Energy = force x distance (Joules)
In chemical reactions, we need energy usually in the
form of heat.
Energy is absorbed to break the bonds of the
reactants and energy is given out when new bonds
are formed in the products.
Exothermic reactions
Endothermic reactions
Heat is the energy transferred between
objects that are at different temperatures.
 The amount of heat transferred depends on
the amount of the substance.
◦ Energy is measured in units called joules
(J).

Temperature is a measure of “hotness” of a
substance and represent the average kinetic
energy of the particles in a substance.
 It does not depend on the amount of the
substance.

Do both beakers contain the same amount of heat?



All chemical reactions are accompanied by some
form of energy change
Exothermic
Energy is given out
Endothermic
Energy is absorbed

Activity : observing exothermic and endothermic reactions


Enthalpy (H) is the heat content that is stored in a
chemical system.
We measure the change in enthalpy ∆H i.e. the amount
of heat released or absorbed when a chemical reaction
occurs at constant pressure, measured in
kilojoules per mole (kJmol-1).
∆H = H(products) – H(reactants)

For exothermic reactions, the reactants have more
energy than the products, and the enthalpy change,
∆H = H(products) - H(reactants)
Enthalpy

∆H is negative since H(products) < H(reactants)

There is an enthalpy decrease and heat is released to the
surroundings


Self-heating cans
◦ CaO (s) + H₂O (l)  Ca(OH)₂ (aq)
Combustion reactions
◦ CH₄ (g) + 2O₂ (g)  CO₂ (g) + 2H₂O (l)


neutralization (acid + base)
◦ NaOH(aq) + HCl(aq)  NaCl(aq) + H₂O(l)
Respiration
◦ C₆H₁₂O₆ (aq) + 6O₂ (g)  6CO₂ (g) + 6H₂O (l)

For endothermic reactions, the reactants have less
energy than the products, and the enthalpy change,
∆H = H(products) - H(reactants)
Enthalpy

∆H is positive since H(products) < H(reactants)

There is an enthalpy increase and heat is absorbed from
the surroundings



Self-cooling beer can
◦ H ₂O (l)  H₂O (g)
Thermal decomposition
 CaCO₃ (s)  CaO (s) + CO ₂ (g)
Photosynthesis
 6CO₂ (g) + 6H₂O (l)  C₆H₁₂O₆ (aq) + 6O₂ (g)


Amount of heat required to raise the temperature of
a unit mass of a substance by 1 degree or 1 kelvin.
Uint : Jg-1 0C-1
The specific heat capacity of alminium is 0.90 Jg-1 0C-1 .
If 0.90J of energy is put into 1g of aluminium, the temperature
will be raised by 10C.
Calculating heat absorbed and released
q = c × m × ΔT
q = heat absorbed or released
c = specific heat capacity of substance
m = mass of substance in grams
ΔT = change in temperature in Celsius
Heat given off by a process is measured
through the temperture change in another
substance (usually water).
 Due to the law of conservation of energy, any
energy given off in a process must be
absorbed by something else, we assume that
the energy given out will be absorbed by the
water and cause a temperature change.
 calculate the heat through the equation
Q = mcΔT

How much heat is required to increase the
temperature of 20 grams of nickel (specific heat
capacity 440Jkg-1 0C-1) from 500C to 700C?
The standard enthalpy change of combustion for a
substance is the heat released when 1 mole of a pure
substance is completely burnt in excess oxygen under
standard conditions.
 Example,
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ΔHƟc=-698 kJmol-1

The heat given out is used to heat another substance,e.g.
water with a known specific heat capacity.
 The experiment set-up can be used to
determine the enthalpy change when 1 mole of

a liquid is burnt.
Example : refer to page 185



Loss of heat to the surroundings (exothermic
reaction); absorption of heat from the surroundings
(endothermic reaction). This can be reduced by
insulating the calorimeter.
Using incorrect specific heat capacity in the calculation
of heat change. If copper can is used, the s.h.c. of
copper must be accounted for.
Others include – e.g incomplete combustion. Some of
the ethanol could be used to produce CO & soot &
water (less heat is given out)
Use bomb calorimeter – heavily insulated & substance is ignited
electronically with good supply of oxygen

