Transcript Standard enthalpy changes

6
1
Energetics
6.1
What is Energetics?
6.2
Enthalpy Changes Related to Breaking and
Forming of Bonds
6.3
Standard Enthalpy Changes
6.4
Experimental Determination of Enthalpy
Changes by Calorimetry
6.5
Hess’s Law
6.6
Calculations involving Standard Enthalpy
Changes of Reactions
New Way Chemistry for Hong Kong A-Level Book 1
6.1 What is energetics? (SB p.136)
What is energetics?
Energetics is the study of energy changes
associated with chemical reactions.
Thermochemistry is the study of heat changes
associated with chemical reactions.
Some terms
Enthalpy(H) = heat content in a substance
Enthalpy change(H)
= heat content of products - heat content of
reactants
= Hp - Hr
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6.1 What is energetics? (SB p.136)
Law of conservation of energy
The law of conservation of energy states that
energy can neither be created nor destroyed.
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6.1 What is energetics? (SB p.137)
Internal energy and enthalpy
e.g.
4
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
New Way Chemistry for Hong Kong A-Level Book 1
6.1 What is energetics? (SB p.138)
Internal energy and enthalpy
Heat change at
constant
=
pressure
Enthalpy
change
5
Change in
internal
+
energy
Work done
on the
surroundings
(Heat change at
constant volume)
New Way Chemistry for Hong Kong A-Level Book 1
6.1 What is energetics? (SB p.138)
Exothermic and endothermic reactions
An exothermic reaction is a reaction that
releases heat energy to the surroundings.
(H = -ve)
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6.1 What is energetics? (SB p.139)
Check Point 6-1
Exothermic and endothermic reactions
An endothermic reaction is a reaction that
absorbs heat energy from the surroundings.
(H = +ve)
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6.2
Energy Changes
Related to Breaking
and Forming of Bonds
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6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)
Enthalpy changes related to breaking and
forming of bonds
e.g. CH4 + 2O2 CO2 + 2H2O
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6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)
Enthalpy changes related to breaking and
forming of bonds
In an exothermic reaction, the energy required in
breaking the bonds in the reactants is less than
the energy released in forming the bonds in the
products (products contain stronger bonds).
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6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)
Enthalpy changes related to breaking and
forming of bonds
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6.2 Enthalpy changes related to breaking and forming of bonds (SB p.140)
Enthalpy changes related to breaking and
forming of bonds
In an endothermic reaction, the energy required
in breaking the bonds in the reactants is more
than the energy released in forming the bonds
in the products (reactants contain stronger
bonds).
Check Point 6-2
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6.3
Standard Enthalpy
Changes
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6.3 Standard enthalpy changes (SB p.141)
Standard enthalpy changes
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H = -802 kJ mol-1
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H = -890 kJ mol-1
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6.3 Standard enthalpy changes (SB p.141)
Standard enthalpy changes
As enthalpy changes depend on temperature
and pressure, it is necessary to define standard
states and conditions:
1.
2.
3.
elements or compounds in their normal
physical states;
a pressure of 1 atm (101325 Nm-2); and
a temperature of 25oC (298 K)
Enthalpy change under standard conditions denoted
by symbol: H
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6.3 Standard enthalpy changes (SB p.142)
Standard enthalpy changes of
neutralization
Standard enthalpy change of neutralization
(Hneut) is the enthalpy change when one mole
of water is formed from the neutralization of an
acid by an alkali under standard conditions.
e.g. H+(aq) + OH-(aq)  H2O(l)
Hneut = -57.3 kJ mol-1
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6.3 Standard enthalpy changes (SB p.142)
Standard enthalpy changes of
neutralization
Acid
HCl
HCl
HCl
HF
Alkali
NaOH
KOH
NH3
NaOH
Hneu
-57.1
-57.2
-52.2
-68.6
H+(aq) + OH-(aq)  H2O(l)
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6.3 Standard enthalpy changes (SB p.142)
Standard enthalpy changes of
neutralization
Enthalpy level diagram for the neutralization of a
strong acid and a strong alkali
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6.3 Standard enthalpy changes (SB p.142)
Standard enthalpy change of solution
Standard enthalpy change of solution (Hsoln)
is the enthalpy change when one mole of a
solute is completely dissolved in a sufficiently
large volume of solvent to form an infinitely
dilute solution under standard conditions.
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6.3 Standard enthalpy changes (SB p.143)
Standard enthalpy change of solution
e.g. NaCl(s) + water  Na+(aq)+Cl-(aq)
Hsoln= +3.9 kJ mol-1
Enthalpy level diagram
for the dissolution of
NaCl
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6.3 Standard enthalpy changes (SB p.143)
Standard enthalpy change of solution
e.g. LiCl(s) + water  Li+(aq) + Cl-(aq)
Hsoln= -37.2 kJ mol-1
Enthalpy level diagram
for the dissolution of
LiCl in water
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6.3 Standard enthalpy changes (SB p.143)
Standard enthalpy change of solution
Salt
NaOH
NaCl
KOH
KBr
22
Hsoln(kJ mol-1)
-44.7
+3.9
-57.8
+20.0
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6.3 Standard enthalpy changes (SB p.144)
Standard enthalpy change of
formation
Standard enthalpy change of formation (Hf )
is the enthalpy change of the reaction when
one mole of the compound in its standard
state is formed from its constituent elements
under standard conditions.
