) BaSO4(s) + NaBr
Na2SO4 + BaBr2 BaSO4(s) + 2NaBr(aq)
driving force = formation of insoluble barium sulfate (precipitate)
Os(NO3)5(aq) + Rb2S(aq) 
Os(NO3)5 + Rb2S  Os2S5(s) + RbNO3(aq)
2 Os(NO3)5 + 5 Rb2S  Os2S5(s) + 10 RbNO3
driving force = form. of insoluble sulfide
(precipitate)
Net Ionic Equations: Spectator Ions
The equation
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
is not quite correct, because three salts are
dissociated in ions while AgCl is a precipitate.

Ag+(aq) + NO3−(aq) + Na+(aq) + Cl−(aq)  AgCl(s) + Na+(aq) + NO3−(aq)
before reaction after reaction

Na+ and NO3− are present on both sides of equation, i.e.,
before and after reaction. They are called spectator ions; do
not participate in net reaction; can be removed from the
equation, but they remain in the solution.

Ag+(aq) + Cl−(aq)  AgCl(s) is the net ionic equation
Net Ionic Equations: Spectator Ions
For two previous examples:
2(NH4)3PO4(aq) + 3MgSO4(aq)  Mg3(PO4)2(s) + 3(NH4)2SO4(aq)

6NH4+(aq) + 2PO43−(aq) + 3Mg2+(aq) + 3SO42−(aq)  Mg3(PO4)2(s) +
before reaction 6NH4+(aq) + 3SO42−(aq)
after reaction
3Mg2+(aq) + 2PO43−(aq)  Mg3(PO4)2(s) is the net equation

spectator ions are eliminated from the equation
=======================
Na2SO4(aq) + BaBr2(aq) BaSO4(s) + 2NaBr(aq)

2Na+(aq) + SO42−(aq) + Ba2+(aq) + 2Br−(aq)  BaSO4(s) + 2Na+(aq) + 2Br−

Ba2+(aq) + SO42−(aq)  BaSO4(s) net ionic equation
Net Ionic Equations: Spectator Ions
For the metathesis reaction
2 KF + Pb(NO3)2  PbF2(s) + 2 KNO3
formula unit equation
spectator ions are eliminated
2K+ + 2F– + Pb2+ + 2 NO3–  PbF2 + 2K+ + 2NO3–
ionic equation

PbF2 is the precipitate
2F–(aq) + Pb2+(aq)  PbF2(s)
net ionic equation
Net Ionic Equations: Spectator Ions

NH4Cl(aq) + KNO3(aq)  NH4NO3(aq) + KCl(aq)

NH4+ + Cl– + K+ + NO3–  NH4+ + NO3– +
K+ + Cl–

all ions are spectators; all can be cancelled
no net ionic equation
no driving force for reaction
N.R. (no reaction)
Acids and Bases
Acid

Arrhenius definition
substance that ionizes in water to produce H+, hydrogen ion, and hence increases the concentration of this ion
HCl(aq)  H+(aq) + Cl–(aq)

Brønsted-Lowry definition
substance capable of donating H+
HCl + H2O  H3O+ + Cl–(aq)
Acids and Bases
Base
Arrhenius definition
substance that increases the concentration of OH– in aqueous solution
KOH(aq)  K+(aq) + OH–(aq)
NH3 + H2O NH4+ + OH–

Brønsted/Lowry definition
substance capable of accepting H+
KOH(aq)  K+(aq) + OH–(aq)
OH– + H+  H2O (OH– from NaOH accepts H+)
NH3 + H2O NH4+ + OH– (NH3 accepts H+)
Water can act as both an acid and a base: it is an amphoteric substance
HClO4 + H2O  H3O+ + ClO4–
acid base
(accepts H+ from HClO4)


NH3 + H2O NH4+ + OH–
base acid
(donates H+ to NH3)
Strong Acids
dissociate ~100%
HX, X = Cl, Br, I hydro…ic acid

HNO3 nitric acid

HClO3 chloric acid

HClO4 perchloric acid

H2SO4 (first proton) sulfuric acid

H2SO4(aq) H+(aq) + HSO4−(aq)

week: HSO4−(aq) H+(aq) + SO42−(aq)
Weak Acids
dissociate <100%
most other acids
HF hydrofluoric acid
HCN hydrocyanic acid
HNO2 nitrous acid
CH3CO2H acetic acid
H2CO3 carbonic acid (both protons)
H3PO4 phosphoric acid (all protons)
H2SO3 sulfurous acid
oxalic acid H2C2O4(aq) H+(aq) + HC2O4−(aq)
Strong Bases
dissociate ~100%
alkali metal hydroxides
LiOH, NaOH, KOH, RbOH
name: lithium hydroxide

hydroxide of
Ca Ca(OH)2 calcium hydroxide
Ba Ba(OH)2
Sr Sr(OH)2
Neutralization Reactions
acid + OH-ctg. base  salt + water
HF(aq) + KOH(aq)  KF(aq) + H2O
HF(aq) + K+(aq) + OH–(aq)  K+(aq) + F–(aq) + H2O
HF(aq) + OH–(aq)  F–(aq) + H2O net ionic
spectator ions are eliminated from equation
HF is a week acid and HCl is a strong acid
acid + non-OH-ctg base  salt
HCl(aq) + NH3(aq)  NH4Cl(aq)
H+(aq) + Cl–(aq) + NH3(aq)  NH4+(aq) + Cl–(aq)
H+(aq) + NH3(aq)  NH4+(aq) net ionic equation
Formation of a Weak Acid or Base as a Driving Force
HNO3(aq) + KCN(aq)  HCN(aq) + KNO3(aq)
H+(aq) + NO3–(aq) + K+(aq) + CN–(aq)  HCN (aq) +
K+(aq) + NO3–(aq)
H+(aq) + CN–(aq)  HCN(aq) (a weak acid)

NH4Cl + NaOH(aq)  NH4OH + NaCl(aq)
NH4+(aq) + Cl–(aq) + Na+(aq) + OH–(aq)  NH4OH
Na+(aq) + Cl–(aq)
NH4+(aq) + OH–(aq)  NH4OH (a weak base)
NH4OH is NH3 in water, i.e., NH3 + H2O
When no Weak Electrolytes are Formed
HNO3(aq) + KCl(aq)  HCl(aq) + KNO3(aq)
H+(aq) + NO3–(aq) + K+(aq) + Cl–(aq) 
H+(aq) + Cl–(aq) K+(aq) + NO3–(aq)
There is no net reaction: N.R. No driving force
All ions are spectators.