If 1g of methanol is burned to heat 100g of water,
raising its temperature by 42K, calculate the
enthalpy change when 1 mole of methanol is burnt.
Note: Specific heat capacity of water is 4.18 Jg-1 0C-1
Practice questions page 187 #1-4
Enthalpy change of neutralisation (ΔHn)
The standard enthalpy change of neutraisation is the enthalpy change that
takes place when 1 mole of H+ is completely neutralised by an alkali under
standard conditions.
Example,
NaOH(g) + HCl(g)  NaCl(g) + H2O(l)
ΔHƟ=-57 kJmol-1



The enthalpy change of neutralisation of a strong acid and a strong alkali is
almost the same as they undergo complete ionisationof ions in water.
Reaction between strong acid and strong base involves
H+(aq) + OH-(aq)  H2O(l)
ΔHƟ=-57 kJmol-1
For sulfuric acid, the enthalpy of neutralisation equation is
½ H2SO4(aq) + KOH(aq)  ½K2SO4(aq) + H2O(l)
ΔHƟ=-57 kJmol-1
Example : refer to page 188
Enthalpy change of neutralisation (ΔHn)


The standard enthalpy change of neutraisation is the enthalpy
change that takes place when 1 mole of H+ is completely
neutralised by an alkali under standard conditions.
Example,
NaOH(g) + HCl(g)  NaCl(g) + H2O(l) ΔHƟ=-57 kJmol-1
The enthalpy change of neutralisation of a strong acid and a
strong alkali is almost the same as they undergo complete
ionisationof ions in water.
Enthalpy change of solution (ΔHsol)
-
The enthalpy change when 1 mol of solute is dissolved in excess solvent to
form a solution of ‘infnite dilution’ under standard conditions.
NH4 NO3(s) in excess water  NH4 + (aq)+ NO 3 -(aq)
Example : refer to page 188
For neutralisation between a weak acid, a weak
base or both, the enthalpy of neutraisation
will be smaller than -57 kJmol-1 (less exothermic)
CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l)

ΔHƟ=-55.2 kJmol-1
Some of the energy released is used to ionise
the acid.

200.0cm3 of 0.150 M HCl is mixed with 100.0cm3 of 0.350 M
NaOH. The temperature rose by 1.360C. If both solutions
were originally at the same temp, calculate the enthalpy
change of neutralisation.
Assume that the density of the solution is 1 gcm-3 and the
specific heat capacity is 4.18J Jg-1 0C-1.
-56.8kJmol-1
The experimental change of neutralisation is
-56.8 kJmol-1
The accepted literature value is -57.2 kJmol-1
(1) Heat loss to the environment.
(2) Assumptions that
(a) the denisty of NaOH and HCl solutions are the same as
water.
(b) the specific heat capacity of the mixture are the same as that
of water
When 3 g of sodium carbonate are added to 50 cm3 of 1.0 M HCl,
the temperature rises from 22.0 °C to 28.5°C. Calculate the
enthalpy change for the reaction. Assume that the density of the
solution is 1 gcm-3 and the specific heat capacity is
4.18J Jg-1 0C-1.
Example : refer to page 189 dissolving
ammonium chloride
The experimental change of solution is
+13.8 kJmol-1
The accepted literature value is 15.2 kJmol-1
(1) Absorption of heat from the environment.
(2) Assumptions that the specific heat capacity of the solution is
the same as that of water
(3) The mass of ammonium chloride is not taken into
consideration when working out the heat energy released.
Example
100.0 cm3 of 0.100 mol dm-3 copper II sulphate solution is placed in a
styrofoam cup. 1.30 g of powdered zinc is added and a single replacement
reaction occurs. The temperature of the solution over time is shown in the
graph below. Determine the enthalpy value for this reaction.
First step
Make sure you understand the
graph.
Extrapolate to determine the
change in temperature.
The extrapolation is necessary to compensate for heat loss while the reaction
is occurring. Why would powdered zinc be used?
100.0 cm3 of 0.100 mol dm-3 copper II sulphate solution is placed in a
styrofoam cup. 1.30 g of powdered zinc is added and a single replacement
reaction occurs. The temperature of the solution over time is shown in the
graph below. Determine the enthalpy value for this reaction.
Determine the limiting reactant
Calculate Q
Calculate the enthalpy for the reaction.
The following measurements are taken:



