Chapter 7: Thermochemistry
General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Chem 1050
Dr. Azra Ghumman
Contents
7-1
7-2
7-3
7-4
7-5
7-6
7-7
7-7
7-8
7-9
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Getting Started: Some Terminology
Heat
Heats of Reaction and Calorimetry
Work
The First Law of Thermodynamics
Heats of Reaction: ∆U and ∆H
The Indirect Determination of ∆H: Hess’s Law
The Indirect Determination of ∆H, Hess’s Law
Standard Enthalpies of Formation
Fuels as Sources of Energy
Focus on Fats, Carbohydrates, and Energy Storage
Thermochemistry
Dr. A. Ghumman
Learning objectives
1. Distinguish between heat and work.
2. Use specific heat to determine temperature changes and
quantities of heat.
3. Interconvert joules and calories. Apply first law of
thermodynamics: U = q+w.
4. Know the definition of work and calculate pressure-volume work.
5. State the meaning of the concept “state function,” especially as
demonstrated by enthalpy.
6. Calculate the heat of a reaction at constant volume, qv from
bomb calorimetry data.
7. Explain how to use a Styrofoam coffee cup calorimeter and
interpret the data obtained.
8. Apply Hess’s Law of constant heat summation.
Chem 1050
Chapt. 7 Thermochemistry
Dr. Azra Ghumman
Learning objectives
State the definitions of “standard state” and standard formation
reaction and write the standard formation reaction for any
substance.
10. Apply Hess’s law in the special case of standard formation
reactions,that is, compute ∆Hrxn° from ∆H°f values.
9.
Chem 1050
Chapt. 7 Thermochemistry
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Thermochemistry: Some Terminology
• System;
closed system ,open system
and isolated system
• Surroundings
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Chapt. 7 Thermochemistry
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Terminology
Energy(U)-The capacity to do work.
• Work -Force acting through a distance.
W= F x d
Different forms of Energy
Kinetic Energy -The energy of motion
– eK= ½ m (kg) x u2(m.s-1)2
Potential Energy
Thermal Energy
Chemical Energy
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Chapt. 7 Thermochemistry
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Energy units
• Kinetic Energy
ek =
1
2
mv2
[ek ] =
kg m2
= J
s2
• Work
w= Fx d
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[w ] =
kg m m
= J
s2
Chapt. 7 Thermochemistry
Dr. Azra Ghumman
Different forms of Energy
• Potential Energy-Energy due to condition, position, or
composition.
– Associated with forces of attraction or repulsion
between objects.
• Energy can change from one form to another form, i.e.
potential to kinetic energy.
Chem 1050
Chapt. 7 Thermochemistry
Dr. Azra Ghumman
Energy of a bouncing ball
•
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Chapt. 7 Thermochemistry
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Energy and Temperature
• Thermal Energy
– Kinetic energy associated with random molecular
motion.
– In general proportional to temperature.
– Thermal E ∝Temperature
Temperature is not thermal energy.
– An intensive property.
• Heat (q )
– Work (w)• Internal Energy-The energy content of a system.
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Intersession 2008
Chapt. 7 Thermochemistry
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Heat
• Energy transferred between a system and its surroundings
as a result of a temperature difference.
• Heat flows from hotter to colder body.
– Temperature may change.
– Phase may change (an isothermal process).
Chem 1050
Chapt. 7 Thermochemistry
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Units of Heat
Calorie (cal)-The quantity of heat required to change the
temperature of one gram of water by one degree Celsius.
A larger unit = kcal
James Joule (1818_1889)
• SI Unit for energy = Joule (J)
• SI unit for heat is the basic SI unit for energy
1 cal = 4.184 J
Chem 1050
Chapt. 7 Thermochemistry
Dr. Azra Ghumman
Heat Capacity
•
Heat Capacity (C)-The quantity of heat required to change the
temperature of a system by one degree.
•
Molar heat Capacity- The quantity of heat required to change the
temperature of one mole of substance by one degree Celsius.