BaCl2(aq) + 2NaOH(aq)  Ba(OH)2(aq) + 2NaCl(aq)
Ba2+(aq) + 2Cl–(aq) + 2Na+(aq) + 2OH–(aq) 
Ba2+(aq) + 2OH–(aq) + 2Na+(aq) + 2Cl–(aq)
There is no net reaction: N.R. No driving force
Gas Forming Reactions (a Driving Force)
Some of the weak acids and bases that are
formed at double replacement reactions
decompose to form a gas and water
CO2
Na2CO3(aq) + 2HCl(aq)  H2CO3(aq) + 2NaCl(aq)
H2CO3(aq)  H2O + CO2(g)
Na2CO3(aq) + 2HCl(aq)  H2O + CO2(g) + 2NaCl(aq)

SO2
Na2SO3(aq) + 2HCl(aq)  H2SO3(aq) + 2NaCl(aq)
H2SO3(aq)  H2O + SO2(g)
Na2SO3(aq) + 2HCl(aq)  H2O + SO2(g) + 2NaCl(aq)
Redox (Oxidation-Reduction) Reactions
involve transfer of electron(s)
oxidation: loss of electron(s)
reduction: gain of electron(s)
some can be identified when an
uncombined element is a reactant or a product

eg. Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Zn  Zn2+
Zn(s)  Zn2+(aq) + 2 e–, oxidation
Cu2+  Cu
Cu2+(aq) + 2e–  Cu(s), reduction
Single Displacement Reactions
Zn(s) + CuCl2(aq)  Cu(s) + ZnCl2(aq)
Zn oxidized to Zn2+
Cu2+ reduced to Cu
occurs because zinc is more active than copper

Cl2(g) + CuBr2(aq)  Br2(l) + CuCl2(aq)
Br oxidized from Br– to Br2
Cl reduced from Cl2 to Cl–
Cl is more active than Br
Oxidation Numbers
also an accounting tool
very useful
oxidation numbers of all atoms in substance add up to charge on substance
e.g.
zero for Al2(SO4)3 and H3PO4

+1 for NH4+

–2 for Cr2O72–
Assigning Oxidation Numbers, ON
ON = 0 for all atoms in any substance in most elemental form, Na(s), Zn(s), Hg(l) H2(g), Cl2(g), I2(s), O2(g), C(s), P4(s), S8(s)
ON = charge for monatomic ions
ON = –1 for F in all compounds
ON = –2 for O in compounds, usually
exceptions: peroxide, O22–, ON = –1
superoxide, O2–, ON = –1/2
ON = +1 for H in compounds, usually
exception: ON = –1 in metallic hydrides
Assigning Oxidation Numbers, ON
ON = +1 for alkali metals in compounds
ON = +2 for alkaline earth metals in compounds
ON = +3 for Al in compounds
ON = −1 for Cl, Br, and I in binary compounds except for those with oxygen
Assign ON to Each Atom in the Following Substances


WCl6

x + 6(–1) = 0
x –6 = 0
x = +6



?
Na2S2O3

+2 + 2x –6 = 0

2x = 6 – 2

x = +2 ON of S


?
Na2S4O8

+2 + 4x –16 = 0

4x = 16 – 2

+14 +7
x = —— = —— ON of S
4 2


Cr2O72–

2x –14 = –2

2x = 14 – 2

+12
x = —— = +6 ON of Cr
2


H2C2O4

+2 + 2x – 8 = 0

2x = 8 – 2

+6
x = —— = +3 ON of C
2


MoBr5+

x – 5 = +1

x = 5 + 1

x = 6 ON of Mo = +6

Oxidizing and Reducing Agents
In every redox reaction there is one reducing
agent (the one that is oxidized) and one
oxidizing agent (the one that is reduced)


ON increases ON decreases

The species is The species is
oxidized reduced





Activity (Electromotive) Series for metals


Li
K
Ba
Ca
Na
Mg
Al
Mn
Zn
Cr
Fe
Co
Ni
Sn
Pb
(H2)
Cu
Ag
Hg
Pt
Au
Examples


Complete and balance each of the following chemical equations

A free and chemically active metal displacing
a less active metal from a compound
Mg + FeCl3 
Mg + FeCl3  Fe + MgCl2
3Mg(s) + 2FeCl3(aq)  2Fe(s) + 3MgCl2(aq)
Mg  Mg2+, oxidized; Mg reducing agent
Fe3+  Fe, reduced; Fe3+ oxidizing agent

Sn + CrF3  Sn is less reactive than Cr
Sn + CrF3  No Reaction

Pb(s) + Au(ClO3)3(aq) 
Pb(s) + Au(ClO3)3(aq)  Au(s) + Pb(ClO3)2(aq)
3Pb(s) + 2Au(ClO3)3(aq)  2Au(s) + 3Pb(ClO3)2(aq)
Pb is oxidized Au+3 is reduced

A free and chemically active metal displacing
a less active metal from a compound

Zn + CrBr3 
Zn + CrBr3  Cr + ZnBr2
3Zn(s) + 2CrBr3(aq)  2Cr(s) + 3 ZnBr2(aq)
Zn oxidized to Zn2+; Zn reducing agent
Cr3+ reduced to Cr; Cr3+ oxidizing agent


Ag(s) + Hg(NO3)2  No Reaction
Ag is less reactive than Hg

A free and chemically active metal displacing
Hydrogen from acids or water
Fe + HBr 
Fe + HBr  H2 + FeBr3
2Fe + 6HBr  3H2 + 2 FeBr3
Fe oxidized to Fe3+; Fe reducing agent
H+ reduced to H2; H+ oxidizing agent

Cu + HBr  Cu less active than H2
Cu + HBr  No Reaction

A free and chemically active metal displacing
Hydrogen from acids or water
K(s) + H2O(l) 
2K(s) + 2H2O(l)  2 KOH(aq) + H2(g)
K oxidized to K+; K reducing agent
H+ reduced to H2; H+ oxidizing agent

Ag(s) + H2O(l)  Ag less active than H2
Ag(s) + H2O(l)  No Reaction
Ni(s) + H2SO4(aq) 
Ni(s) + H2SO4(aq)  NiSO4(aq) + H2(g)
Ni oxidized to Ni2+; Ni reducing agent
H+ reduced to H2; H+ oxidizing agent


Which one of the following metals could
be used safely for lining a tank intended
for storage of sulfuric acid?
H2SO4
aluminum
iron
chromium
mercury
copper
tin
Nonmetal Activity Series
F
Cl
Br
I
same order as in periodic table
Complete and Balance
Cl2 + FeBr3 
Cl2(g) + FeBr3(aq)  Br2(l) + FeCl3(aq)
3Cl2(g) + 2FeBr3(aq)  3 Br2(l) + 2 FeCl3(aq)
Cl2 oxidizing agent; Br– reducing agent

I2(s) + NaF(aq)  I2 is less active than F2
I2(s) + NaF(aq)  No Reaction

F2(g) + NaCl(aq) 
F2(g) + 2 NaCl(aq)  Cl2(g) + 2 NaF(aq)
F2 oxidizing agent; Cl– reducing agent

Br2 + FeCl3  Br less active than Cl
Br2 + FeCl3  No Reaction
Identifying Oxidizing and Reducing Agents
0 +3 0 +2
3Zn(s) + 2CrBr3(aq)  2Cr(s) + 3 ZnBr2(aq)
Zn oxidized to Zn2+; Zn reducing agent
Cr3+ reduced to Cr; Cr3+ oxidizing agent
0 +1 +2 0
Ni(s) + H2SO4(aq)  NiSO4(aq)) + H2(g)
Ni oxidized to Ni2+; Ni reducing agent
H+ reduced to H2; H+ oxidizing agent

