Transcript CLOVERLICIOUS

Thermochemistry
Chapter 6
Notes
Thermochemistry
Part 1: Energy,
Heat and
Work
Energy is the capacity to do
work or to produce heat. The
law of conservation of energy
states that energy can be
converted from one form to
another, but can be neither
created nor destroyed.
Potential energy is energy due to
position or composition.
Kinetic energy is energy
due to the motion of the
object and depends on
the mass and velocity of
2
the object. KE = ½mv
**Mass must be in kilograms and velocity must be in
meters/second!!!!! The unit kgm2 = J
s2
Heat and temperature are different.
Temperature is a measure of the kinetic
energy of the molecules.
Heat refers to the transfer of energy
between two objects due to a
temperature difference.
Heat is not a substance contained by
an object, although we often talk of
heat as if this were true.
The pathway is likened to a “path” or
“route.” For instance, I can get to the
stadium by walking out the front door
or the back door.
A state function or state property
depends only on the characteristics of
the present state – not on the pathway.
The universe is divided into two parts:
a. The system is the part of the
universe on which we focus.
b. The surroundings include
everything else in the universe.
For a reaction, the system includes the
reactants and products.
The surroundings includes the
reaction container, the room, etc.
(i.e. anything else other than
reactants and products.)
The study of energy and its
interconversions is called
thermodynamics. The law of
conservation of energy is often
called the first law of
thermodynamics. It states: The
energy of the universe is constant.
Thermodynamic quantities
always consist of two
parts: a number, giving
the magnitude of the
change, and a sign,
indicating the direction of
the flow. The sign reflects
the system’s point of view.
q = heat
*If energy flows INTO the
system via heat
(endothermic), then q = +.
*If energy flows OUT OF
the system via heat
(exothermic), then q = .
w = work
*If the surroundings do work
on the system (energy flows
into the system), then w = +.
* If the system does work on
the surroundings (energy
flows out of the system), then
w = .
Work is defined as force acting over a
distance. The formula used to solve for
work is: W = PΔV
Pressure is measured in
atmospheres and volume is
measured in liters. Convert to
joules using 101.3 J = 1 Latm
The sign changes depending
on: compressed gas = +PΔV
expanding gas = PΔV
Memorize!!!
sublimation
deposition
gas
vaporization
endothermic
condensation
liquid
melting
exothermic
freezing
solid
The internal energy, E, of a system is
defined as the sum of the kinetic and
potential energies. The formula is
ΔE = q + w
where q is heat and w is work.
The sign convention is that anything that
leaves the system is negative.
*q is negative: system releases heat
*q is positive: system absorbs heat
*w is negative: system does work
*w is positive: surroundings do work
Thermochemistry
Part 2:
Properties of
Enthalpy
Enthalpy, H, is defined as:
H = E + PV
where
E = internal energy
P = pressure
V = volume
Internal energy, pressure and
volume are all state functions
(independent of the pathway)
therefore enthalpy is also a
state function.
At constant pressure, the change in
enthalpy (ΔH) of the system is equal to
the energy flow as heat.
Therefore, ΔH = q at constant P.
At constant pressure, exothermic
means that ΔH is negative and
endothermic means that ΔH is positive.
Stoichiometric Calculations
When a mole of methane (CH4) is burned at constant pressure,
890 kJ of energy are released as heat. Calculate H when 5.8
grams of methane are burned at constant pressure.
16.0 g/mol
CH4 + 2 O2  CO2 + 2 H2O
5.8 g
H = –890 kJ
x kJ
– 890. kJ
1 mol CH4
= – 320 kJ
1 mol CH4
16.0 g CH4
5.8 g CH4
Thermochemistry
Part 3:
Calorimetry and
Heat Capacity
Calorimetry
Calorimetry is the
study of heat flow
and heat
measurement.
Calorimetry experiments
determine the heats (enthalpy
changes) of reactions by
making accurate
measurements of temperature
changes produced in a
calorimeter.