•
Specific heat capacity or specific heat (c )- The quantity of heat
required to change the temperature of one gram of substance by one
degree Celsius .
•
Specific heat of water 4.18 J.g-1.°C-1
Quantity of heat
q = m . c . ∆T
Temperature of a system increases Tf>Ti
Temperature of a system decreases Ti >Tf
• Heat capacity
q = C ∆T
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where ∆ T = Tf -Ti
∆ T is = + ve and q = +ve
∆ T is = - ve and q = ?
Chapt. 7 Thermochemistry
Dr. Azra Ghumman
Law of conservation of Energy and experimental
determination of specific heat
• In interactions between a system and its surroundings, the
total energy remains constant.
qsystem + qsurroundings = 0
or
qsystem = - qsurroundings
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Chapt. 7 Thermochemistry
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Determination of specific heat
Determining Specific Heat from Experimental Data.
Use the data presented on the last slide to calculate the
specific heat of lead.
• qlead = -qwater
qwater = mc∆T = (50.0 g)(4.184 J/g °C)(28.8 - 22.0)°C
qwater = 1.4x103 J
qlead = -1.4x103 J = mc∆T = (150.0 g)(c)(28.8 - 100.0)°C
clead = 0.13 Jg-1°C-1
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Chapt. 7 Thermochemistry
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Importance of Specific Heat Values
• Why frozen food thaw more quickly in Aluminum
containers compared to other metals?
– (cAl = 0.903 J g-1 °C-1)
• Higher specific heats for compounds.
Why frozen food can thaw much faster in room
temperature water than if just exposed to room
temperature air ?
• Effect of larger lakes on local climate.
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Chapt. 7 Thermochemistry
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Heats of Reaction and Calorimetry
• Chemical Energy- contributes to internal energy of the
system.
– Energy associated with the bonds and intermolecular
attractions
• C3H8 (g) + 5O2 (g)→ 3CO2 (g) + 4H2O (l) + heat (-2219.9kJ)
• Heat of reaction, qrxn.
– The quantity of heat exchanged between a system and
its surroundings when a chemical reaction occurs
within the system, at constant temperature.
Chem 1050
Chapt. 7 Thermochemistry
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Heats of Reaction
• Exothermic reactions.
– release heat, qrxn < 0.
• Endothermic reactions.
– absorb heat, qrxn > 0.
• Calorimeter
– A device for measuring quantities of heat.
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Bomb Calorimeter
• qrxn = -qcal
• qcal = qbomb + qwater + qwires +…
Heat capacity of the calorimeter
heat
qcal = alliΣmici∆T = C∆T
•
Must be determined experimentally
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Chapt. 7 Thermochemistry
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Example 7-3
Using Bomb Calorimetry Data to Determine a Heat of
Reaction.
The combustion of 1.010 g sucrose, in a bomb calorimeter,
causes the temperature to rise from 24.92 to 28.33°C.
The heat capacity of the calorimeter assembly is 4.90 kJ
°C-1.
(a) What is the heat of combustion of sucrose, (C12H22O11)
expressed in kJ mol-1.
(b) Verify the claim of sugar producers that one teaspoon of
sugar (about 4.8 g) contains only 19 Calories.
Chem 1050
Chapt. 7 Thermochemistry
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Example 7-3
• Solution:
Calculate qcalorimeter:
qcal = C∆T = (4.90 kJ °C-1)(28.33-24.92)°C
= (4.90)(3.41) kJ
= 16.7 kJ
Calculate qrxn:
qrxn = -qcal = -16.7 kJ per 1.010 g
Chem 1050
Chapt. 7 Thermochemistry
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Example cont’d
Calculate qrxn in the required units:
qrxn = -qcal = 16.7 kJ = -16.5 kJ g-1
1.010 g
qrxn = -16.5 kJ g -1 = 343.3 g/1.00 mol
= -5.65 x 103 kJ mol-1
(a)
Calculate qrxn for one teaspoon:
4.8 g 1.00 kcal
)= -19 Cal tsp-1
qrxn = (-16.5 kJ g-1)(
)(
1 tsp 4.184 kJ
(b)
19 Calorie = 19 kcal tsp-1
Chem 1050
Chapt. 7 Thermochemistry
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Coffee Cup Calorimeter
• A simple calorimeter.