0 -1 0 -1
F2(g) + 2 NaCl(aq)  Cl2(g) + 2 NaF(aq)
F2 reduced; Cl– oxidized
F2 oxidizing agent; Cl– reducing agent
Identifying Oxidizing and Reducing Agents
The device for testing breath for the presence of alcohol is based on the following reaction. Identify the oxidizing and reducing agents
ON: –1 +6
3CH3CH2OH(aq) + 2Cr2O72–(aq) + 16H+ 
ethanol orange-red
+3
3CH3CO2H(aq) + 4 Cr3+(aq) + 11H2O
acetic acid green
ethanol oxidized (reducing agent)
Cr2O72– (dichromate ion) reduced (oxidizing agt.)
Balancing redox equations
ON: +6 in acidic media (H+)
Cr2O72–(aq) + Fe2+(aq)  Cr3+(aq) + Fe3+(aq)

Cr2O72– + 6e−  2Cr3+ (this the reduction)
Fe2+  Fe3+ + e− (this is the oxidation)

Cr2O72– + 6e− + 14H+  2Cr3+ (to have +6 = +6, charge)

Cr2O72– + 6e− + 14H+  2Cr3+ + 7H2O (to balance H & O)

Cr2O72– + 6e− + 14H+  2Cr3+ + 7H2O to equal # of
6(Fe2+  Fe3+ + e−) electrons

Cr2O72– + 14H+ + 6Fe2+  2Cr3+ + 7H2O + 6Fe3+
Balancing redox equations
ON: +7 +4 in basic media (OH−) +4 +6
MnO4–(aq) + SO32−(aq)  MnO2(s) + SO42−(aq)

MnO4– + 3e−  MnO2 (this the reduction)
SO32−  SO42− + 2e− (this is the oxidation)

MnO4– + 3e−  MnO2 + 4OH− (to equal charges, −4)
SO32− + 2OH−  SO42− + 2e− (to equal charges, −4)

2(MnO4– + 3e− +2H2O  MnO2 + 4OH−) to equal H, O,
3(SO32− + 2OH−  SO42− + 2e− + H2O) and electrons

2MnO4– + H2O + 3SO32−  2MnO2 + 2OH− + 3SO42−
Measuring Concentrations of Compounds in Solutions


Concentration Terms
Parts Per
Hundred (percent, %)
weight/weight, %(w/w), % (most common)

mass solute
——————— x 100
mass solution

volume/volume, %(v/v)

V solute
——————— x 100
V solution
Parts Per
Hundred (percent, %)
weight/volume, %(w/v)

mass solute
——————— x 100
V solution (mL)


Learning Check
A solution is prepared by mixing 15.0 g of Na2CO3 and 235 g of H2O. The final V of solution is 242 mL. Calculate the %w/w and %w/v concentration of the solution.
g solution = 15.0 g Na2CO3 + 235 g H2O = 250. g

15.0 g solute
%w/w = ——————— x 100 = 6.00% Na2CO3
250. g solution

15.0 g solute
%w/v = ——————— x 100 = 6.20% Na2CO3
242 mL solution
Molarity, M
The Molarity, M, usually known as the molar
concentration of a solute in a solution, is the
number of moles of solute per liter (1000 mL)
of solution/ or mmoles per mL of solution.

To calculate it we need moles of solute
and V(in liters) of solution (or mmol and mL):

mol solute mmol solute
M = —————— = ———————
V(L) solution V(mL) solution
Calculation of Molarity, M
What is the molarity of 500. mL NaOH solution if
it contains 6.00 g NaOH?
FW (NaOH) = 40.0 g/mol (from periodic table)
How many moles of NaOH?

1 mol NaOH
6.00 g x —————— = 0.150 mol NaOH
40.0 g NaOH

mol solute 0.150 mol
M = —————— = ————— = 0.300 M NaOH
V(L) solution 0.500 L
(0.300 mol in 1 L)
Formality, F
is the same as molarity, but referred to ionic
compounds in aqueous solution

FW (formula weights) of solute per L of solution

1 FW
# FW = g solute x ————— = # of FW
g solute

# FW (# of formula weights)
F = ——————
V(L) solution (volume in liters)

Formality = Molarity
Molality, m
is the amount (moles) of solute per kg of
solvent (usually but not necessarily water).
What is the m of a solution prepared by
dissolving 25.3 g Na2CO3 in 458 g water?
458 g = 0.458 kg (after dividing by 1000)
1 mol Na2CO3
25.3 g Na2CO3 x —————— = 0.239 mol Na2CO3
106.0 g Na2CO3

mol solute 0.239 mol
m = —————— = ————— = 0.522 m Na2CO3
kg H2O 0.458 kg
(0.522 mol in 1 kg H2O)
Mole Fraction, X
is the amount (mol) of a given component of a
solution per mol of solution
Here we need # moles of every component
(solute(s) and solvent) and the total
e.g. for a solution with n1, n2, n3, … mol

ni ni
mol fraction Xi = —————— = —— (no units)
n1+ n2+ n3 + … nt

ni: moles of component i (1, 2, 3, …)
nt: total number of moles
How many g of NaCl are contained in 250.0 mL of 0.2193 M NaCl solution?

250.0 mL 1000 = 0.2500 L

Now, M as a conversion factor
0.2193 mol
0.2500 L x ————— = 0.0548 mol NaCl
1 L sol’n

grams out of moles and formula weight (58.44)
58.44 g NaCl
0.0548 mol x —————— = 3.204 g NaCl
1 mol NaCl
Making Solutions
Consider the making of 1.00 L of 1.00 M NaCl solution.
need 1.00 mol NaCl or 58.5 g NaCl
dissolve 58.5 g NaCl in 1.00 L water?
NO!!
dissolve 58.5 g NaCl in water and dilute to a total volume of 1.000 L


volumetric flask: calibrated to contain, tc


pipet: calibrated to deliver, td
Example: Describe the preparation of 300.0 mL of 0.4281 M silver nitrate solution.
300.0 mL 1000 = 0.3000 L AgNO3 FW = 169.97 g/mol
0.4281 mol
0.3000 L x ————— = 0.1284 mol AgNO3
1 L sol’n
169.97 g
0.1284 mol x ————— = 21.82 g AgNO3
1 mol AgNO3

Dissolve 21.82 g AgNO3 in 300.0 mL water?
NO!!
Dissolve 21.82 g AgNO3 in water and dilute to 300.0 mL.
How many milliliters of 2.00 M HNO3 contain 24.0 g HNO3?
HNO3 FW = 63.0 g/mol How many moles?