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6.3 Standard enthalpy changes (SB p.144)
Standard enthalpy change of
formation
e.g.
2Na(s) + Cl2(g)  2NaCl(s)
H = -822 kJ mol-1
Na(s) + ½Cl2(g)  NaCl(s)
Hf = -411 kJ mol-1
Standard enthalpy change of
formation of NaCl is -411 kJ mol-1.
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6.3 Standard enthalpy changes (SB p.144)
Standard enthalpy change of
formation
What is Hf [N2(g)] ?
N2(g)  N2(g)
Hf [N2(g)] = 0
The enthalpy change of formation of an
element is always zero.
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6.3 Standard enthalpy changes (SB p.146)
Standard enthalpy change of combustion
e.g. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
H1 = -2220 kJ
2C3H8(g) + 10O2(g) 6CO2(g) + 8H2O(l)
H2 = -4440 kJ
 It is more convenient to report enthalpy
changes per mole of the main reactant
reacted/product formed.
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H2 = ?
6.3 Standard enthalpy changes (SB p.146)
Standard enthalpy change of combustion
Standard enthalpy change of combustion (Hc )
of a substance is the enthalpy change when one
mole of the substance burns completely under
standard conditions.
e.g. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
1 mole
Hc = -2220 kJ mol-1
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New Way Chemistry for Hong Kong A-Level Book 1
6.3 Standard enthalpy changes (SB p.147)
Standard enthalpy change of combustion
Substance
Hc (kJ mol-1)
H2(g)
C (diamond)
C (graphite)
CO(g)
CH4(g)
-285.8
-395.4
-393.5
-283.0
-890.4
Check Point 6-3
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6.4
Experimental
Determination of
Enthalpy Changes
by Calorimetry
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6.4 Experimental determination of enthalpy changes by calorimetry (SB p.148)
Experimental determination of enthalpy
changes by calorimetry
Calorimeter = a container used for measuring
the temperature change of solution
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6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149)
Determination of enthalpy change of
neutralization
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6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149)
Heat evolved = (m1c1 + m2c2) ΔT
where m1 is the mass of the solution,
m2 is the mass of calorimeter,
c1 is the specific heat capacity of the
solution,
c2 is the specific heat capacity of
calorimeter,
ΔT is the temperature change of
the reaction
Example 6-4A
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6.4 Experimental determination of enthalpy changes by calorimetry (SB p.150)
Determination of enthalpy change of
combustion
The Philip Harris
calorimeter used for
determining the enthalpy
change of combustion of
a liquid fuel
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New Way Chemistry for Hong Kong A-Level Book 1
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151)
Determination of enthalpy change of
combustion
A simple apparatus used to determine the enthalpy
change of combustion of ethanol
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New Way Chemistry for Hong Kong A-Level Book 1
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151)
Heat evolved = (m1c1 + m2c2) ΔT
Where m1 is the mass of water in the calorimeter,
m2 is the mass of the calorimeter,
c1 is the specific heat capacity of the water,
c2 is the specific heat capacity of
calorimeter,
ΔT is the temperature change of the
reaction
Example 6-4B
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New Way Chemistry for Hong Kong A-Level Book 1
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.152)
Determination of enthalpy change of
solution
• By measuring the temperature change when
a known mass of solute is added to a known
volume of solvent in a calorimeter
• Heat change = (m1c1 + m2c2) T
Example 6-4C
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6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)
Determination of enthalpy change of
formation
• The enthalpy change of formation of a
substance can be quite high
• Found out by applying Hess’s law of
constant heat summation
Check Point 6-4
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6.5
Hess’s Law
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6.5 Hess’s law (SB p.153)
Hess’s Law
Route 1
A+B
H1
H2
C+D
H3
E
Route 2
H1 = H2 + H3
Hess’s law of constant heat summation states
that the total enthalpy change accompanying a
chemical reaction is independent of the route by
which the chemical reaction takes place.
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6.5 Hess’s law (SB p.154)
Enthalpy level diagram
• Relate substances together in terms of
enthalpy changes of reactions
Enthalpy level
diagram for the
oxidation of
C(graphite) to CO2(g)
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6.5 Hess’s law (SB p.155)
Enthalpy cycle (Born-Haber cycle)
• Relate the various equations involved in a
reaction
Enthalpy cycle for the
oxidation of C(graphite)
to CO2(g)
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6.5 Hess’s law (SB p.155)
Importance of Hess’s law
The enthalpy change of some chemical reactions
cannot be determined directly because:
•
•
•
the reactions cannot be performed in the
laboratory
the reaction rates are too slow
the reactions may involve the formation of side
products
But the enthalpy change of such reactions
can be determined indirectly by applying
Hess’s Law.