– Well insulated and therefore isolated.
– Measure temperature change.
qrxn = -qcal
See example 7-4 for a sample calculation.
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Chapt. 7 Thermochemistry
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Example
• Two solutions, 100.0 mL of 1.020M HCl(aq) and 50.0 mL
of 1.988 M NaOH both initially at 24.52°C are mixed in a
styrofoam cup calorimeter, what will be the final
temperature of the mixture? Make the same assumptions
• No heat loss to surroundings, solution has same density
and specific heat as water (d=1.00g mL-1of water, 4.18J g-1
°C-1) and use the heat of neutralization –56 kJ mol-1.
Chem 1050
Chapt. 7 Thermochemistry
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Work
• In addition to heat changes chemical reactions may also do
work
– Pressure-Volume work
• Gas formed pushes against
the atmosphere.
• Volume changes.
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Chapt. 7 Thermochemistry
Dr. Azra Ghumman
Pressure-Volume Work
W=Fxd
= (P x A) x h
= P∆V
W = -Pext∆V
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Chapt. 7 Thermochemistry
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Calculating Pressure-Volume Work
Calculating Pressure-Volume Work.
Suppose the gas in the previous figure is 0.100 mol He at 298 K. How
much work, in Joules, is associated with its expansion at constant
pressure?
Assume an ideal gas and calculate the volume change:
Vi = nRT/P
= (0.100 mol)(0.08201 L atm mol-1 K-1)(298K)/(2.40 atm)
= 1.02 L
Vf = 1.88 L
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∆V = 1.88-1.02 L = 0.86 L
Chapt. 7 Thermochemistry
Dr. Azra Ghumman
Example 7-5
Calculate the work done by the system:
w = -P∆V
= -(1.30 atm)(0.86 L)(
= -1.1 x 102 J
101 J )
1 L atm
Hint: If you use
pressure in kPa you
get Joules directly.
Where did the conversion factor come from?
Compare two versions of the gas constant and calculate.
8.3145 J mol-1 K-1 ≡ 0.082057 L atm mol-1K-1
1 ≡ 101.33 J L-1 atm-1
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Chapt. 7 Thermochemistry
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The First Law of Thermodynamics
• Internal Energy, U.
– Total energy (potential and kinetic).
Translational kinetic energy.
Molecular rotation.
Bond vibration.
Intermolecular attractions.
Chemical bonds.
Electrons
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Chapt. 7 Thermochemistry
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First Law of Thermodynamics
• A system contains only internal energy.
– A system does not contain heat or work.
– These only occur during a change in the system.
∆U = q + w
• Law of Conservation of Energy
– The energy of an isolated system is constant
– ∆Uisolated system = 0
Chem 1050
Chapt. 7 Thermochemistry
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First Law of Thermodynamics
∆U = q + w
When q is negative
i.e. qrxn< 0 and w is negative
work is done by the system
(expansion)
When q is positive, i.e. qrxn> 0 w is
positive
Work is done on the system
(contraction)
Chem 1050
Chapt. 7 Thermochemistry
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State Functions
• State Function- Any property that has a unique value for
a specified state of a system.
Water at 293.15 K and 1.00 atm is in a specified
state.
d = 0.99820 g mL-1
This density is a unique function of the state.
It does not matter how the state was established.
Chem 1050
Chapt. 7 Thermochemistry
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Functions of State
• U is a function of state.
– Not easily measured.
• ∆U has a unique value between two states (state function).
– can easily measured.
• The quantity of energy
(as heat) that must be
transferred from surrounding
to system and vice versa
during the the change of states.
