### 15 Energetics (ahl) - slider-dpchemistry-11

```15 ENERGETICS (AHL)
YR 11 DP CHEMISTRY
ROB SLIDER
STANDARD ENTHALPY CHANGES
Let’s define some terms:
Standard State
This refers to reactants and products
being at 298K and 101.3kPa
Standard enthalpy change of formation ΔHfθ
This is the enthalpy change associated with
the formation of 1 mole of a compound from
its elements under standard conditions
Standard enthalpy change of combustion ΔHcθ
This is the enthalpy change associated with the
combustion of a compound under standard
conditions
STANDARD ENTHALPIES OF
FORMATION
Standard molar enthalpies of
formation can be looked up
in external resources. The
table on the right is one such
Chemistry Data Booklet
contains many more.
These can be used to
determine standard
enthalpies of reaction – see
next slide.
Notice that standard
enthalpies of formation for
stable elements is zero (0).
USING HESS’S LAW
We can use the standard enthalpies of formation to determine standard
enthalpies of reaction by using Hess’s Law
Reactants
ΔH
Products
ΣΔHf
(products)
ΣΔHf
(reactants)
Elements
Construct a formula for ΔH using this the above enthalpy cycle
STANDARD ENTHALPIES OF REACTION FROM
ENTHALPIES OF FORMATION
Systematic way to compare energy changes of reactions.
Can be written in terms of the standard enthalpies of formation of
products minus reactants (Hess's Law).
ΔHθrxn = Σ ni ΔHθf (products) – Σ nj ΔHθf (reactants)
Calculate from tables of reference data, such as the IB Chemistry Data Booklet.
Example: CaCO3(s) + heat –––> CaO(s) + CO2(g)
ΔHθrxn = ΔHθf (CaO, s) + ΔHθf (CO2, g) – ΔHθf (CaCO3, s)
= [(1mol)(-635)] +[(1mol)(-393.51)]- [(1mol)(-1207.6)]
= +179 kJ/mol
(note: this should be a +ve value as heat is absorbed in the rxn)
STANDARD ENTHALPIES OF REACTION FROM
ENTHALPIES OF COMBUSTION
Standard enthalpies of combustion ΔHcθ can also be used to solve
enthalpy problems using Hess’s Law.
Example:
The enthalpy of combustion for H2, C(graphite) and CH4 are -285.8, 393.5, and -890.4 kJ/mol respectively. Calculate the standard enthalpy
of formation ΔHfθ for CH4.
First, write out the equations:
1) H2(g) + 0.5 O2(g) -> H2O(l)
2) C(graphite) + O2(g) -> CO2(g)
3) CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)
From the above equations, derive C + 2H2 -> CH4
Answer: C + 2H2 -> CH4 -74.7 Hint: 2*(1) + (2) - (3),
Thus,
ΔHfθ = [2 * (-285.8)] + (-393.5) - (-890.4) = -74.7 kJ/mol
ΔHcθ /(kJ/mol)
-285.8
-393.5
-890.4
ANOTHER EXAMPLE
ΔHfθ for ethanol can be determined using the enthalpy cycle below:
2C(s)
2X ΔHfθ
(CO2)
+
2O2(g)
3H2(g)
3X ΔHfθ
(H2O)
+
1/2O2(g)
ΔHfθ (C2H5OH)
C2H5OH
1½O2(g)
2O2(g)
ΔHcθ (C2H5OH)
2CO2(g)
+
3H2O(l)
First, add the appropriate reactants to the cycle with correct coefficients
Next, add the correct enthalpy change values multiplied by the appropriate coefficients.
Finally, set up and solve the equation using Hess’s Law
ΔHfθ (C2H5OH) = [2X ΔHfθ(CO2)] + [3X ΔHfθ(H2O)] - ΔHcθ (C2H5OH)
ΔHfθ (C2H5OH) = [2X -393.5] + [3X -285.8] – (-1371) = -273.4 kJ mol-1
States of matter are very important when looking up enthalpy change
values. You must make sure you are choosing the correct value.
For example:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
ΔHcθ = -890.4 kJ/mol
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
ΔHcθ = -802.4 kJ/mol
Notice that the production of water vapour instead of liquid results in a
lower enthalpy change of combustion. This is because liquid water must
absorb energy (endothermic) to become vapour:
2H2O(l)  2H2O(g)
ΔHcθ = +88 kJ/mol
EXERCISES
1. Given the standard enthalpy change of formation values for
ammonia (g), hydrogen bromide (g) and ammonium bromide (s) are 46, -36 and -271 kJ/mol respectively, calculate ΔHrxnθ for the reaction
between ammonia and hydrogen bromide.