Mass of cold water (g)
Temperature rise of the water (0C)
The loss of mass of the fuel (g)
We know that it takes 4.18J of energy to raise the
temperature of 1g of water by 10C. This is called the
specific heat capacity of water, c, and has a value of
4.18Jg-1K-1.
Hence, energy transferred can be calculated using: Energy
transfer = mcΔT (joules)
 If one mole of the fuel has a mass of M grams, then:
 Enthalpy transfer = m x 4.18 x T x M/y
 where y is mass loss of fuel.
Given that:
Vol of water = 100 cm3
Temp rise = 34.50C
Mass of methanol burned = 0.75g
Specific heat capacity of water = 4.18 Jg-10C-1
Calculate the molar enthalpy change of the combustion of
methanol.
What is the big assumption made with this type
of experiment?
States that
 If a reaction consists of a number of steps, the
overall enthalpy change is equal to the sum of
enthalpy of individual steps.
 the overall enthalpy change in a reaction is
constant, not dependent on the pathway take.

measured under standard conditions: pressure of 1
atmosphere (1.013 x 105 Pa), temperature of 250C
(298K) and concentration of 1 moldm-1.
e.g.
N2(g) + 3H2(g)  2NH3(g)
ΔHƟ = -92 kJmol-1
The enthalpy change of reaction is -92 kJmol-1
92 kJ of heat energy are given out when 1 mol of nitrogen
reacts with 3 mols of hydrogen to form 2 mols of ammonia.
Calculate the enthalpy change for the formation of
sodium chloride solution from solid sodium hydroxide.
Indirect path
+ H2O(l)
NaOH(s)
NaOH(aq)
ΔH2
ΔH1
Direct path
+ HCl(aq)
+ HCl(aq)
NaCl(s) + H2O(l)
1. Indirect path: NaOH(s) + (aq)  NaOH(aq)
ΔHƟ1=-43kJmol-1
2.
NaOH(aq) + HCl (aq)  NaCl(aq) + H2O(l) ΔHƟ2=-57kJmol-1
3.
NaOH(s) + HCl (aq)  NaCl(aq) + H2O(l).
Calculate the enthalpy change for the combustion of
carbon monoxide to form carbon dioxide.
C(s) + O2(g)  CO2(g)
ΔHƟ =-394 kJmol-1
2C(s) + O2(g)  2CO(g)
ΔHƟ = -222kJmol-1
2CO(s) + O2(g)  2CO2(g)
2CO(g) + O2(g)
ΔHƟ
2CO2(g)
ΔHƟ = -(-222)+2(-394) = -566kJmol-1
Example : refer to page 196 evaporation
of water & 197 formation of ethanol
from ethene
Calculate the enthalpy change for the thermal
decomposition of calcium carbonate.
CaCO3(s)  CaO(s) + CO2(g)
CaCO3(s) +2HCl(aq)  CaCl2(aq) + H2O(l)
CaO(s) +2HCl(aq)  CaCl2(aq) + H2O(l)
CaCO3(s)
-17
kJmol-1
ΔH
Direct path
+ 2HCl(aq)
ΔHƟ1=-17 kJmol-1
ΔHƟ1=-195kJmol-1
CaO(s) +CO2(g)
+ 2HCl(aq)
-195kJmol-1
CaCl2(aq) + H2O(l) +CO2(g)
Indirect path
Calculate the enthalpy of hydration of anhydrous
copper(II)sulfate change.
CuSO4(s) +5H2O(l)  CuSO4.5H2O (s)
CuSO4(s) +5H2O(l)
ΔH
Direct pathway
ΔH1
CuSO4.5H2O (s)
ΔH2
Cu2+(aq) + SO42- (aq)
Indirect pathway
From the following data at 250C and 1 atmosphere
pressure:
Eqn 1: 2CO2(g) 2CO(g) + O2(g) ΔHƟ=566 kJmol-1
Eqn 2: 3CO(g) + O3(g) 3CO2(g) ΔHƟ=-992 kJmol-1
Calculate the enthalpy change calculated for the
conversion of oxygen to 1 mole of ozone,i.e. for the
reaction 3 O2(g) O3 (g)
2
Calculate the enthalpy change for the conversion of
graphite to diamond under standard thermodynamic
conditions.
C (s,graphite) + O2(g) CO2 (g)
ΔHƟ=-393 kJmol-1
C (s, diamond) + O2(g) CO2(g) ΔHƟ=-395 kJmol-1
Practice questions page 199 #7-9