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6.5 Hess’s law (SB p.153)
Enthalpy change of formation of CO(g)
Given: Hf [CO2(g)] = -393.5 kJ mol-1;
Hc [CO(g)] = -283.0 kJ mol-1
C(graphite) + ½O2(g)
+ ½O2(g)
Hf [CO(g)]
+ ½O2(g)
H1
CO2(g)
H2
Hf [CO(g)] + H2 = H1
Hf [CO(g)] = H1 - H2
= -393.5 - (-283.0 )
= -110.5 kJ mol-1
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CO(g)
New Way Chemistry for Hong Kong A-Level Book 1
6.5 Hess’s law (SB p.153)
Enthalpy change of formation of CaCO3(s)
Ca(s) + C(graphite) + 3 O2
2
Hf [CaCO3(s)]
CaCO3(s)
H2
H1
CaO(s) + CO2(g)
Hf [CaCO3(s)] = H1 + H2
= -1028.5 kJ mol-1 + (-178.0) kJ mol-1
= -1206.5 kJ mol-1
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6.5 Hess’s law (SB p.153)
Enthalpy change of hydration of MgSO4(s)
MgSO4(s) + 7H2O(l)
ΔH1
ΔH
MgSO4·7H2O(s)
aq
aq
ΔH2
Mg2+(aq) + SO42-(aq) + 7H2O(l)
ΔH = enthalpy of hydration of MgSO4(s)
ΔH1 = molar enthalpy change of solution of anhydrous
magnesium sulphate(VI)
ΔH2 = molar enthalpy change of solution of magnesium
sulphate(VI)-7-water
ΔH = ΔH1 - ΔH2
45
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Check Point 6-5
6.6
Calculations involving
Standard Enthalpy
Changes of Reactions
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6.6 Calculations involving standard enthalpy changes of reactions (SB p.159)
Calculation of standard enthalpy change of
reaction from standard enthalpy changes
of formation
Hreaction =  Hf [products] -  Hf [reactants]
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New Way Chemistry for Hong Kong A-Level Book 1
6.6 Calculations involving standard enthalpy changes of reactions (SB p.159)
48
Example 6-6A
Example 6-6B
Example 6-6C
Example 6-6D
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6.6 Calculations involving standard enthalpy changes of reactions (SB p.162)
Calculation of standard enthalpy change of
formation from standard enthalpy changes
of combustion
Hf =  Hc [products] -  Hc [reactants]
49
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6.6 Calculations involving standard enthalpy changes of reactions (SB p.162)
Example 6-6E
Example 6-6F
Check Point 6-6
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6.7 Entropy change (SB p.164)
6.7
Entropy Change
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6.7 Entropy change (SB p.164)
Entropy
A process is said to be spontaneous
• If no external forces are required to keep the
process going
• The process may be physical change or a
chemical change
• Example of spontaneous physical change:
cooling of hot water
• Example of spontaneous chemcial change:
burning of wood once the fire is started
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6.7 Entropy change (SB p.164)
Entropy
• Exothermicity is the reason for the spontaneity
of a process
• Some spontaneous changes are endothermic
• Examples: Melting of ice, dissolution of
ammonium nitrate in water
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6.7 Entropy change (SB p.164)
Melting of ice
Dissolution of ammonium
nitrate in water
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6.7 Entropy change (SB p.165)
Entropy
• Entropy is a measure of the randomness or
the degree of disorder of a system
Solid
Liquid
Entropy increases
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Gas
6.7 Entropy change (SB p.166)
Entropy change
• Entropy change means the change in the
degree of disorder of a system
S = Sfinal - Sinitial
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6.7 Entropy change (SB p.166)
Positive entropy
• Increase in entropy
• Final state has a larger entropy that the initial
state
• Example:
Ice (less entropy)  Water (more entropy)
S = Swater – Sice = +ve
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6.7 Entropy change (SB p.166)
Negative entropy
• Decrease in entropy
• Initial state has a larger entropy that the final
state
• Example:
Water (more entropy)  Ice (less entropy)
S = Sice – Swater = -ve
Check Point 6-7
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6.8
59
Free Energy
Change
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6.8 Free energy change (SB p.168)
Free energy change
• Entropy is temperature dependent
• At a higher temp, the entropy of the system is
higher
• At a lower temp, the entropy of the system is
lower
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6.8 Free energy change (SB p.168)
Free energy
• Another driving force for a process
• Called free energy (G)
G = H – TS
where H is the enthalpy
T is Kelvin temperature
S is the entropy
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6.8 Free energy change (SB p.168)
Free energy change
G = H – TS
• At a given temp, there are two driving forces
for a process to occur
• Overall enthalpy of the system tends to be low
• Overall entropy of the system tends to be high
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6.8 Free energy change (SB p.168)
Significance of the equation
• Process favoured by:
H = -ve; S = +ve
• Process not favoured by:
H = +ve; S = -ve
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6.8 Free energy change (SB p.168)
Significance of the equation
• A process is spontaneous or favourable
when G is negative
• A process is not spontaneous or favourable
as indicated when G is positive, but is
spontaneous in the opposite direction
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6.8 Free energy change (SB p.169)
How H and S affect the
spontaneity of a process
H
S
G = H - TS
Result
-ve
+ve
-ve
Process is spontansous
at all temepratures
+ve
-ve
+ve
Process is not
spontaneous at any
temperature (reverse
process is spontaneous
at all temperatures)
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6.8 Free energy change (SB p.170)
Effects of relative magnitudes of H and
S on the spontaneity of a process
H
S
Condition
+ve
+ve
At high temp,
TS > H
+ve
+ve
At low temp,
H > TS
G = +ve
-ve
-ve
At high temp,
TS > H
G = +ve
Process is not
spontaneous
-ve
-ve
At low temp,
H > TS
G = -ve
Process is
spontaneous at
low temp
66
G = H TS
G = -ve
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Result
Process is
spontaneous at
high temp
Process is not
spontaneous
6.8 Free energy change (SB p.170)
Check Point 6-8
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The END
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6.1 What is energetics? (SB p.140)
Back
State whether the following processes are exothermic or
endothermic.