The formula used in
calorimetric calculations is:
q = mcT
where
q = heat (J)
m = mass (g)
c = specific heat capacity (J/g C)
T = change in temperature (C)
The heat capacity of an object
is the amount of heat needed
to raise the temperature of the
object by 1C.
The heat capacity of one gram
of a substance is called its
specific heat.
The specific heat is a
physical property of the
substance, like its color and
melting point. Substances
have different specific heat
capacities.
The specific heat
capacity of water
= 4.18 J/gC
When calculating T, always
subtract the smaller temperature
FROM the larger temperature:
T = Tlarger  Tsmaller
**If the temperature rises,
(ex: from 25C to 30C) then q will be
negative and the reaction is exothermic.
**If the temperature drops,
(ex: from 40C to 30C) then q will be
positive and the reaction is endothermic.
Example 1:
What is the specific heat capacity of
iron if the temperature of a 12.3-g
sample of iron is increased
by 10.2C when 56.7 J
of heat is added?
Example 1: What is the specific heat capacity of iron if
the temperature of a 12.3-g sample of iron is increased by
10.2C when 56.7 J of heat is added?
q = mcT
56.7 J = (12.3 g) (c)(10.2 C)
56.7 J = (12.3 g)( c)(10.2C)
(12.3 g)(10.2C) (12.3 g)(10.2C)
0.452 J/gC = c
Example 2: When a 13.7-g sample
of solid Pb(NO3)2 dissolves in 85.0 g
of water in a calorimeter, the
temperature drops from 23.4C to
19.7C. Calculate H for the
solution process.
Pb(NO3)2 (s)  Pb+2 (aq) + 2 NO31 (aq)
H=?
Example 2: When a 13.7-g sample of solid Pb(NO3)2 dissolves in 85.0 g of
water in a calorimeter, the temperature drops from 23.4C to 19.7C.
Calculate H for the solution process.
Pb(NO3)2 (s)  Pb+2 (aq) + 2 NO31 (aq)  H = ?
q = mcT
q = mc(Tlarger  Tsmaller)
q= (85.0 g) (4.18 J/gC )(23.4 C19.7C )
q = 1310 J
**Since the temperature dropped, q will be
positive and the reaction is endothermic.
Example 2: When a 13.7-g sample of solid Pb(NO3)2 dissolves in 85.0
g of water in a calorimeter, the temperature drops from 23.4C to
19.7C. Calculate H for the solution process.
Pb(NO3)2 (s)  Pb+2 (aq) + 2 NO31 (aq)  H = ?
Calculate the molar mass
of Pb(NO3)2:
MM = 331 g/mol
Remember that from the first
part, q = +1310 J
Example 2: When a 13.7-g sample of solid Pb(NO3)2 dissolves in 85.0
g of water in a calorimeter, the temperature drops from 23.4C to
19.7C. Calculate H for the solution process.
Pb(NO3)2 (s)  Pb+2 (aq) + 2 NO31 (aq)  H = ?
+1310 J
331 g
Pb(NO3)2
1 mol
Pb(NO3)2
13.7 g
1 mol
Pb(NO3)2 Pb(NO3)2
= +31,700 J = 31.7 kJ
Energy
Calculations
Important Information:
q = mcΔT
q = ΔHfusion  moles
q = ΔHvaporization  moles
specific heat capacity of water = 4.18 J/gC
specific heat capacity of ice = 2.1 J/gC
specific heat capacity of steam = 1.8 J/gC
ΔHfusion of water = 6.0 kJ/mole
ΔHvaporization of water = 40.7 kJ/mole
How much energy
does it take to convert
130. grams of ice at
40.0C to steam at
160.C?