1 mol HNO3
24.0 g HNO3 x —————— = 0.381 mol HNO3
63.0 g HNO3

Now the M with the volume on top to get mL
1 L sol’n 1000 mL
0.381 mol x —————— x ———— = 191 mL
2.00 mol HNO3 1 L sl’n

How many grams of AlCl3 are needed to 
prepare 25 mL of a 0.150 M solution?
25 mL 1000 = 0.025 L FW (AlCl3) = 133.5 g/mol

All at once:

V(L) of sol’n and mol and FW
M to calculate to calculate
mol of AlCl3 g of AlCl3


0.025 L x 0.150 mole x 133.5 g = 0.500 g AlCl3
1 L 1 mole
Dilution
the process of decreasing the concentration
of solutes in a solution by addition of solvent
or another solution that does not contain the
same solutes ChemNow 5.16 final excercise
volume increases and concentration decreases.
concentrated diluted solution
Example: Calculate the concentration of a solution made by diluting 25.0 mL of 0.200 M methanol, CH3OH, solution to 100.0 mL.
Key for calculations: moles of solute taken
from the concentrated solution are the same
in the diluted solution (only solvent is added.)
moles = concentration x V = M x V

Cc x Vc = Cd x Vd Mc x Vc = Md x Vd
one of the four is unknown
c: concentrated d: diluted
L or mL can be used for volume
Example: Calculate the concentration of a solution made by diluting 25.0 mL of 0.200 M methanol, CH3OH, to 100.0 mL.
Mc x Vc = Md x Vd
Md is the unknown
25.0 mL x 0.200 M = 100.0 mL x Md

25.0 mL x 0.200 M
Md = ————————— = 0.0500 M
100.0 mL
0.0500 mole/L
How many mL of a 0.515 M NaBr solution must be diluted to produce 500.0 mL of a 0.103 M NaBr solution?
Mc x Vc = Md x Vd Vc is the unknown

Vc x 0.515 M = 500.0 mL x 0.103 M

500.0 mL x 0.103 M
Vc = ————————— = 100.0 mL
0.515 M

Serial Dilutions
Example: Calculate the M of NaOH in a solution
made by diluting 25.0 mL of 0.928 M NaOH to
200.0 mL and then diluting 50.0 mL of the
second solution to 100.0 mL. Mc x Vc = Md x Vd
First dilution:
25.0 mL x 0.928 M
Md = ————————— = 0.116 M
200.0 mL

Second dilution:
50.0 mL x 0.116 M
Md = ————————— = 0.0580 M
100.0 mL
Solution Stoichiometry
Use of M, V, and coefficients in equations to
calculate any amount of reagent or product.
Example: What volume of 0.273 M potassium chloride solution is required to react with exactly 0.836 mmol of silver nitrate?
KCl(aq) + AgNO3(aq) 
KCl(aq) + AgNO3(aq)  KNO3(aq) + AgCl(s)
driving force, formation of precipitate AgCl
Key: work with mmol of KCl and AgNO3
Solution Stoichiometry
What volume of 0.273 M KCl solution is required
to react with exactly 0.836 mmol of AgNO3 ?
KCl(aq) + AgNO3(aq)  KNO3(aq) + AgCl(s)
First, calculate mmol of KCl:
1 mmol KCl
0.836 mmol AgNO3 x—————— = 0.836 mmol
1 mmol AgNO3 KCl

Second, calculate volume of KCl solution:
1 mL
0.836 mmol KCl x ———— = 3.06 mL of solution
0.273 mmol (KCl)
Titration
Titration
buret: calibrated td, fine tip to deliver small volumes accurately, stopcock for flow control
Erlenmeyer flask: sloped walls allow swirling of solutions without spilling or splashing
Titrant: solution containing reagent that will react with sample in well known manner
Titrate: solution containing the sample
Titration
equivalence point: point in a titration at which the exact stoichiometric amount of titrant has been added to react with the titrate
end point: the point in a titration at which the indicator changes, titration stopped here and volume of titrant read (ChemNow 5.19 Exercise: Titration)
ideally, end point = equivalence point
reality, not so, error introduced, hopefully small error
Four parameters: V, M of titrant and V, M of titrate. Usually one is unknown.
Titration
Example: Calculate the concentration of
hydrochloric acid in a solution if 35.0 mL of
it required 28.9 mL of 0.178 M potassium
hydroxide solution for titration.

HCl: titrated, V known, M unknown
(in this problem)
KOH: titrant, V and M known
Titration
Titration
Strategy: mmol of KOH are calculated first.
Second, by using stoichiometry coefficients
mol of HCl are calculated.
Finally, M of HCl is calculated with mmol and V(mL)
0.178 mmol KOH 1 mmol HCl
28.9 mLx———————— x————— = 5.14 mmol
1 mL 1 mmol KOH HCl

Based on the end point concept (mol ratio)

5.14 mmol
MHCl = ————— = 0.147 M (mol/L)
35.0 mL (mmol/mL)
Example: Calculate the concentration of arsenic acid in a solution given that a 25.0 mL sample of that solution required 42.2 mL of 0.274 M potassium hydroxide for titration.
H3AsO4 + KOH  K3AsO4 + H2O
H3AsO4 + 3 KOH  K3AsO4 + 3 H2O

M of H3AsO4 is the unknown.

1 mol H3AsO4 reacts with 3 mol KOH. That is
the mole ratio.

We can use mL and mmol instead of L and mol
Example: Calculate the concentration of arsenic acid in a solution given that a 25.0 mL sample of that solution required 42.2 mL of 0.274 M potassium hydroxide for titration.
H3AsO4 + 3 KOH  K3AsO4 + 3 H2O

0.274 mmol KOH 1 mmol H3AsO4
42.2 mLx———————— x————— = 3.85 mmol
1 mL 3 mmol KOH H3AsO4


3.85 mmol
M H3AsO4 = ————— = 0.154 M (mol/L)
25.0 mL (mmol/mL)
Purity Analysis: A 1.034 g-sample of impure oxalic acid is dissolved in water and an acid-base indicator added. The sample requires 34.47 mL of 0.485 M NaOH solution to reach the equivalence point. What is the mass of H2C2O4 and what is its mass percent in the sample?
H2C2O4 + 2 NaOH  Na2C2O4 + 2 H2O
Strategy: mol of H2C2O4 out of mol of NaOH (by
using V and M). Then, mol of H2C2O4 will be
used to calculate g of H2C2O4 and, hence, % of it
in the sample.
Coefficients are 1 for H2C2O4 and 2 for NaOH.
Oxalic acid: Oxac M W = 90.04 g/mol
0.485 mol 1 mol Oxac
34.47 mLx————— x—————— = 0.00836 mol
1000 mL 2 mol NaOH Oxac

Based on the end point concept (mol ratio)

90.04 g
0.00836 mol x ———— = 0.753 g oxalic acid
1 mol

0.753 g oxac
%Oxac = ——————— x 100 = 72.8%
1.034 g of sample
Molar Mass of an Acid: A 1.056 g of a pure acid, HA, is dissolved in water and an acid-base indicator added. The solution requires 33.78 mL of 0.256 M NaOH solution to reach the equivalence point. What is the molar mass of the acid? We don’t know what A is
HA + NaOH  NaA + H2O
Strategy: mol of HA out of mol of NaOH (by
using V and M). Then, mol of HA and g of HA
will be used to calculate the molar mass.

Coefficients are 1 for HA and 1 for NaOH.
Molar Mass of HA
33.78 mL NaOH solution = 0.03378 L

0.256 mol NaOH 1 mol HA
0.03378 Lx———————— x————— = 8.65x10-3
1 L sol’n 1 mol NaOH mol HA



1.056 g HA
Molar mass, M W = ——————— = 122 g/mol
8.65x10-3 mol HA

122 grams per mol
Vinegar: A 25.0-mL sample of vinegar (which contains the weak acetic acid, CH3CO2H) requires 28.33 mL of a 0.953 M solution of NaOH for titration to the equivalence point. What mass of the acid, in grams, is in the vinegar sample and what is the M of acetic acid in the vinegar?
CH3CO2H + NaOH  NaCH3CO2 + H2O
Strategy: mol of CH3CO2H (HOAc) out of mol of
NaOH (by using V and M). Then, g of HOAc will
be calculated with the molar mass. The M will
be calculated by using the volume of vinegar.