(a) Melting of ice.
(a) Endothermic
(b) Dissolution of table salt.
(c) Condensation of steam.
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(b) Endothermic
(c) Exothermic
New Way Chemistry for Hong Kong A-Level Book 1
Answer
6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141)
(a) State the difference between exothermic and endothermic
reactions with respect to
(i) the sign of H;
(ii) the heat change with the surroundings;
Answer
(iii) the total enthalpy of reactants and products.
(a) (i) Exothermic reactions: H = -ve; endothermic reactions: H
= +ve
(ii) Heat is given out to the surroundings in exothermic
reactions whereas heat is taken in from the surroundings
in endothermic reactions.
70
(iii) In exothermic reactions, the total enthalpy of products is less
than that of the reactants. In endothermic reactions, the total
enthalpyNew
is greater
than that of the reactants.
Way Chemistry for Hong Kong A-Level Book 1
6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141)
(b) Draw an enthalpy level diagram for a reaction which is
(i) endothermic, having a large activation energy.
(ii) exothermic, having a small activation energy.
Answer
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6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141)
(b) (i)
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6.2 Enthalpy changes related to breaking and forming of bonds (SB p.141)
Back
(ii)
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6.3 Standard enthalpy changes (SB p.147)
(a) Why must the condition “burnt completely in oxygen” be
emphasized in the definition of standard enthalpy change
of combustion?
(a) If the substance is not completely burnt in
excess oxygen, other products such as
C(s) and CO(g) may be formed. The
enthalpy change of combustion
measured will not be accurate.
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Answer
6.3 Standard enthalpy changes (SB p.147)
(b) The enthalpy change of the following reaction under
standard conditions is –566.0 kJ.
2CO(g) + O2(g)  2CO2(g)
What is the standard enthalpy change of combustion of
carbon monoxide?
Answer
(b) Standard enthalpy change of combustion of CO
1
=
 (-566.0) kJ
2
= -283.0 kJ
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6.3 Standard enthalpy changes (SB p.147)
(c) What terms may be given for the enthalpy change of the
following reaction?
1
N2(g) + O2(g)  NO2(g)
2
(c) Enthalpy change of combustion of nitrogen or
enthalpy change of formation of nitrogen dioxide.
Back
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Answer
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149)
Determine the enthalpy change of neutralization of 25 cm3 of
1.25 M hydrochloric acid and 25 cm3 of 1.25 M sodium
hydroxide solution using the following data:
Mass of calorimeter = 100 g
Initial temperature of acid = 15.5 oC (288.5 K)
Initial temperature of alkali = 15.5 oC (288.5 K)
Final temperature of the reaction mixture
= 21.6 oC (294.6 K)
The specific heat capacities of water and calorimeter
are 4200 J kg-1 K-1 and 800 J kg-1 K-1 respectively.
Answer
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6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149)
Assume that the density of the reaction mixture is the same as that of water,
i.e. 1 g cm-3.
Mass of the reaction mixture = (25 + 25) cm3  1 g cm-3 = 50 g = 0.05 kg
Heat given out = (m1c1 + m2c2) T
= (0.05 kg  4200 J kg-1 K-1 + 0.1 kg  800 J kg-1 K-1) 
(294.6 – 288.5) K
= 1769 J
H+(aq) + OH-(aq)  H2O(l)
Number of moles of HCl = 1.25 mol dm-3  25  10-3 dm3 = 0.03125 mol
Number of moles of NaOH = 1.25 mol dm-3  25  10-3 dm3 = 0.03125 mol
Number of moles of H2O formed = 0.03125 mol
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6.4 Experimental determination of enthalpy changes by calorimetry (SB p.149)
Heat given out per mole of H2O formed
1769 J
=
0.03125 mol
= 56608 J mol-1
The enthalpy change of neutralization is –56.6 kJ mol-1.