Convert grams to moles of
water:
1 mol H2O
130. g H2O
18.0 g H2O
= 7.22 mol H2O
Plan:
a. Heat ice from 40.0C to
0.00C.
q = mcT
q = (130. g)( 2.1 J/gC)(40.0C)
q = 10,920 J
q  10,900 J
q = 10.9 kJ
b. Add heat to convert ice to liquid
water at 0C.
q = Hfusion  moles
q = (6.0 kJ/mol)(7.22 mol)
q = 43.32 kJ
q  43.3 kJ
c. Heat liquid water from 0.00C to
100.C.
q = mcT
q = (130. g)( 4.18 J/gC)(100.0C)
q = 54,340 J
q  54,300 J
q = 54.3 kJ
d. Add heat to convert liquid water to
steam at 100C.
q = Hvaporization  moles
q = (40.7 kJ/mol)(7.22 mol)
q = 293.854 kJ
q  294 kJ
e. Heat steam from 100.C to 160.C.
q = mcT
q = (130. g)( 1.8 J/gC)(60.0C)
q = 14,040 J
q  14.0 kJ
Add the energy
values:
Total energy =
a + b + c + d + e
Total energy =
10.9 kJ
43.3 kJ
54.3 kJ
294 kJ
14.0 kJ
416.5 kJ  417 kJ
Thermochemistry
Part 4:
Hess’ Law
The amount of heat that
a reaction absorbs or
releases depends on the
conditions under which
the reaction is carried out
(temperature, pressure,
and physical states of the
reactants and products.)
To make comparing enthalpy changes
easier, chemists chose a pressure of 1
atmosphere and a temperature of 25C
as conditions to carry out reactions.
These are called standard states.
An enthalpy change under these
conditions is called a standard enthalpy
change. It is denoted with a superscript.
It is shown as H.
Conventional Definitions of Standard States
For a compound:
*The standard state of a gaseous substance is a
pressure of exactly 1 atmosphere.
*For a pure substance in a condensed state (liquid
or solid), the standard state is the pure liquid or
solid.
*For a substance present in a solution, the standard
state is a concentration of exactly 1 M.
For an element:
*The standard state of an element is the form in
which the element exists under conditions of 1
atmosphere and 25C. (The standard state for
oxygen is O2 (g) at a pressure of 1 atm; the
standard state for sodium is Na(s); the standard
state for mercury is Hg(ℓ).
In the 19th century, a Swiss chemist
named G.H. Hess proposed a way of
finding the enthalpy change for a
reaction (even if the reaction could not
be performed directly.)
In 1840, Hess demonstrated
experimentally that the heat
transferred during a given reaction is
the same whether the reaction occurs
in one step or several steps.
His method is now
called Hess’ law of heat
summation
or simply Hess’ Law.
The method is
analogous to solving
simultaneous equations
in algebra.
In solving problems
using Hess’ Law, there
are some basic rules
that must be
memorized.
Rules for manipulating reactions:
1. If a reaction is reversed, the sign of
H must be reversed.
2. If a reaction is multiplied or
divided by a coefficient, H must
also be multiplied or divided by
that coefficient.
Rules for adding reactions:
1. Identical substances on the
same side of a reaction are
added together.
2. Identical substances on
opposite sides of a reaction
are cancelled.
3. Simply add the H’s of each
reaction to get the H of the
final reaction.
Example 1:
Use:
A+BC+D
2E+CD+2F
E + AF
H =  10 kJ
H =  20 kJ
H =  30 kJ
Determine H for
3 A + B  2C
A+BC+D
H =  10 kJ
2E + C  D + 2F H =  20 kJ
E +AF
H =  30 kJ
A+BC+D
2F+D2E+C
2E + 2 A  2 F
Determine H for
3A + B  2C
H =  10 kJ
H = + 20 kJ
H = 2( 30 kJ )
Write the reactions by comparing
the substances with the desired
result.
A+BC+D
H =  10 kJ
2E + C  D + 2F H =  20 kJ
E +AF
H =  30 kJ
A+BC+D
2F+D2E+C
2 E + 2A  2 F
Determine H for
3A + B  2C
H =  10 kJ
H = + 20 kJ
H = 2( 30 kJ)
Cancel identical items on opposite
sides of the arrow.