Coefficients are 1 for HOAc and 1 for NaOH.
CH3CO2H: HOAc M W = 60.05 g/mol
0.953 mol 1 mol HOAc
28.33 mLx————— x—————— = 0.0270 mol
1000 mL 1 mol NaOH HOAc

60.05 g
0.0270 mol x ——— = 1.62 g acetic acid in vinegar
1 mol

For the molarity, Vvinegar = 25.0 mL = 0.0250 L (sol’n)

0.0270 mol HOAc
M HOAc = ————————— = 1.08 M
0.0250 L sol’n
Problem 75 of textbook: To analyze an iron-containing compound, you convert all the iron into Fe2+ in aqueous solution and then titrate the solution with standardized KMnO4. The balanced-net ionic equation is
MnO4− + 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 4H2O

A 0.598-g sample of the iron compound requires 22.25 mL of 0.0123 M KMnO4 solution for titration to the equivalence point. What is the mass % of iron in the sample? Strategy:
mL and M of MnO4− to mol of MnO4− and Fe2+

Coefficients are 1 and 5 for MnO4− and Fe2+ respectively
Fe2+ and MnO4− (mole ratio is 5 to 1):
0.0123 mol MnO4− 5 mol Fe2+
22.25 mLx ———————— x————— = 1.37x10−3
1000 mL 1 mol MnO4− mol Fe2+

55.85 g Fe
g of iron: 1.37 x 10−3 mol Fe2+ x ————— = 0.0765
1 mol Fe2+ g Fe


Now, for the mass % of iron:

0.0765 g Fe
———————— x 100 = 12.8% Fe
0.598 g sample
Normality
equivalents (eq) solute per liter solution
milliequivalents (meq) solute per mL solution

eq solute meq solute
N = —————— = —————— (calculated by dividing)
V(L) sol’n V(mL) sol’n

equivalent = 1 equivalent weight

Equivalent Weight (EW); given in g/eq

acid/base: the mass of a substance required to furnish or react with exactly 1 mol H+

redox reactions: the mass of substance able to gain or lose 1 mol e−
Equivalent weight
HCl
HCl  H+ + Cl–
1 mol HCl  1 mol H+,  EW = MW
H2SO4
H2SO4  2H+ + SO42–
1 mol H2SO4  2 mol H+, EW = MW/2
HnA (in general)
EW = MW/n, n = # H+ in molecule of acid
KOH
KOH  K+ + OH–
OH– + H+  H2O
, 1 mol KOH reacts with 1 mol H+,  EW = FW

continued with polyhydroxides…

Equivalent weight
Fe(OH)3
1 mol Fe(OH)3  3OH–
1 mol reacts with 3 mol H+,  EW = FW/3
M(OH)n (in general)
EW = FW/n, n = # OH– in formula unit of base


Redox:
+7 0
MnO4– + 5e–  Mn2+ Zn  Zn2+ + 2 e–

FW MnO4– AW Zn
EW = ————— EW = ———
5 2
Example: Calculate the normality of barium hydroxide in a solution made by dissolving 0.991 g of barium hydroxide in water and diluting to 100.0 mL.
Ba(OH)2
FW 171.32 g mg
EW = —— = ———— = 85.66 —— = 85.66 ——
2 2 eq meq


1 eq Ba(OH)2
0.991 g Ba(OH)2 x —————— = 0.0116 eq Ba(OH)2
85.66 g Ba(OH)2

0.0116 eq Ba(OH)2
N Ba(OH)2 = ————————— = 0.116 N Ba(OH)2
0.1000 L sol’n
Example: Describe the preparation of 250.0 mL of 0.100 N oxalic acid solution from the solid. Oxalic acid is H2C2O4.
1 mol oxalic acid  2 mol H+
EW = 1/2 MW = 1/2(90.04) = 45.03g/eq
0.100 N indicates 0.100 eq OA/L sol’n

0.100 eq OA
0.2500 L x —————— = 0.0250 eq OA
1 L sol’n

45.03 g OA
0.0250 eq OA x —————— = 1.13 g OA
1 eq OA

Dissolve 1.13 g of oxalic acid in water and dilute to a total volume of 250.0 mL
Example: What is the normality of a 0.300 M arsenic acid solution?
H3AsO4
MW
1 mol H3AsO4  3 mol H+ EW = ——
3
g H3AsO4 g H3AsO4 g H3AsO4
eq = ————— = ———— = 3 x ————— = 3 mol
EW H3AsO4 MW MW H3AsO4 H3AsO4
3

eq mol
N = —— M = ——— Then,
V(L) V(L)

N H3AsO4 = 3 x M H3AsO4 = 3 x 0.300 = 0.900 N
CONCLUSION: For an Acid HnA or a Base M(OH)n

N = n x M

n = # H+ in molecule of acid or # OH– in formula unit of base
By definition, 1 eq of anything will react with exactly 1 eq of anything else

Equivalence Point
The point in a titration at which
eq titrant = eq titrate
meq titrant = meq titrate
Example: Calculate the concentration of phosphoric acid in a solution given that a 25.0 mL sample of that solution required 42.2 mL of 0.274 N potassium hydroxide for titration.
H3PO4 + 3 KOH  K3PO4 + 3 H2O
at Eq. Pt., meq H3PO4 = meq KOH

meq H3PO4 = meq KOH
(mL H3PO4)(N H3PO4) = ( mL KOH)(N KOH)
(25.0 mL)(X ) = (42.2 mL)(0.274 N )
X = 0.463 N
Now, N = 3x M, then M = N/3
X = 0.463/3 = 0.154 M
Example: Calculate the % oxalic acid (H2C2O4) in a solid given that a 1.709 g sample of the solid required 24.9 mL of 0.0998 N potassium hydroxide for titration.
H2C2O4 + 2KOH  K2C2O4 + 2H2O

Due to the two protons of H2C2O4 (OA),
EW = MW/2 = 45.03g/eq = 45.03mg/meq

g OA
% OA = ————— x 100
g sample

Strategy: calculate eq of NaOH that are the same
for OA; then, calculate g of OA with eq and EW
of OA

At equivalence point,

eq OA = eq KOH data of KOH sol’n

0.0998 eq KOH
eq OA = 0.0249 L x ——————— = 0.00249 eq
1 L sol’n


45.03 g OA
0.00249 eq OA x ——————— = 0.112 g OA
1 eq OA


0.112 g OA
% OA = ——————— x 100 = 6.55%
1.709 g sample

Chapter 6
Principles of Reactivity:
Energy and 
Chemical Reactions

Thermochemistry
Goals of Chapter
Assess heat transfer associated with changes in temperature and changes of state.

Apply the First Law of Thermodynamics.

Define and understand the state functions enthalpy (H) and internal energy (E).