Back
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6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151)
Determine the enthalpy change of combustion of ethanol
using the following data:
Mass of spirit lamp before experiment = 45.24 g
Mass of spirit lamp after experiment = 44.46 g
Mass of water in copper calorimeter = 50 g
Mass of copper calorimeter without water = 380 g
Initial temperature of water = 18.5 oC (291.5 K)
Final temperature of water = 39.4 oC (312.4 K)
The specific heat capacities of water and copper
calorimeter are 4200 J kg-1 K-1 and 2100 J kg-1 K-1
respectively.
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Answer
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151)
Heat evolved by the combustion of ethanol
= Heat absorbed by the copper calorimeter
= (m1c1 + m2c2) T
= (0.05 kg  4200 J kg-1 K-1 + 0.38 kg  2100 J kg-1 K-1) 
(312.4 – 291.5)K
= 21067 J
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
Mass of ethanol burnt = (45.24 – 44.46) g = 0.78 g
0.78 g
Number of moles of ethanol burnt =
= 0.017 mol
1
46.0 g mol
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New Way Chemistry for Hong Kong A-Level Book 1
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.151)
Heat given out per mole of ethanol
21067 J
=
0.017 mol
= 1239235 J mol-1
Back
= 1239 kJ mol-1
The enthalpy change of combustion of ethanol is –1239 kJ mol-1.
There was heat loss by the system to the surroundings, and incomplete
combustion of ethanol might occur. Also, the experiment was not carried out
under standard conditions. Therefore, the experimentally determined value
(-1239 kJ mol-1) is less than the theoretical value of the standard enthalpy
change of combustion of ethanol (-1371 kJ mol-1).
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6.4 Experimental determination of enthalpy changes by calorimetry (SB p.152)
0.02 mol of anhydrous ammonium chloride was added to
45 g of water in a polystyrene cup to determine the enthalpy
change of solution of anhydrous ammonium chloride. It is
found that there was a temperature drop from 24.5 oC to
23.0 oC in the solution.
Given that the specific heat capacity of water is 4200 J kg-1 K-1
and
NH4Cl(s) + aq  NH4Cl(aq)
Calculate the enthalpy change of solution of anhydrous
ammonium chloride.
(Neglect the specific heat capacity of the polystyrene cup.)
Answer
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New Way Chemistry for Hong Kong A-Level Book 1
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.152)
Heat absorbed = m1c1T ( c2  0)
= 0.045 kg  4200 J kg-1 K-1  (297.5 – 296) K
= 283.5 J (0.284 kJ)
0.284 g
Heat absorbed per mole of ammonium chloride =
0.02 mol
= 14.2 kJ mol-1
The enthalpy change of solution of anhydrous ammonium chloride is
+14.2 kJ mol-1.
Back
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6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)
(a) A student tried to determine the enthalpy change of
neutralization by putting 25.0 cm3 of 1.0 M HNO3 in a
polystyrene cup and adding 25.0 cm3 of 1.0 M NH3 into it.
The temperature rise recorded was 3.11 oC. Given that the
mass of the polystyrene cup is 250 g, the specific heat
capacities of water and the polystyrene cup are
4200 J kg-1 K-1 and 800 J kg-1 K-1 respectively. Determine
the enthalpy change of neutralization of nitric acid and
aqueous ammonia. (Density of water = 1 g cm-3)
Answer
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New Way Chemistry for Hong Kong A-Level Book 1
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)
(a) Heat evolved = m1c1T + m2c2 T
= 0.050 kg  4200 J kg-1 K-1  3.11 K + 0.25 kg 
800 J kg-1 K-1  3.11 K
= (653.1 + 622) J
= 1275.1 J
No. of moles of HNO3 used = 1.0 M  25  10-3 m3
= 0.025 mol
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New Way Chemistry for Hong Kong A-Level Book 1
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)
(a) No. of moles of NH3 used = 1.0 M  25  10-3 dm3
= 0.025 mol
No. of moles of H2O formed = 0.025 mol
Heat evolved per mole of H2O formed
1275.1 J
=
0.025 mol
= 51.004 kJ mol-1
The enthalpy change of neutralization of nitric acid and aqueous
ammonia is –51.004 kJ mol-1.
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6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)
(b) When 0.05 mol of silver nitrate was added to 50 g of water
in a polystyrene cup, a temperature drop of 5.2 oC was
recorded. Assuming that there was no heat absorption by
the polystyrene cup, calculate the enthalpy change of
solution of silver nitrate.