A+BC+D
H =  10 kJ
2E + C  D + 2F H =  20 kJ
E +AF
H =  30 kJ
A+BC+D
2F+D2E+C
2 E + 2A  2 F
Determine H for
3A + B  2C
H =  10 kJ
H = + 20 kJ
H = 2( 30 kJ)
3 A + B  2C
Add identical items on the same side of the arrow.
A+BC+D
H =  10 kJ
2E + C  D + 2F H =  20 kJ
E +AF
H =  30 kJ
A+BC+D
2F+D2E+C
2 E + 2A  2 F
Determine H for
3A + B  2C
H =  10 kJ
H = + 20 kJ
H = 2( 30 kJ)
3 A + B 2C
Verify that the final reaction matches the
reaction given in the original problem.
A+BC+D
H =  10 kJ
2E + C  D + 2F H =  20 kJ
E +AF
H =  30 kJ
A+BC+D
2F+D2E+C
2 E + 2A  2 F
3 A + B  2C
Determine H for
3A + B  2C
H =  10 kJ
H = + 20 kJ
H = 2( 30 kJ)
H = 50 kJ
Combine the H’s.
When dealing with actual
substances in the reactions, the
physical state of each substance
must be written in parentheses.
For example:
(s) is solid, (ℓ) is liquid, (g) is gas,
and (aq) is aqueous – meaning
that it is dissolved in water.
Example 2:
From the following enthalpy changes,
2PbS (s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g) H =  124 kJ
Pb (s) + CO (g)  PbO (s) + C (s)
H =  106.8 kJ
calculate the value of H for the following reaction:
2PbS (s) + 3 O2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO2 (g)
Find H for:
2PbS (s) + 3 O2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO2 (g)
Using:
2PbS (s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g) H =  124 kJ
Pb (s) + CO (g)  PbO (s) + C (s)
H =  106.8 kJ
Leave the first reaction as written:
2PbS (s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g)
H =  124 kJ
Find H for:
2PbS (s) + 3 O2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO2 (g)
Using:
2PbS (s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g) H =  124 kJ
Pb (s) + CO (g)  PbO (s) + C (s)
H =  106.8 kJ
Reverse the second reaction and change the sign:
2PbS (s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g)
PbO (s) + C (s)  Pb (s) + CO (g)
H =  124 kJ
H = + 106.8 kJ
Find H for:
2PbS (s) + 3 O2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO2 (g)
Using:
2PbS (s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g) H =  124 kJ
Pb (s) + CO (g)  PbO (s) + C (s)
H =  106.8 kJ
Multiply the second reaction by 2:
2PbS (s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g) H =  124 kJ
2 PbO (s) + 2C (s)  2Pb (s) + 2 CO (g)
H = 2(+ 106.8 kJ)
Find H for:
2PbS (s) + 3 O2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO2 (g)
Using:
2PbS (s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g) H =  124 kJ
Pb (s) + CO (g)  PbO (s) + C (s)
H =  106.8 kJ
Cancel identical items on opposite sides of the arrow.
2PbS (s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g) H =  124 kJ
2 PbO (s) + 2C (s)  2Pb (s) + 2 CO (g)
H = 2(+ 106.8 kJ)
Find H for:
2PbS (s) + 3 O2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO2 (g)
Using:
2PbS (s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g) H =  124 kJ
Pb (s) + CO (g)  PbO (s) + C (s)
H =  106.8 kJ
Add identical items on the same side of the arrow.
2PbS (s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g) H =  124 kJ
2 PbO (s) + 2C (s)  2Pb (s) + 2 CO (g)
H = 2(+ 106.8 kJ)
2PbS (s) + 3 O2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO2 (g)
Find H for:
2PbS (s) + 3 O2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO2 (g)
Using:
2PbS (s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g) H =  124 kJ
Pb (s) + CO (g)  PbO (s) + C (s)
H =  106.8 kJ
2PbS (s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g) H =  124 kJ
2 PbO (s) + 2C (s)  2Pb (s) + 2 CO (g)
H = 2(+ 106.8 kJ)
2PbS (s) + 3 O2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO2 (g)
Verify that the final reaction matches the reaction given in the
original problem.