Calculate the energy changes in chemical reactions and learn how these changes are measured.
Thermochemistry
study of the relationships between energy changes and chemical processes
Energy
The capacity to do work or to transfer heat
Kinetic Energy
energy of motion; KE = ½ mv2
Potential Energy
stored energy: fuel of motor-cars, trains, jets.
It is converted into heat and then to work.

due to relative position: water at the top of a
water wheel. It is converted to mechanical E

electrostatic: lightning converts it to light and
heat
Joule
SI unit for energy
the energy of a 2 kg mass moving at 1 m/s

KE = ½ mv2 = ½(2 kg)(1 m/s)2 = 1 kgm2/s2 = 1 J

1 cal is the amount of energy required to raise the temperature of 1 g water 1°C
1 cal = 4.184 J 1 cal = 1 calorie

1 Cal = 1000 cal = 1 kcal
1 Cal = dietary Calorie (nutritional calorie)
System
the part of the universe under study

the substances involved in the chemical and physical changes under investigation
in chemistry lab, the system may be the chemicals inside a beaker
Surroundings
the rest of the universe

in chemistry lab, the surroundings are outside the beaker where the chemicals are

The system plus the surroundings is the universe.
System and Surroundings
SYSTEM
The object under study
SURROUNDINGS
Everything outside the system
Thermodynamic State
The set of conditions that specify all of the properties of the system is called the thermodynamic state of a system.
For example the thermodynamic state could include:
The number of moles and identity of each substance.
The physical states of each substance.
The temperature of the system.
The pressure of the system.
The volume of the system.
The height of a body relative to the ground.
First Law of Thermodynamics
law of conservation of energy
during any process, energy is neither created nor destroyed, it is merely converted from one form to another*

the mass of a substance is a form of energy
E = mc2 (Albert Einstein)
e.g. in nuclear reactions mass is not conserved,
it is transformed into heat (E)

* “The combined amount of energy in the universe is constant.”
Internal Energy (E)
the total energy of a system: Σ of kinetic and
potential E of all atoms, molecules, or ions in the
system
E cannot be measured exactly
E is a state function; change in E does not depend on how change of state happens
E: change in E. E can be measured
E = Efinal – Einitial (of final and initial states)
E > 0 (+) indicates system gains energy during process (E increases, )
E < 0 (−) indicates system loses energy during process (E decreases, )
E = q + w
first law of thermodynamics
q = heat
w = work done on the system
w > 0 (+)  work done on system by surroundings (eg. compressing gas); E of system increases
w < 0 (–)  work done by system on surroundings (expanding gas); E decreases
q > 0 (+)  heat flows into system; E
q < 0 (–)  heat flows out of system; E
q and w are not state functions
Exothermic reactions give off energy in the form of heat (they give off heat).
Endothermic reactions absorb heat.
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) + 890 kJ
exothermic
Directionality of Heat Transfer
Heat always transfers from hotter object to cooler one.
EXOthermic: heat transfers from SYSTEM to SURROUNDINGS.
Directionality of Heat Transfer
Heat always transfers from hotter object to cooler one.
ENDOthermic: heat transfers from SURROUNDINGS to the SYSTEM.
Calculate E of a system that absorbs 35 J of heat and does 44 J of work on the surroundings.
q = +35 J (absorbed)
w = –44 J (the system did it)
E = q + w
E = 35 J + (–44 J)
E = –9 J (internal E decreases)
Note that Efinal and Einitial are not calculate, just E
Work symbol: w
w = force x distance
Expansion/compression work at constant P,
w = –PV V = Vfinal – Vinitial
E = q – PV
under conditions of constant volume, PV = 0, w=0, because V = 0 (no work done on or by the system)
, E = q – 0
E = q
E = qV This provides a way of measuring E; in a reactor at constant V.
Heat Capacity (C)
the amount of heat energy required to raise the temperature of an object 1 K (1°C),
units = J/K, cal/°C, ...
q = C T T = Tfinal – Tinitial
The amount of heat can be calculated from T
Specific Heat (c)
the amount of heat energy required to raise the temperature of 1 g of something 1 K (1°C)
units = J/gK, J/g°C, cal/g°C, ...
q = c m  T m: mass (grams)
Molar Heat Capacity
the specific heat of water is
1 cal/g°C = 4.184 J/g°C
KNOW THIS!!!
The molar heat capacity is the amount of heat energy required to raise the temperature of 1 mol of a substance 1 K (1°C).
units = J/K mol, cal/°C mol
For water, it is:

cal 18.0 g cal
C = 1 ——— ——— = 18.0 ————
g °C 1 mol mol °C or K
Heat/Energy Transfer
No Change in State
q transferred = (sp. ht.)(mass)(∆T)
Law Zero of Thermodynamics
When two bodies, liquids, solutions, solid-liquid, etc.,(*) initially at different temperatures, are put in contact or mixed, the amount of heat absorbed and given off by the two samples have the same absolute value, but one is >0 and the other is <0

q1 + q2 = 0 q1 = −q2

If more than two components(*)
q1 + q2 + q3 + … = 0
Example: Calculate the amount of heat energy given off when 45.3 g water cools from 77.9 °C to 14.3 °C
T = 14.3 °C – 77.9 °C = – 63.6 °C
q = c m T (know this formula)

1 cal
q = —— 45.3 g  (–63.6 °C) = –2.88 103 cal
g °C

T is negative because T lowers,
hence q is negative (it is given off)

J
Cu specific heat c = 0.385 ───
g K

q = m c  T = m c  (Tfinal − Tinitial)

1000 J 0.385 J
1.850 kJ───── = 500. g ─────(37°C − Ti)
1 kJ g °C

1.850 1000
37°C − Ti = ─────────= 9.6 °C Ti = 27.4°C
500. 0.385
A 182-g sample of Au at some temperature is added to 22.1 g of water. The initial water T is 25.0 °C, and the final T of the whole is 27.5 °C. If the specific heat of gold is 0.128 J/g.K, what is the initial T of the gold?
T of H2O increased, hence it absorbed heat. Then,
Au gave off heat, i.e., its temperature decreased.
qwater(absorbed) + qAu(given off) = 0 q = m c T
>0 <0 T = Tf – Ti

22.1 g 4.184 J/g.°C (27.5 – 25.0)°C +
182 g 0.128 J/g.°C  (27.5°C – Ti) = 0
231 °C + 23.3 (27.5 °C – Ti) = 231 °C + 641°C – 23.3Ti = 0

231 + 641
Ti(Au) = ————— = 37.4 °C
23.3
One beaker contains 156 g of water at 22 °C and a second contains 85.2 g of water at 95 °C. If the water in the two beakers is mixed, what is the final temperature?
Water in beaker 1 (w1) will absorb heat, its T will .
Water in beaker 2 will give off heat, its T will .
q1(absorbed) + q2(given off) = 0 q = m c T
>0 <0 T = Tf – Ti

156 g 4.184 J/g.°C (Tf – 22°C) +
+ 85.2 g 4.184 J/g.°C (Tf – 95 °C)= 0
156 Tf – 3432 + 85.2 Tf – 8094 = 0