(Specific heat capacity of water = 4200 J kg-1 K-1)
Answer
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New Way Chemistry for Hong Kong A-Level Book 1
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)
(b) Energy absorbed = mcT
= 0.05 kg  4200 J kg-1 K-1  5.2 K
= 1092 J
No. of moles of AgNO3 used = 0.05 mol
1092 J
0.05 mol
= 21.84 kJ mol-1
Energy absorbed per mole of AgNO3 used =
The enthalpy change of solution of silver nitrate is +21.84 kJ mol-1.
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New Way Chemistry for Hong Kong A-Level Book 1
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)
(c) A student used a calorimeter as shown in Fig. 6-15 to
determine the enthalpy change of combustion of methanol.
In the experiment, 1.60 g of methanol was used and 50 g of
water was heated up, raising the temperature by 33.2 oC.
Given that the specific heat capacities of water and copper
calorimeter are 4200 J kg-1 K-1 and
2100 J kg-1 K-1 respectively and the mass of the calorimeter
is 400 g, calculate the enthalpy change of combustion of
methanol.
Answer
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New Way Chemistry for Hong Kong A-Level Book 1
6.4 Experimental determination of enthalpy changes by calorimetry (SB p.153)
Back
(c) Heat evolved = m1c1T + m2c2 T
= 0.05 kg  4200 J kg-1 K-1  33.2 K + 0.4 kg 
2100 J kg-1 K-1  33.2 K
= (6972 + 27888) J
= 34860 J
1.60 g
No. of moles of methanol used =
32.0 g mol -1
= 0.05 mol
34860 J
Heat evolved per mole of methanol used =
0.05 mol
= 697.2 kJ mol-1
The enthalpy change of combustion of methanol is –697.2 kJ mol-1.
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6.5 Hess’s law (SB p.158)
(a) Given the following thermochemical equation:
2H2(g) + O2(g)  2H2O(l)
(i) Is the reaction endothermic or exothermic?
(ii) What is the enthalpy change for the following
reactions?
(1) 2H2O(l)  2H2(g) + O2(g)
1
(2) H2(g) + O2(g)  H2O(l)
2
(iii) If the enthalpy change for the reaction H2O(l) 
H2O(g) is +41.1 kJ mol-1, calculate the H for 2H2(g) +
O2(g)  2H2O(g).
Answer
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New Way Chemistry for Hong Kong A-Level Book 1
6.5 Hess’s law (SB p.158)
(a) (i) Exothermic
(ii) (1) +571.6 kJ mol-1
(2) –285.8 kJ mol-1
(iii)
H = [-571.6 + 2  (+41.1)] kJ mol-1 = -489.4 kJ mol-1
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6.5 Hess’s law (SB p.158)
(b) Given the following information about the enthalpy
change of combustion of allotropes of carbon:
Hc [C(graphite)] = -393.5 kJ mol-1
Hc [C(diamond)] = -395.4 kJ mol-1
(i) Which allotrope of carbon is more stable?
(ii) What is the enthalpy change for the following process?
C(graphite)  C(diamond)
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Answer
6.5 Hess’s law (SB p.158)
(b) (i) Graphite
(ii)
H = [-393.5 – (-395.4)] kJ mol-1 = +1.9 kJ mol-1
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6.5 Hess’s law (SB p.158)
(c) The formation of ethyne (C2H2(g) can be represented by
the following equation:
2C(graphite) + H2(g)  C2H2(g)
(i) Draw an enthalpy level diagram relating the above
equation to carbon dioxide and water.
(ii) Calculate the standard enthalpy change of formation
of ethyne.
(Given: Hc [C(graphite)] = -393.5 kJ mol-1;
Hc [H2(g)] = -285.8 kJ mol-1;
Hc [C2H2(g)] = -1299 kJ mol-1)
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New Way Chemistry for Hong Kong A-Level Book 1
Answer
6.5 Hess’s law (SB p.158)
Back
(c) (i)
(ii) Hf [C2H2(g)] = Hc [C(graphite)]  2 + Hc [H2(g)] – Hc [C2H2(g)]
= [(-393.5)  2 + (-285.8) – (-1299)] kJ mol-1
= +226.2 kJ mol-1
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New Way Chemistry for Hong Kong A-Level Book 1
6.6 Calculations involving standard enthalpy changes of reactions (SB p.159)
Given the following information, find the standard enthalpy
change of the reaction:
C2H4(g) + H2(g)  C2H6(g)
Hf [C2H4(g)] = +52.3 kJ mol-1
Hf [C2H6(g)] = -84.6 kJ mol-1
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Answer
6.6 Calculations involving standard enthalpy changes of reactions (SB p.159)
Back
Note:
H1 = [Hf (reactants)] = Hf [C2H4(g)] + Hf [H2(g)]
H2 = [Hf (products)] = Hf [C2H6(g)]
Applying Hess’s law,
H1 + H = H2
H = H2 - H1
= Hf [C2H6(g)] – (Hf [C2H4(g)] + Hf [H2(g)])
= [-84.6 – (+52.3 + 0)] kJ mol-1 =-136.9 kJ mol-1
The
99
standard enthalpy
ofHong
theKong
reaction
–136.9
kJ mol-1.