Enthalpy changes can also be calculated
directly from bond enthalpies.
The bond enthalpy is the amount of energy
required to break one mole of a specified
covalent bond in the gaseous state.
For diatomic molecule the bond enthalpy is
defined as the enthalpy change for the
process X-Y(g)
X(g) + Y(g) [gaseous state]
Bond enthalpy can only be calculated for
substances in the gaseous state.
atomisation
Br2(l)  2Br(g)
ΔHƟ= 224 kJmol-1

Br2(l)
2 x ΔH Ɵat
ΔH Ɵvap
2Br(g)
Br-Br
bond enthalpy
enthalpy change
of vaporisation
Br2(g)
Energy must be supplied to break the van der Waals’ forces between the Bromine
molecules and to break the Br-Br bonds. Endothermic process

Ave bond enthalpies are enthalpies calculated
from a range of compounds,eg C-H bond
enthalpy is based on the ave bond energies in
CH4 , alkanes and other hydrocarbons.
Bond
Ave bond enthalpy,
ΔHƟ (Kjmol-1)
Bond length
(nm)
H-H
436
0.07
C-C
348
0.15
C-H
412
0.11
O-H
463
0.10
N-H
388
0.10
N-N
163
0.15
C=C
612
0.13
O=O
496
0.12
C ΞC
837
0.12
NΞN
944
Refer to page 201
When a hydrocarbon e.g. methane (CH4) burns,
CH4 + O2  CO2 + H2O
What happens?
Enthalpy Level (KJ)
Bond Breaking
O
ENERGY
H
+
C
H
H
O
O
O
O
H
H
O
H
O
ENERGY
H
Bond Forming
H
4 C-H
O
C
2 O=O
H
O
O
C
O
H
4 H-O
CH4 + 2O2  CO2 + 2H2O
Progress of Reaction
O
H
H
2 C=O
CH4 + 2O2  CO2 + 2H2O
H
+
C
H
H
H
O
O
O
O
H
O
O
C
H
O
H
H
Why is this an exothermic reaction (produces heat)?
O
Break
Form