Find H for:
2PbS (s) + 3 O2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO2 (g)
Using:
2PbS (s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g) H =  124 kJ
Pb (s) + CO (g)  PbO (s) + C (s)
H =  106.8 kJ
Combine the H’s.
2PbS (s) + 3 O2 (g)  2 PbO (s) + 2 SO2 (g) H =  124 kJ
2 PbO (s) + 2C (s)  2Pb (s) + 2 CO (g)
H = 2(+ 106.8 kJ)
2PbS (s) + 3 O2 (g) + 2 C (s)  2 Pb (s) + 2 CO (g) + 2 SO2 (g)
H =
+ 89.6 kJ
Example 3:
From the following enthalpy changes,
C (graphite) + O2 (g)  CO2 (g)
H2 (g) + ½O2 (g)  H2O (ℓ)
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (ℓ)
H =  394 kJ
H =  286 kJ
H =  890.3 kJ
calculate the value of H for the
following reaction:
C (graphite) + 2 H2 (g)  CH4 (g)
Find H for:
C (graphite) + 2 H2 (g)  CH4 (g)
Using:
C (graphite) + O2 (g)  CO2 (g)
H =  394 kJ
H2 (g) + ½O2 (g)  H2O (ℓ)
H =  286 kJ
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (ℓ)
H =  890.3 kJ
Leave the first reaction as written:
C (graphite) + O2 (g)  CO2 (g)
H =  394 kJ
Find H for:
C (graphite) + 2 H2 (g)  CH4 (g)
Using:
C (graphite) + O2 (g)  CO2 (g)
H =  394 kJ
H2 (g) + ½O2 (g)  H2O (ℓ)
H =  286 kJ
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (ℓ)
H =  890.3 kJ
Leave the second reaction as written:
C (graphite) + O2 (g)  CO2 (g)
H =  394 kJ
½O2 (g)  H2O (ℓ)
H =  286 kJ
H2 (g) +
Find H for:
C (graphite) + 2 H2 (g)  CH4 (g)
Using:
C (graphite) + O2 (g)  CO2 (g)
H =  394 kJ
H2 (g) + ½O2 (g)  H2O (ℓ)
H =  286 kJ
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (ℓ)
H =  890.3 kJ
Reverse the third reaction and change the sign:
C (graphite) + O2 (g)  CO2 (g)
H2 (g) +
½O2 (g)  H2O (ℓ)
CO2 (g) + 2 H2O (ℓ)  CH4 (g) + 2 O2 (g)
H =  394 kJ
H =  286 kJ
H = + 890.3 kJ
Find H for:
C (graphite) + 2 H2 (g)  CH4 (g)
Using:
C (graphite) + O2 (g)  CO2 (g)
H =  394 kJ
H2 (g) + ½O2 (g)  H2O (ℓ)
H =  286 kJ
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (ℓ)
H =  890.3 kJ
Multiply the second reaction by 2:
C (graphite) + O2 (g)  CO2 (g)

2 H2 (g) + 2 ½O2 (g) 
2 H2O (ℓ)
CO2 (g) + 2 H2O (ℓ)  CH4 (g) + 2 O2 (g)
H =  394 kJ
H = 2(  286 kJ)
H = + 890.3 kJ
Find H for:
C (graphite) + 2 H2 (g)  CH4 (g)
Using:
C (graphite) + O2 (g)  CO2 (g)
H =  394 kJ
H2 (g) + ½O2 (g)  H2O (ℓ)
H =  286 kJ
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (ℓ)
H =  890.3 kJ
Cancel identical items on opposite sides of the arrow.