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Chemistry for
A-Level is
Book
1
6.6 Calculations involving standard enthalpy changes of reactions (SB p.160)
Given the following information, find the standard enthalpy
change of the reaction:
6PbO(s) + O2(g)  2Pb3O4(s)
Hf [PbO(g)] = -220.0 kJ mol-1
Hf [Pb3O4(g)] = -737.5 kJ mol-1
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New Way Chemistry for Hong Kong A-Level Book 1
Answer
6.6 Calculations involving standard enthalpy changes of reactions (SB p.160)
Back
Note:
H1 = [Hf (reactants)] = 6  Hf [PbO(s)] + Hf [O2(g)]
H2 = [Hf (products)] = 2  Hf [Pb3O4(s)]
Applying Hess’s law,
H1 + H = H2
H = H2 - H1
= 2  Hf [Pb3O4(s)] – (6  Hf [PbO(s)] + Hf [O2(g)])
= [2  (-737.5) – 6  (-222.0) – 0] kJ mol-1 =-155.0 kJ mol-1
The
101
standard enthalpy
ofHong
theKong
reaction
–155.0
kJ mol-1.
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Chemistry for
A-Level is
Book
1
6.6 Calculations involving standard enthalpy changes of reactions (SB p.160)
Given the following information, find the standard enthalpy
change of the reaction:
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
Hf [Fe2O3(s)] = -822.0 kJ mol-1
Hf [CO(g)] = -110.5 kJ mol-1
Hf [CO2(g)] = -393.5 kJ mol-1
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New Way Chemistry for Hong Kong A-Level Book 1
Answer
6.6 Calculations involving standard enthalpy changes of reactions (SB p.160)
Back
H1 = [Hf (reactants)] = Hf [Fe2O3(s)] + 2  Hf [CO(g)]
H2 = [Hf (products)] = 2  Hf [Fe(s)] + 3  Hf [CO2(g)]
Applying Hess’s law,
H1 + H = H2
H = H2 - H1
= 2  Hf [Fe(s)] + 3  Hf [CO2(g)] - Hf [Fe2O3(s)] - 3 
Hf [CO(g)]
= [2  (0) + 3  (-393.5) –(-822.0) – 3  (-110.5)] kJ mol-1
=-27.0 kJ mol-1
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New Way Chemistry for Hong Kong A-Level Book 1
The standard enthalpy change of the reaction is –27.0 kJ mol-1.
Note:
6.6 Calculations involving standard enthalpy changes of reactions (SB p.161)
Given the following information, find the standard enthalpy
change of the reaction:
4CH3 · NH · NH2(l) + 5N2O4(l)
 4CO2(g) + 12H2O(l) + 9N2(g)
Hf [CH3 · NH · NH2(l)] = +53 kJ mol-1
Hf [N2O4(l)] = -20 kJ mol-1
Hf [CO2(g)] = -393.5 kJ mol-1
Hf [H2O(l)] = -285.8 kJ mol-1
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Answer
6.6 Calculations involving standard enthalpy changes of reactions (SB p.161)
Back
Note:H1 = [Hf (reactants)] = 4  Hf [CH3·NH ·NH2(l)] + 5  Hf [N2O4(l)]
H2 = [Hf (products)] = 4  Hf [CO2(g)] + 12  Hf [H2O(l)] + 9  Hf [N2(g)]
Applying Hess’s law,
H1 + H = H2
H = H2 - H1
= (4  Hf [CO2(g)] + 12  Hf [H2O(l)] + 9  Hf [N2(g)] – (3 
Hf [CH3·NH ·NH2(l)] + 5 Hf [N2O4(l)]
= [4  (-393.5) + 12  (-285.8) + 9  (0) – 4  (+53) – 5  (-20)] kJ mol-1
105
=- 5115.6 kJ mol-1
New Way Chemistry for Hong Kong A-Level Book 1
The standard enthalpy change of the reaction is –5115.6 kJ mol-1.
6.6 Calculations involving standard enthalpy changes of reactions (SB p.162)
Given the following information, find the standard enthalpy
change of formation of methane gas.
C(graphite) + O2(g)  CO2(g)
Hc [C(graphite)] = -393.5 kJ mol-1
1
H2(g) + O2(g)  H2O(l)
Hc [H2(g)] = -285.8 kJ mol-1
2
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
Hc [CH4(g)] = -20 kJ mol-1
Answer
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New Way Chemistry for Hong Kong A-Level Book 1
6.6 Calculations involving standard enthalpy changes of reactions (SB p.162)
Direct measurement of ΔHf [CH4(g)] is impossible because carbon(graphite)
and hydrogen do not combine directly, and methane does not decompose
directly to form carbon(graphite) and hydrogen. Since methane contain
carbon and hydrogen only, they can be related to carbon dioxide and water
by the combustion of methane and its constituent elements as shown in the
diagram below.