CH4 + 2O2  CO2 + 2H2O
H
H
C
H
O O
+
H O O
O H
H
H
O C O
O
H
Energy absorbed when bonds are broken
= (4 x C-H + 2 x O=O)
= 4 x 412 + 2 x 496
= 2640 kJ/mol
Energy given out when bonds are formed
= ( 2 x C=O + 4 x H-O)
= 2 x 803 + 4 x 464
= 3338 kJ/mol
Bond
Ave Bond
Enthalpy (kJ/mol)
C-H
412
H-O
463
O=O
496
C=O
743
Energy absorbed when bonds are broken (a)
= 2640 kJ/mol
Energy released when bonds are formed (b)
= 3338 kJ/mol
Enthalpy change, ΔH
= ∑(bonds broken) - ∑(bonds made)
= a + (-b)
= 2640 – 3338
= -698 kJ/mol
Why is this an exothermic reaction (produces heat)?
What can be said about the hydrogenation
reaction of ethene?
H
H
C=C
(g) + H-H (g) 
H
H
H H
H-C-C-H (g)
H H
What can be said about the combustion of
hydrazine in oxygen?
H
H
N-N
H
(g) + O=O (g) 
H
NΞN (g) + 2 O
H
(g)
H
Example
Calculate the mean Cl-F bond enthalpy given that
Cl2(g) + 3F2(g)  2ClF3(g) ΔHƟ= -164 kJmol-1
Bond enthalpy for Cl-Cl = 242 kJmol-1 and F-F = 158 kJmol-1
Standard enthalpy change of atomisation (ΔH Ɵat ) is the
enthalpy change when 1 mole of gaseous atoms is
formed from the element under standard conditions.
Example
C(s)  C(g)
ΔH Ɵat = 715 kJmol-1
Calculate the enthalpy change for the process
3 C(s) + 4H2(g)  C3H8(g) ΔHƟ= -164 kJmol-1
Bond enthalpy for C-H = 412 kJmol-1 , H-H = 436 kJmol-1 and C-C = 348 kJmol-1
Practice questions page 206 #10,12,13
The combustion of both C and CO to form CO2 can
be measured easily but the combustion of C to CO
cannot. This can be represented by the energy
cycle.
C(s)+ ½O2(g)
-393kJmol-1
½O2(g)
CO2(g)
ΔHx = -393 – (-283)
= - 110 kJmol-1
CO(g)
ΔHx
½O2(g)
-283kJmol-1
Calculate the standard enthalpy change when one
mole of methane is formed from its elements in
their standard states. The standard enthalpies of
combustion of carbon, hydrogen and methane are
-393, -286 and -890 kJmol-1 respectively.
Dissolving sugar .
Sugar molecules are dispersed
throughout the solution and
are moving around.
More disordered or random.
Other examples:
• melting ice
H2O(s)  H2O(l)
•
evaporating water
H2O(l)  H2O(g)
Increasing entropy
Entropy (S) : amount of disorder
Unit : JK-1mol-1
SƟ : standard entropy
Δ SƟ : entropy change
If Δ SƟ > 0 => increase in entropy => increase in disorder
E.g. H2O(l )  H2O(g) Δ SƟ =+119JK-1mol-1
If Δ SƟ < 0 => decrease in entropy => decrease in disorder
E.g. NH3(g ) + HCl(g)  NH4Cl(s) Δ SƟ = - 285JK-1mol-1
What are the factors that affect
ENTROPY?
(1) State of matter
◦ Gas particle motion is more random in a gas
◦ Liquid particle motion is less random than in a
liquid than a gas but more than a solid
◦ Solid particle motion is restricted.
◦ Examples
 (changing state) H2O(l)  H2O(g)
 (changing state) H2O(s)  H2O(l)
ΔS(gas) > ΔS(liquid) > ΔS(solid)
(2) Temperature
◦ Comparing two gasses, one at 20 C and one at
80 C
◦ Molecules in the 80 C gas have more kinetic
energy, they are moving more and colliding more
(3) The number of molecules
◦ More molecules means more possible positions
relative to the other molecules
 (more moles and change of state)
Li2CO3(s)  Li2O(s) + CO2(g)
 (more moles)
MgSO48H2O  Mg2+(aq) + SO42-(aq) + 8H2O(l)
(4) More complex molecules have higher
entropy values
Is there an increase or decrease in disorder of the system?
Is there an increase or decrease in the no. of moles of gas?
Reaction
Entropy
(increase/
decrease)
ΔSƟ
(+/-)
N2(g) + 3H2(g)  2NH3(g)
decrease
-
4 moles of gas to 2
moles of gas
increase
+
1 mole of solid to 1 mole
of solid + 1 mole of gas
CH4(g) + 2O2(g)  CO2(g) +2H2O(l) decrease
-
3 moles of gas to 1 mole
of gas
C2H4(g) + H2(g)  C2H6(g)
-
2 moles of gas to 1 mole
of gas
CaCO3(s)  CaO(s) + CO2(g)
decrease
Explanation
For reaction where the no. of moles of gas is the same on both sides,
the ΔS = 0.
E.g.
F2(g) + Cl2(g)  2ClF(g)
Practice Qn 24from pg 230 (textbook
Which of the following reactions has the largest
ΔS value?




CO2(g) + 3H2(g)  CH3OH(g) + H2O(g)
2Al(s) + 3S(s)  Al2S3(s)
CH4(g) + H2O(l)  3H2(g) + CO(g)
2S(s) + 3O2(g)  2SO3(g)
Entropy change =
total entropy of products – total entropy of reactants
ΔSƟ =∑ ΔSƟproducts - ∑ ΔSƟreactants
E.g. Calculate the standard entropy change for the
reaction CH4 (g) + 2O2(g)  CO2(g) + 2H2O(l)
Use standard entropy from pg 230 (textbook)
Some reactions are spontaneous because
they give off energy in the form of heat
(H < 0).
 Others are spontaneous because they lead to
an increase in the disorder of the system
(S > 0).
 Calculations of H and S can be used to probe
the driving force behind a particular reaction.

Spontaneous reaction – one that occurs
without any outside influence (no input of
energy)
 A spontaneous reaction does not have to
happen quickly.
 E.g.
4Na(s) + O2(g)  2Na2O(s) can happen by itself.