2. Given the standard enthalpy of formation of water vapour and
carbon dioxide are -286 and -394 kJ/mol respectively and the
standard enthalpy of combustion of methanol (l) (CH3OH) is -715
kJ/mol, construct an enthalpy cycle and calculate the ΔHfθ for
methanol.
SOLUTIONS
1. NH3(g) + HBr (g)  NH4Br(s)
ΔHθrxn = Σ ni ΔHθf (products) – Σ nj ΔHθf (reactants)
= -271 – (-46+ -36)
= -189 kJ mol-1
2.
2C(s)
2X ΔHfθ
(CO2)
2O2(g)
2CO2(g)
+
4H2(g)
4X ΔHfθ
(H2O)
+
+
2 x ΔHfθ
(CH3OH)
O2(g)
2CH3OH(l)
2O2(g)
3O2(g)
4H2O(l)
2 x ΔHcθ
(C2H5OH)
[2 x ΔHfθ (CH3OH)] + [2 x ΔHcθ (C2H5OH)] = [2X ΔHfθ (CO2)] + [4X ΔHfθ (H2O)]
[2 x ΔHfθ (CH3OH)] = - [2 x ΔHcθ (C2H5OH)] + [2X ΔHfθ (CO2)] + [4X ΔHfθ (H2O)]
2 x ΔHfθ (CH3OH) = -(2 X -715) + (2 X -394) + (4 X -286)
ΔHfθ (CH3OH) = [1430 – 788 – 1144]/2 = -502/2 = -251 kJ mol-1
LATTICE ENTHALPY
Recall that ionic compounds are
made up of repeating positive and
negative ions in a lattice.
Different ionic compounds have different
stabilities based on the strength of the
interactions between the ions in the lattice.
The relative stabilities are measured by the
lattice enthalpy values
Lattice enthalpy – the energy required to split 1mole
of a solid ionic substance into its gaseous ions
Note: Lattice enthalpies cannot be measured directly. However, we can calculate them using other values such
as ionisation energies and electron affinities
ELECTRON AFFINITY
Recall electron affinity is a
measure of how strongly an
atom attracts electrons to
itself
Electron affinity ΔHEAθ – the energy change
that occurs when 1mole of electrons is
accepted by one mole of atoms in the
gaseous state forming one mole of
negative ions.
Recall the pattern of electron affinity
values on the Periodic Table
PROCESSES INVOLVED IN FORMING AN
IONIC LATTICE
Electron Affinity ΔHEAθ – the enthalpy change when 1mole of gaseous ions is formed from one
mole of atoms (these are generally exothermic, at least for the first electron)
Ionisation Energy ΔHIEθ – the energy required to remove electrons from atoms/ions in the
gaseous state (this process is endothermic as the process requires energy)
Enthalpy of Atomisation ΔHatθ – the enthalpy change when 1mole of gaseous atoms is formed
from its elements in the standard state. For a diatomic molecule (e.g. Cl 2), this is ½ the bond
enthalpy (this process occurs prior to ionisation and is endothermic)
Enthalpy of Dissociation ΔHDθ (aka bond enthalpy) – the enthalpy change when 1mole of
gaseous atoms is formed from the dissociation of a covalent bond in the standard state. (this
process occurs prior to ionisation and is endothermic)
Lattice Enthalpy ΔHlattθ – the enthalpy change when 1mole of an ionic lattice is formed
(exothermic) or gaseous ions are formed from the lattice (endothermic)
THE GREATER THE LATTICE ENTHALPY, THE MORE STABLE THE COMPOUND
Enthalpy change of formation for NaCl
using a Born-Haber cycle
If we want to find the ΔHfθ of
sodium chloride, we can use a
Born-Haber cycle and consider
all of the steps that we would
need to get from the reactants
to the products
-349
-787
Step 1 – convert Na from a solid to a gas (atomisation)
Step 2 – convert diatomic Cl2 to the atom Cl (½ dissociation or atomisation)
Step 3 – convert Na to Na+ (ionisation)
Step 4 – convert Cl to Cl- (electron affinity)
Step 5 – formation of the NaCl lattice (lattice energy – exothermic)
Step 6 – solve for ΔHfθ = ΔHatθ + ½ΔHDθ + ΔHIθ + ΔHEAθ + ΔHlattθ