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6.6 Calculations involving standard enthalpy changes of reactions (SB p.163)
Back
Note: H1 = Hc [C(graphite)]
H2 = 2  Hc [H2(g)]
H3 = Hc [CH4(g)]
Applying Hess’s law,
Hf [CH4(g)] + H3 = H1 + H2
Hf [CH4(g)] = H1 + H2 - H3
= Hc [C(graphite)] + 2  Hc [H2(g)] - Hc [CH4(g)]
= [-393.5 + 2  (-285.8) –(-890.4)] kJ mol-1
= -74.7 kJ mol-1
The standard enthalpy change of formation of methane gas is –74.7 kJ mol-1.
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6.6 Calculations involving standard enthalpy changes of reactions (SB p.163)
Given the following information, find the standard enthalpy
change of formation of methanol.
C(graphite) + O2(g)  CO2(g)
Hc [C(graphite)] = -393.5 kJ mol-1
1
H2(g) + O2(g)  H2O(l)
Hc [H2(g)] = -285.8 kJ mol-1
2
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
Hc [C2H5OH(l)] = -1371 kJ mol-1
Answer
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6.6 Calculations involving standard enthalpy changes of reactions (SB p.163)
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6.6 Calculations involving standard enthalpy changes of reactions (SB p.163)
Back
Note: H1 = 2  Hc [C(graphite)]
H2 = 3  Hc [H2(g)]
H3 = Hc [C2H5OH(l)]
Applying Hess’s law,
Hf [C2H5OH(l)] + H3 = H1 + H2
Hf [C2H5OH(l)] = H1 + H2 - H3
= 2  Hc [C(graphite)] + 3  Hc [H2(g)] - Hc [C2H5OH(l)]
= [2  (-393.5) + 3  (-285.8) –(-1371)] kJ mol-1
= -273.4 kJ mol-1
The standard enthalpy change of formation of ethanol is –273.4 kJ mol-1.
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6.6 Calculations involving standard enthalpy changes of reactions (SB p.164)
(a) Find the standard enthalpy change of formation of
butane gas (C4H10(g)).
Given: Hc [C(graphite)] = -393.5 kJ mol-1
Hc [H2(g)] = -285.8 kJ mol-1
Hc [C4H10(g)] = -2877 kJ mol-1
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Answer
6.6 Calculations involving standard enthalpy changes of reactions (SB p.164)
Hf [C4H10(g)]
= Hc [C(graphite)]  4 + Hc [H2(g)]  5 - Hc [C4H10(g)]
= [(-393.5)  4 + (-285.8)  5 – (-2877)] kJ mol-1
=113-126 kJ mol-1
New Way Chemistry for Hong Kong A-Level Book 1
6.6 Calculations involving standard enthalpy changes of reactions (SB p.164)
(b) Find the standard enthalpy change of the reaction:
Br2(l) + C2H4(g)  C2H4Br2(l)
Given: Hf [C2H4(g)] = +52.3 kJ mol-1
Hf [C2H4Br2(l)] = -80.7 kJ mol-1
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Answer
6.6 Calculations involving standard enthalpy changes of reactions (SB p.164)
Back
H
= [Hf (products)] - [Hf (reactants)]
= [-80.7 – (+52.3) – 0)] kJ mol-1
= -133 kJ mol-1
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6.7 Entropy change (SB p.167)
Back
Predict whether the following changes or reactions involve an
increase or a decrease in entropy.
• Dissolving salt in water to form salt solution
• Condensation of steam on a cold mirror
• Complete combustion of carbon
(a) Increase
(b) Decrease
(c) Increase
• Complete combustion of carbon monoxide
(d) Decrease
• Oxidation of sulphur dioxide to sulphur trioxide
(e) Decrease
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Answer
6.8 Free energy change (SB p.170)
Back
In the process of changing of ice to water, at what
temperature do you think G equals 0?
G equals 0 means that neither the forward nor
the reverse process is spontaneous. The system
is therefore in equilibrium. Melting point of ice is
0 oC (273 K) at which the process of changing
ice to water and the process of water turning to
ice are at equilibrium. At 0 oC, G of the
processes equals 0.
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Answer
6.8 Free energy change (SB p.170)
(a) At what temperatures is the following process
spontaneous at 1 atmosphere?
Water  Steam
(a) 100 oC
(b) What are the two driving forces that determine the
spontaneity of a process? (b) Enthalpy and entropy
Answer
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6.8 Free energy change (SB p.170)
Back
(c) State whether each of the following cases is spontaneous at
all temperatures, not spontaneous at any temperature,
spontaneous at high temperatures or spontaneous at low
temperatures.
(i) positive S and positive H
(ii) positive S and negative H
(iii) negative S and positive H
(iv) negative S and negative H
(i)
Spontaneous at high temperatures
(ii) Spontaneous at all temperatures
(iii) Not spontaneous at any temperature
(iv) Spontaneous at low temperatures
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Answer