Calculate H and S for the following reaction and
decide in which direction each of these factors
will drive the reaction.
N 2 ( g) + 3 H2( g )
2 NH3(g)
Using a standard-state enthalpy of formation and
absolute entropy data table, we find the following
information:
Compound
Hfo(kJ/mol)
S°(J/mol-K)
N 2 ( g)
0
191.61
H2(g)
0
130.68
NH3(g)
-46.11
192.45
What happens when one of the potential
driving forces behind a chemical reaction is
favorable and the other is not? We can answer
this question by defining a new quantity
known as the Gibbs free energy (G) of the
system, which reflects the balance between
these forces.
ΔG= ΔH – TΔS
ΔG : free energy change
ΔGƟ : standard free energy change
Temperature, T should be in Kelvin, K
0°C = 273K (273.15K)
 Check the units of ΔS, entropy is often given
in JK–1mol–1 but must be converted to
kJK–1mol–1
ΔGө = ΔHө – TΔSө
 ө = standard conditions, 25°C (298K) and 1
atm (101.3 kPa)




measures the balance between the two
driving forces that determine whether a
reaction is spontaneous.
the enthalpy and entropy terms have different
sign conventions.
Favourable
Unfavourable
ΔHƟ < 0
ΔHƟ > 0
ΔSƟ > 0
ΔSƟ < 0
When heat is released in a chemical reaction,
the surrounding is hotter and particles move
around more => entropy increases.


Because of the way the free energy of the
system is defined, Go is negative for any
reaction for which Ho is negative and So is
positive.
For a Favorable, or spontaneous reactions:
Go < 0


When a reaction is favored by both enthalpy
(Ho < 0) and entropy (So > 0), there is no
need to calculate the value of Go to decide
whether the reaction should proceed.
Similarly, for reactions favored by neither
enthalpy (Ho > 0) nor entropy (So < 0). Free
energy calculations become important for
reactions favored by only one of these
factors.
Given that the changes in enthalpy and entropy are
-139 kJ and 277 J/K respectively for the reaction
given below, calculate the change in Gibbs energy.
Then state whether the reaction is spontaneous at
25 C
C6H12O6(aq)  2C2H5OH(aq) + 2CO2(g)
Calculate H and S for the following reaction:
NH4NO3(s) + H2O(l)  NH4+ (aq) + NO3- (aq)
Use the results of this calculation to determine the
value of Go for this reaction at 25o C, and explain
why NH4NO3 spontaneously dissolves is water at
room temperature.
Using a standard-state enthalpy of formation and
absolute entropy data table, we find the following
information:
Compound
Hfo(kJ/mol)
S°(J/mol-K)
NH4NO3(s)
-365.56
151.08
NH4+ (aq)
-132.51
113.4
NO3- (aq)
-205.0
146.4


The balance between the contributions from
the enthalpy and entropy terms to the free
energy of a reaction depends on the
temperature at which the reaction is run.
Predict whether the following reaction is
spontaneous at 250C:
N2(g) + 3 H2(g)
2 NH3(g)




The equation suggests that the entropy term will
become more important as the temperature
increases.
Go = Ho - TSo
Since the entropy term is unfavorable, the
reaction should become less favorable as the
temperature increases.
Predict whether the following reaction is still
spontaneous at 500C:
N 2( g ) + 3 H 2( g )
2 NH3(g)
Assume that the values of Ho and S used in the
previous example are still valid at this
temperature.

What does the value of Go tell us about the following reaction?
N2(g) + 3 H2(g)



2 NH3(g) Go = -32.96 kJ
The value of Go for a reaction
◦ measures the difference between the free energies of the reactants and
products when all components of the reaction are present at standardstate conditions.
◦ describes this reaction only when all three components are present at 1
atm pressure.
The fact that Go is negative for this reaction at 25oC means that a
system under standard-state conditions at this temperature
would have to shift to the right, converting some of the reactants
into products, before it can reach equilibrium.
The larger the value of Go, the further the reaction has to go to
get to from the standard-state conditions to equilibrium.
ΔGƟ =∑ ΔGfƟproducts - ∑ ΔGfƟreactants
standard free energy of formation : free energy change for the formation of 1 mole
of substance from its elements in their standard states & under standard
conditions.
Calculate ΔGƟ for the reaction
CaCO3(s)  CaO(s) + CO2(g) given that
Substance
ΔGfƟ /kJmol -1
CaCO3
- 1129
CaO
- 604
CO2
- 395
Practice Qn from pg 235 (textbook
ΔH : Enthalpy Change
ΔS : Entropy Change
For a spontaneous reaction ΔG is negative (–)
For a non–spontaneous reaction ΔG is positive (+)