C (graphite) + O2 (g)  CO2 (g)

2 H2 (g) + 2 ½O2 (g) 
2 H2O (ℓ)
CO2 (g) + 2 H2O (ℓ)  CH4 (g) + 2 O2 (g)
H =  394 kJ
H = 2(  286 kJ)
H = + 890.3 kJ
Find H for:
C (graphite) + 2 H2 (g)  CH4 (g)
Using:
C (graphite) + O2 (g)  CO2 (g)
H =  394 kJ
H2 (g) + ½O2 (g)  H2O (ℓ)
H =  286 kJ
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (ℓ)
H =  890.3 kJ
Add identical items on the same side of the arrow.
C (graphite) + O2 (g)  CO2 (g)

2 H2 (g) + 2 ½O2 (g) 
2 H2O (ℓ)
CO2 (g) + 2 H2O (ℓ)  CH4 (g) + 2 O2 (g)
C (graphite) + 2 H2 (g)  CH4 (g)
H =  394 kJ
H = 2(  286 kJ)
H = + 890.3 kJ
Find H for:
C (graphite) + 2 H2 (g)  CH4 (g)
Using:
C (graphite) + O2 (g)  CO2 (g)
H =  394 kJ
H2 (g) + ½O2 (g)  H2O (ℓ)
H =  286 kJ
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (ℓ)
H =  890.3 kJ
C (graphite) + O2 (g)  CO2 (g)

2 H2 (g) + 2 ½O2 (g) 
2 H2O (ℓ)
CO2 (g) + 2 H2O (ℓ)  CH4 (g) + 2 O2 (g)
H =  394 kJ
H = 2(  286 kJ)
H = + 890.3 kJ
C (graphite) + 2 H2 (g)  CH4 (g)
Verify that the final reaction matches the reaction given in the
original problem.
Find H for:
C (graphite) + 2 H2 (g)  CH4 (g)
Using:
C (graphite) + O2 (g)  CO2 (g)
H =  394 kJ
H2 (g) + ½O2 (g)  H2O (ℓ)
H =  286 kJ
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (ℓ)
H =  890.3 kJ
Combine the H’s.
C (graphite) + O2 (g)  CO2 (g)

2 H2 (g) + 2 ½O2 (g) 
2 H2O (ℓ)
CO2 (g) + 2 H2O (ℓ)  CH4 (g) + 2 O2 (g)
C (graphite) + 2 H2 (g)  CH4 (g)
H =  394 kJ
H = 2(  286 kJ)
H = + 890.3 kJ
H =  75.7 kJ
Thermochemistry
Part 5:
Standard
Enthalpies of
Formation
The standard enthalpy of formation (ΔHf)
of a compound is defined as the change in
enthalpy that accompanies the formation
of one mole of a compound from its
elements with all substances in their
standard states.
The ΔHf values for some common
substances are shown in Table 6.2.
More values are found in Appendix 4.
**ΔHf for an element in
its standard state is zero.
The change in enthalpy for a given reaction
can be calculated from the enthalpies of
formation of the reactants and products:
ΔH˚rxn = Σ npΔHf˚products  Σ nrΔHf˚reactants
Use the standard enthalpies of formation in Appendix 4 to
calculate the standard enthalpy change for the overall reaction
that occurs when ammonia is burned in air to form nitrogen
dioxide and water.
4 NH3 (g) + 7 O2 (g) → 4 NO2 (g) + 6 H2O (ℓ)
ΔH˚rxn = Σ npΔHf˚products  Σ nrΔHf˚reactants
ΔH˚rxn = [(4 mol NO2) (34 kJ/mol) +
(6 mol H2O) (286 kJ/mol)]
 [(4 mol NH3) (46 kJ/mol) +
(7 mol O2) (0 kJ/mol)]
ΔH˚rxn = [(136 kJ) + (1716 kJ)]  [(184 kJ) + (0 kJ)]
ΔH˚rxn = [1580 kJ]  [184 kJ]
ΔH˚rxn =  1396 kJ
It is to your benefit to work every
assigned homework problem because
you won’t be able to memorize how to
work all of the problems.
The test over this chapter will consist
of a multiple choice section (75%)
and an essay section (25%). The test
will be given over a period of two
days. This format is similar to the AP
Chemistry exam. On the AP exam,
each section is worth 50%.