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Chapt. 7 Thermochemistry
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Path Dependent Functions: work and heat
• Changes in heat and work are not functions of state.
– Remember example 7-5, w = -1.1 x 102 J in a one step
expansion of gas:
– Consider the expansion in two steps
– 2.40 atm →1.80 atm → 1.30 atm.
w = (-1.80 atm)(1.30-1.02)L – (1.30 atm)(1.88-1.36)L
= -0.61 L atm – 0.68 L atm = -1.3 L atm x
101 J
1 Latm
w = -1.3 x 102 J in two step expansion of gas
Chem 1050
Chapt. 7 Thermochemistry
Dr. Azra Ghumman
Pressure-Volume work
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Chapt. 7 Thermochemistry
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Heats of Reaction: ∆U and ∆H
Reactants → Products
•
Ui
Uf
∆U = Uf – Ui
•
∆U = qrxn + w
In a system at constant volume:
∆U = qrxn + 0 = qrxn = qv
Internal energy is the heat of reaction at constant volume.
But we live in a constant pressure world!
How does qp relate to qv?
Chem 1050
Chapt. 7 Thermochemistry
Dr. Azra Ghumman
Heats of Reaction
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Chapt. 7 Thermochemistry
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Heats of Reaction
qV = qP + w
using w = - P∆V and ∆U = qv (at constant volume), we obtain
∆U = qP - P∆V
and rearranging terms qP = ∆U + P∆V
These are all state functions ( U, P and V), so define a new state function.
Enthalpy of a system
H = U + PV
Then
∆H = Hf – Hi = ∆U + ∆PV
If we work at constant P and T, the heat flow for the process under these
conditions: qP = ∆U + P∆V
and
∆H = ∆U + P∆V
so ∆H = qP
Enthalpy change is the heat of reaction at constant pressure.
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Chapt. 7 Thermochemistry
Dr. Azra Ghumman
Comparing Heats of Reaction
qP
= -566 kJ mol-1
= ∆H
P∆V = P(Vf – Vi)
= RT(nf – ni)
= -2.5 kJ
∆U = ∆H - P∆V
= -563.5 kJ mol-1
= qV
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Chapt. 7 Thermochemistry
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Comparing ∆H and ∆U
•
a.
Three types of reaction.
There is no formation of gaseous products
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Then ∆H ≈ ∆U ( as ∆V=0, so -P∆V = 0)
b. Formation of the gas with no change in the number of moles of the
reactants and products.
N2(g) +O2(g) → 2NO(g)
∆H ≈ ∆U (as ∆V = 0, so -P∆V = 0)
c. Formation of gas and change in the number of moles of product.
2NH3 → N2(g) + 3H2(g) (2mol → 4 mol)
∆H = - 92.22 kJ at 298.5K
P∆V = nRT and n = nf –ni = 4 –2 = 2 mol
-1
-1
= 2.0 mol x 8.314 J mol K x 298.15K = 4975 J = 5.0kJ
and ∆U = ∆H - P∆V = - 92.22 kJ – (+ 5.0 kJ)
= -97.22 kJ
∆U and ∆H are still very close.
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Enthalpy change and the changes in state of
Matter
• Molar enthalpy of vaporization:
H2O (l) → H2O(g)
∆H = 44.0 kJ at 298 K
• Molar enthalpy of fusion:
H2O (s) → H2O(l)
∆H = 6.01 kJ at 273.15 K
• Enthalpy for one mole of a substance is called molar
enthalpy and its units are kJ mol-1
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Chapt. 7 Thermochemistry
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Practice Example A
Enthalpy Changes Accompanying Changes in States of Matter.
•
Calculate ∆H for the process in which 50.0 g of water is converted
from liquid at 10.0°C to vapor at 25.0°C.
Solution: Break the problem into two steps: Raise the temperature of
the liquid first then completely vaporize it. The total enthalpy change
is the sum of the changes in each step.