= 109 + 121 + 494 + (-349) + (-787)
= -412 kJ mol-1
Another Example: Construct a Born-Haber
cycle for the formation of MgO
Note: Mg forms Mg2+, so ΔHIθ will be the sum of the first two ionisation enthalpy changes
Mg2+(s) + O2-(g)
ΔHIθ = ΔHIθ(1)+ ΔHIθ(2)
ΔHEAθ
ΔHlattθ
Mg(g) + O(g)
ΔHatθ
½ΔHDθ
Mg(s) + ½O2(g)
ΔHfθ
MgO(s)
EXERCISE
Construct a Born-Haber cycle and find the enthalpy change for the
following reaction: Mg(s) + Cl2(g)  MgCl2(s)
Use the following data:
ΔHatθ = +148 kJ/mol
ΔHIEθ (Mg1)= +736 kJ/mol
ΔHIEθ (Mg2)= + 1451kJ/mol
ΔHDθ (Cl2)= +244 kJ/mol
ΔHEAθ (Cl)= -349 kJ/mol
ΔHlattθ = -2542 kJ/mol
Mg2+(s) + 2Cl-(g)
ΔHIEθ = ΔHIEθ(1)+ ΔHIEθ(2)
= +2187kJ/mol
2ΔHEAθ=
-698kJ/mol
ΔHlattθ=
-2542 kJ/mol
Mg(g) + 2Cl(g)
ΔHfθ =+148+244+2187-698-2542
= -661kJ mol-1
ΔHatθ=
+148kJ/mol
ΔHDθ=
+244kJ/mol
Mg(s) + Cl2(g)
ΔHf
θ
MgCl2(s)
FACTORS AFFECTING LATTICE
ENTHALPIES
Lattice enthalpy depends on two factors:
1. Ion Size
Small ions are close to each other and exhibit stronger attraction.
When charges are same, smaller ions have higher L.E.
LiCl (845 kJ/mol) versus NaCl (787 kJ/mol)
The charges are the same (+1 & -1)
Since Li+1 is smaller than Na+1, LiCl has greater L.E.
2. Ion Charge
Ion with higher charges exhibit stronger attraction.
When size is similiar, higher charges lead to higher L.E.
KCl (709 kJ/mol) versus CaCl2 (2258 kJ/mol)
K+1 and Ca+2 are similiar in size (same period next to each other).
Since Ca+2 has higher charge, CaCl2 has greater L.E.
THEORETICAL VS. EXPERIMENTAL
Lattice enthalpies can be determined experimentally as we have seen with the BornHaber cycles. We can also determine lattice enthalpies theoretically using
Coulomb’s Law:
F=k
q1q2
r2
Where:
• F is the force of attraction on one ion on another,
• q1&q2 are the charges on the ions,
• r is the distance between ions and
• k is a constant.
This equation shows mathematically what we stated in the previous slide – increasing
charge and decreasing radii increases lattice enthalpy.
See next slide for comparative data
Comparing ΔHlattθ for some ionic compounds
Compound
Lattice Enthalpy (kJ mol-1)
Experimental
Theoretical
Difference (%)
NaF
902
891
1.2
NaCl
771
766
0.6
NaBr
733
732
0.1
NaI
684
686
0.3
AgF
955
870
8.9
AgCl
905
770
14.9
AgBr
890
758
14.8
AgI
876
736
16.0
What do the values of %difference tell us about the ionic/covalent character of the
bonds in these substances?
Firstly, notice that the experimental values are always
higher than the theoretical values.
Secondly, notice that the silver halides have a much
greater % difference than sodium halides.
Since theoretical values are based on charge
interactions (ionic bonding), this implies that there is
some other bonding influencing the lattice enthalpy
values.
The experimental values that are closest to the
theoretical value fit the ionic model more closely. For
example, the ionic model for NaCl fits quite well, so
we say it has a large ionic character.
AgCl on the other hand, has a greater % difference
which is due to more covalent character, which we
would expect due to the smaller electronegativity
difference.
ENTROPY
When a tyre is punctured with a
nail, why does the air escape
spontaneously?
When perfume is sprayed on
one side of the room, why does
where you can smell it?
ENTROPY (S)
Air escapes from the tyre spontaneously as energy tends to disperse
naturally unless it is hindered from doing so (in the unpunctured tyre).