8094 + 3432
Tf = —————— = 47.8 ≈ 48 °C
156 + 85.2
Bomb Calorimeter: constant V
Example: A 1.50g sample of methane was burned in excess oxygen in a bomb calorimeter with a heat capacity of 11.3kJ/°C. The temperature of the calorimeter increased from 25.0 to 32.3°C. Calculate the E in kJ per gram of methane for this reaction.
T = (32.3 – 25.0) °C = 7.3 °C
CH4(g) + 2O2(g)  CO2(g) + 2 H2O(l)
In a bomb calorimeter V is constant  E = q
qcalorim = C T = (11.3 kJ/°C) 7.3°C = 83 kJ

q + qcalorim = 0 Then, E = q = −qcalorim, E = – 83 kJ

E is negative because the rxn gives off heat that the
calorimeter absorbs
– 83 kJ kJ kJ
E = ———— = – 55.3 —— = – 55 ——— (two SF)
1.50 g g g CH4
Example: A bomb calorimeter was heated with a heater that supplied a total of 8520 J of heat. The temperature of the calorimeter increased 2.00°C. A 0.455g sample of sucrose, C12H22O11, was then burned in excess oxygen in that calorimeter causing the temperature to increase from 24.49°C to 26.25°C. Calculate the E for this reaction in kJ/mol sucrose and the dietary calories per gram of sucrose.
2C12H22O11 + 35 O2 24CO2 + 22H2O

We will need MW of sucrose = 342.3 g/mol
Calorimeter heat capacity (C)
heat supplied q 8520 J J
C = —————— = —— = ———— = 4260 ——
T T 2.00 °C °C

For the reaction, T = (26.25 – 24.49)°C

V = 0, then, E = q = –qcalor (reaction gives off heat)

qcalorim = C T = (4260 J/°C) 1.76°C = 7.50 x103 J

– 7.50 x103 J 342.3 g sucrose 1 kJ
E = ——————— ——————— ———
0.455 g sucrose 1 mol sucrose 103 J

kJ
E = – 5.64x103 ——————
mol sucrose
Nutritional (dietary) Calories (Cal)
Strategy: divide J by grams of sucrose.

Convert J to cal and cal to Cal.

1 Cal = 1000 cal = 1 kcal (see slide # 5)

– 7.50 x103 J 1 cal 1 Cal –3.94 Cal
—————— ———— ————= —————
0.455 g sucro 4.184 J 1000 cal g sucrose
Heat Transfer 
with Change of State
Changes of state involve energy (at const T)
Ice + 333 J/g (heat of fusion) Liquid water
q = (heat of fusion)(mass)
Heat Transfer and 
Changes of State
Requires energy (heat).
This is the reason
a) you cool down after swimming
b) you use water to put out a fire
Heating/Cooling Curve for Water

What quantity of heat is required to melt
500. g of ice and heat the water to steam
at 100 oC?

What quantity of heat is required to melt 500.
g of ice and heat the water to steam at 100 oC?
1. To melt ice
q1 = (500. g)(333 J/g) = 1.67 x 105 J
2. To raise water from 0 oC to 100 oC
q2 = (500. g)(4.2 J/g•K)(100 - 0)K = 2.1 x 105 J
3. To evaporate water at 100 oC
q3 = (500. g)(2260 J/g) = 1.13 x 106 J
4. Total heat energy = q1 + q2 + q3 =
1.51 x 106 J = 1510 kJ


13.6 g
q = q1 + q2 1mL ──── = 13.6 g
1 mL
q = m c  T + m (−qfus) negative for freezing

0.140 J (−11.4 J)
q = 13.6 g ─────(−38.8 −23)°C + 13.6 g ─────
g °C g

q = −117 J −155 = −272 J (given off)

Exothermic process
Enthalpy, H
Chemistry is commonly done in open beakers or flasks on a desk top at atmospheric pressure.
Because atmospheric pressure only changes by small amounts, this is almost at constant pressure.

Because heat at constant pressure is so frequent in chemistry and biology, it is helpful to have a measure of heat transfer under these conditions. That is the enthalpy change.
Enthalpy, H
heat content
state function; change in H does not depend on how change of state happens
H = E + PV (We do not measure
H = Hfinal – Hinitial or calculate Hf,i but H)
at constant pressure, H = qP
H > 0 (+)  heat flows into system; H ;
endothermic
H < 0 (–)  heat flows out of system; H ;
exothermic
Calorimetry
A coffee-cup
calorimeter is used to
measure the amount of
heat produced (or
absorbed) in a reaction
at constant P. That is qp or
H. The cup is under
constant atmospheric
pressure (P) because it is
not completely sealed. It is
‘isolated’: no heat transfer
between system and
surroundings.
Product- or Reactant-Favored Reactions
nature favors processes that decrease energy of the system
, nature favors exothermic processes
4 Fe(s) + 3 O2(g)Fe2O3(s) H = –1651 kJ
Exothermic. The formation of product is favored.
CaCO3(s)CaO(s) + CO2(g) H = 179.0 kJ
Heat is required for the reaction to occur. The
reaction is endothermic. Reagent is favored.

Is it always like that? Is H the only factor that
matters? No, entropy change counts too…
Enthalpy of Reaction, Heat of Reaction

enthalpy of reaction, heat of reaction
Hreaction when a reaction takes place
Hfusion when a solid is melted
Hcrystallization when a compound is crystallized
from a solution
Hess’ Law
if a reaction is the sum of two or more other reactions, the H for the overall reaction is equal to the sum of the H values of those reactions.
also applies to E. E and H are state functions

The H of some reactions can not be measured in a
calorimeter, because some other reactions take
place at the same time in the reactor.
C(s) + 2H2(g)  CH4(g); C2H2, C2H4, C2H6, C3H6, etc.,
are also produced. In a case like this, the H of other
related reactions can be employed to calculate H
C(s) + 2H2(g)  CH4(g) Hrxn = ?
C(s) + O2(g)  CO2(g) H = –393.5 kJ
2H2(g) + O2(g)  2H2O(l) H = –571.6 kJ
CO2(g) + 2H2O(l)  CH4(g) + 2O2(g) H = 890.4 kJ
_______________________________________
C(s) + 2H2(g)  CH4(g) Hrxn = – 74.7 kJ

Hr = –393.5 kJ + (–571.6 kJ) + 890.4 kJ = – 74.7 kJ
Calculate H of the fourth equation out of Hs of the first three: 
 C(s) + O2(g)  CO2(g) H = –393.5 kJ Eq. (1)
 H2(g) + 1/2O2(g)  H2O(l) H = –285.8 kJ Eq. (2) 
 2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O H = –2598.8 kJ Eq. (3) 
2C(s) + H2(g)  C2H2(g) Hrxn =? Eq. (4)
We need to: multiply Eq. (1) by 2; leave Eq. (2) as it is; multiply Eq. (3) by −½, that is to reverse it and times ½.
2 { C(s) + O2(g)  CO2(g) H = –393.5 kJ }
 H2(g) + 1/2O2(g)  H2O(l) H = –285.8 kJ −½ { 2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O H =–2598.8 kJ}
2C(s) + 2O2(g)  2CO2(g) H = 2(–393.5 kJ)
H2(g) + 1/2O2(g)  H2O(l) H = –285.8 kJ
2CO2(g) + H2OC2H2(g) + 5/2O2(g) H = 1/2(+2598.8 kJ)
————————————————————
2C(s) + H2(g)  C2H2(g) Hrxn = 226.6 kJ
Hrxn = 2(–393.5 kJ) –285.8 kJ + 1/2(+2598.8 kJ)
2C(s) + H2(g)  C2H2(g) Hrxn =?