Set up the equation and calculate:
qP = mcH2O∆T + n∆Hvap
50.0 g
= (50.0 g)(4.184 J g-1 °C-1)(25.0-10.0)°C +
44.0 kJ mol-1
18.0 g mol-1
= 3.14 kJ + 122 kJ = 125 kJ
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Chapt. 7 Thermochemistry
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Standard States and Standard Enthalpy Changes
• Define a particular state as a standard state to compare
enthalpy changes.
• Standard enthalpy of reaction, ∆H°
– The enthalpy change of a reaction in which all reactants
and products are in their standard states.
• Standard State
– The pure element or compound at a pressure of 1 bar
(105 Pa ) and at the temperature of interest.
• Standard State for gas
– The pure gas behaving as an ideal gas at a pressure of 1
bar and at the temperature of interest.
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Enthalpy Diagrams
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Indirect Determination of ∆H:
Hess’s Law
• ∆H is an extensive property.
– Enthalpy change is directly proportional to the amount of
substance in a system.
N2(g) + O2(g) → 2 NO(g)
1
1
N 2 ( g ) + O2 ( g ) → NO ( g )
2
2
∆H = +180.50 kJ at 25°C
∆H = +90.25 kJ
• ∆H changes sign when a process is reversed
– NO(g) → ½N2(g) + ½O2(g) ∆H = -90.25 kJ
Chem 1050
Chapt. 7 Thermochemistry
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Indirect Determination of ∆H: Hess’s Law
•
Hess’s law of constant heat summation
– If a process occurs in stages or steps (even hypothetically), the
enthalpy change for the overall process is the sum of the enthalpy
changes for the individual steps.
½N2(g) + ½O2(g) → NO(g)
∆H = +90.25 kJ
NO(g) + ½O2(g) → NO2(g)
∆H = -57.07 kJ
½N2(g) + O2(g) → NO2(g)
∆H = +33.18 kJ
Regardless of the path taken to reach from initial state to
final state, ∆H has the same value.
Chem 1050
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Hess’s Law Schematically
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Applying Hess’s Law
• Determine the value of ∆H for the following reaction:
2C(gr) + 3H2(g) → C2H6(g) ∆H = ?
• Given the following data:
2C2H6(g) + 7O2(g) → 4CO2 +6 H2O(l)
C(gr) + O2(g) → CO2(g)
2H2(g) + O2(g) → 2H2O(l)
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∆H = -3120 kJ
∆H = -394 kJ
∆H = -572 kJ
Standard Enthalpies of Formation
Standard enthalpy of formation (∆Hf°)
• The enthalpy change that occurs in the formation of one
mole of a substance in the standard state from the
reference forms of the elements in their standard states.
• The standard enthalpy of formation of a pure element in its
reference state is 0.
∆Hf°(element in reference form) = 0
The formation of an element from itself is no change at all.
Chem 1050
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Standard Enthalpies of Formation
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Chapt. 7 Thermochemistry
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Standard Enthalpies of Formation
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Standard Enthalpies of Reaction
∆Hoverall = -2∆Hf°NaHCO3+ ∆Hf°Na2CO3
+ ∆Hf°CO2 + ∆Hf°H2O
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Enthalpy of Reaction
∆ H°rxn = ∑vp∆Hf°products- ∑vr ∆Hf°reactants
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Calculating ∆H°rxn from ∆Hf° data
Example: Large quantities of ammonia are used to prepare
nitric acid.The first step consists of the catalytic oxidation
of ammonia to nitric oxide NO by the following reaction,
• 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) ∆H°rxn= ?
Given data
Substances
∆Hf° kJ mol-1
NO(g)
90.3
H2O(g)
-241.6
NH3(g)
-45.9
O2(g)
0
What is the standard enthalpy change for this reaction?
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Ionic Reactions in Solutions
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Fuels as Sources of Energy
•Fossil fuels.
–Combustion is exothermic.
–Non-renewable resource.
–Environmental impact
Chem 1050
Chapt. 7 Thermochemistry
Dr. Azra Ghumman