 C(s) + O2(g)  CO2(g) H = –393.5 kJ
 H2(g) + 1/2O2(g)  H2O(l) H = –285.8 kJ 
 2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O 
 H = –2598.8 kJ
2C(s) + 2O2(g)  2CO2(g) H = 2(–393.5 kJ)
H2(g) + 1/2O2(g)  H2O(l) H = –285.8 kJ
2CO2(g) + H2OC2H2(g) + 5/2O2(g)
H = 1/2(+2598.8 kJ)
————————————————————
2C(s) + H2(g)  C2H2(g) Hrxn = 226.6 kJ
Hrxn = 2(–393.5 kJ) –285.8 kJ + 1/2(+2598.8 kJ)
P4(s) + 6 Cl2(g)  4 PCl3(l) Hrxn = ?

P4(s) + 10 Cl2(g)  4 PCl5(s) H = –1774.0kJ
PCl3(l) + Cl2(g)  PCl5(s) H = –123.8 kJ

P4(s) + 10 Cl2(g)  4 PCl5(s) H = –1774.0kJ
–4x(PCl3(l)  PCl5(s) + Cl2(g) H = –123.8 kJ)

P4(s) + 10 Cl2(g)  4 PCl5(s) H = –1774.0kJ
4PCl5(s)  4PCl3(l) + 4Cl2(g) H = 4(+123.8 kJ)
———————————————————————
P4(s) + 6 Cl2(g)  4 PCl3(l) Hrxn = –1278.8 kJ

Hrxn = –1774.0 kJ + 495.2 kJ = –1278.8 kJ
Formation Reaction
Reaction in which 1 mol of a substance is
formed from its elements in their most stable,
natural states.
eg. formation reaction for C2H6SO(l)

C(s) + H2(g) + S8(s) + O2(g)C2H6SO(l)

2C(s) + 3H2(g) + 1/8S8(s) + 1/2O2(g)C2H6SO(l)

Heat (Enthalpy) of Formation

Hf
enthalpy change associated with a formation reaction
Standard Conditions

T = 25°C = 298 K

P = 1 atm
Standard Heat of Formation or
Standard Molar Enthalpy of Formation

Hf° is the enthalpy change for the formation of 1 mol of a compound directly from its elements in their standard states

Hf° may be >0 or <0 for a compound (tables)

Hf° = 0 for any element in its most stable form,
e.g. Na(s), Hg(l), Cl2(g), Br2(l), H2(g), P4(s), C(s)
Enthalpy Change for a Reaction
a A + b B  c C + d D

H°rxn = Σ(Hf° products) – Σ(Hf° reagents)

“means take the sum”

H°rxn = cHf°(C) + dHf°(D)
– aHf°(A) – bHf°(B)
Hf°(element) = 0
Example: Calculate the H° in 
kJ/mol B5H9 for the following reaction.
2B5H9 + 12O2  5B2O3 + 9H2O
Compound Hf° (kJ/mol)
B5H9 73.2
B2O3 –1272.8
H2O –241.83
H°Rxn = [(5(Hf° B2O3) + 9(Hf° H2O)]
– [(2 (Hf° B5H9) + 12(Hf° O2)]

H°Rxn= [5mol(–1272.8 kJ/mol) + 9mol(–241.83 kJ/mol)]
–[2 mol(73.2 kJ/mol) +12 mol(0 kJ/mol)] = –8686.9 kJ

–8686.9 kJ/2 mol B5H9 = –4343.5 kJ/mol B5H9
Calculate the H° in kJ/mol Mg for the following reaction.
3Mg + SO2  MgS + 2MgO
Compound Hf° (kJ/mol)
MgO –601.7
MgS –598.0
SO2 –296.8
H°Rxn = [1 mol(Hf° MgS) + 2(Hf° MgO)]
– [3(Hf° Mg) + 1 mol(Hf° SO2)]

H°Rxn= [1mol(–598.0 kJ/mol) + 2 mol(–601.7 kJ/mol)]
–[3 mol(0 kJ/mol) + 1 mol(–296.8 kJ/mol)] = –1504.6 kJ
–1504.6 kJ/3 mol Mg = –501.5 kJ/mol Mg
The enthalpy change for the combustion of styrene, C8H8, is measured by calorimetry:
 C8H8(l) + 10 O2(g)  8 CO2(g) + 4 H2O(l)
 H°rxn = –4395.0 kJ
Use this, along with the Hf° of CO2 and H2O, to
calculate the Hf° C8H8, in kJ/mol
Hf° CO2(g) = –393.51 kJ/mol
H2O(l) = –285.83 kJ/mol
H°Rxn = [8(Hf° CO2) + 4(Hf° H2O(l)] – (Hf° C8H8)

Hf° C8H8 = [8(Hf° CO2) + 4(Hf° H2O(l)] – H°rxn

Hf° C8H8 = 8(–393.51) + 4(–285.83) – (–4395.0)

Hf° C8H8 = 103.6 kJ/mol
Nitroglycerin is a powerful explosive that forms four different gases when detonated

2C3H5(NO3)3(l) 3N2(g) + ½O2(g) + 6CO2(g) + 5H2O(g)
Calculate the enthalpy change when 10.0 g of
nitroglycerin is detonated. Hf°(kJ/mol) are:
CO2(g) = –393.5 H2O(g) = –241.8
NTG, C3H5(NO3)3(l) = –364

H°rxn = [6(Hf° CO2) + 5(Hf°H2O(g)] – 2(Hf°NTG)


H°rxn= 6(–393.5) + 5(–241.8) – 2(–364) =

= –2842 kJ for 2 mol of NTG

For 10.0 g of NTG we need its MW = 227.1 g/mol
–2842 kJ for 2 mol of NTG

for 10.0 g of NTG we need its MW = 227.1 g/mol


– 2842 kJ 1 mol NTG
————— —————— 10.0 g NTG= –62.6 kJ
2 mol NTG 227.1 g NTGJ


– 1421 kJ 1 mol NTG
————— —————— 10.0 g NTG= –62.6 kJ
1 mol NTG 227.1 g NTGJ
Calculate the amount of heat released in the complete combustion of 8.17 grams of Al to form Al2O3(s) at 25°C and 1 atm. ΔHf° for Al2O3(s) = ­1676 kJ/mol
4Al(s) + 3O2(g) 2Al2O3(s)

1 mol Al
8.17 g Al—————= 0.303 mol Al
26.98 g Al

2 moles mol Al2O3 are produced out of 4 moles of Al

−1676 kJ 2 mol Al2O3
————— ————— 0.303 mol Al = −254 kJ
1 mol Al2O3 4 mol Al
How much heat energy is liberated when 11.0 grams of manganese is converted to Mn2O3 at standard state conditions?
4Mn(s) + 3O2(g) 2Mn2O3(s) ΔH= −1924.6 kJ

1 mol Mn
11.0 g Mn—————= 0.200 mol Mn
54.94 g Mn

The given ΔH corresponds to the reaction of 4 moles Mn

− 1924.6 kJ
—————— 0.200 mol Mn = −96.2 kJ
4 mol Mn

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