Similarly, the perfume droplets will also diffuse throughout the room to
your nose, again due to natural dispersion of the particles.
This natural dispersion of energy is known as entropy,
symbol S and is often described as a measure of the
degree of randomness or disorder in a system.
“Entropy: A measure of the amount of energy in a physical system not available to do work. As a physical system becomes
more disordered, and its energy becomes more evenly distributed, that energy becomes less able to do work. For example,
a car rolling along a road has kinetic energy that could do work (by carrying or colliding with something, for example); as
friction slows it down and its energy is distributed to its surroundings as heat, it loses this ability. The amount of entropy is often
thought of as the amount of disorder in a system.”
Source: The American Heritage® Science Dictionary Copyright © 2005 by Houghton Mifflin Company.
ENTROPY (S)
http://contrarianinconsistent.wordpress.com/2011/05/06/entropy-a-dish-best-served-hot/
http://www.cartoonstock.com/directory/e/entropy.asp
ENTROPY CHANGE (ΔS)
Because energy naturally tends to become more dispersed (i.e. more random). This would
represent a positive change in entropy over time (ΔS = +). The opposite is true for a negative
change in entropy (ΔS = -). The standard units are J K-1 or J K-1mol-1
Identify whether the examples below are positive or negative entropy changes:
Process
Ethanol evaporating
Water freezing
A salt dissolving in water
Decomposition: CuCO3(s)  CuO(s) + CO2(g)
Heating water from 150C to 450C
2Al(s) + 3S(s)  Al2S3(s)
CH4(g) + H2O(g)  3H2(g) + CO2(g)
ΔS (+/-)
ENTROPY CHANGE (ΔS)
Because energy naturally tends to become more dispersed (i.e. more random). This would
represent a positive change in entropy over time (ΔS = +). The opposite is true for a negative
change in entropy (ΔS = -). The standard units are J K-1 or J K-1mol-1
Identify whether the examples below are positive or negative entropy changes:
Process
ΔS (+/-)
Ethanol evaporating
+
Water freezing
-
A salt dissolving in water
+
Decomposition: CuCO3(s)  CuO(s) + CO2(g)
+
Heating water from 150C to 450C
+
2Al(s) + 3S(s)  Al2S3(s)
-
CH4(g) + H2O(g)  3H2(g) + CO2(g)
+
STANDARD ENTROPY CHANGE
The standard entropy change, ΔSrxnθ is calculated just as the standard
enthalpy change using standard entropy values (all positive) at 101.3kPa
and 298K.
Many of these values can be found in the IB Chemistry Data Booklet.
ΔSrxnθ = Σ ni ΔSθ (products) – Σ nj ΔSθ (reactants)
Example:
Evaluate the entropy change for the reaction: CO + 3 H2 -> CH4 + H2O in
which all reactants and products are gaseous. Entropy values are
respectively 198, 131, 186, 189 J (K mol)-1
Solution
Standard entropies of reaction, ΔSrxnθ , equals the entropy of products minus the entropy of
reactants.
The standard entropies of the reactants and products have been given above:
CO + 3 H2 -> CH4 + H2O
ΔSrxnθ = ((186 + 189) - (198 + 3*131)) J (K mol)-1
= -216 J (K mol)-1
ENTROPY AND PROBABILITY
Entropy is often equated with
probability:
If the boxes below represent gases in a closed system, which of
the two configurations do you think is more probable? Which
has more entropy?
There is a higher probability
that a system will be in disorder
rather than ordered. (2nd Law
of Thermodynamics)
The greater the probability a
state exists, the higher it’s
entropy.
So, we can conclude that a positive entropy change is more probable
ΔS = + (more favourable)
SPONTANEITY
We have now looked at two values that describe whether a reaction is
spontaneous or not. The following values predict spontaneity:
ΔHrxnθ = - (negative enthalpy)
Consider the reaction:
ΔSrxnθ = + (positive entropy)
N2(g) + 3H2(g)  2NH3(g)
ΔHrxnθ
ΔSrxnθ
-92.6 kJ mol-1
-198.5 J K-1
Predicted spontaneous
Predicted non-spontaneous
Which is correct? Is it spontaneous or not??
GIBBS FREE ENERGY
In reality, spontaneity is determined by Gibbs Free Energy (ΔG) and
depends on:
• Enthalpy (kJ mol-1)
ΔG>0 non-spontaneous
• Entropy (kJ K-1 mol-1)
ΔG<0 spontaneous
• Temperature (K)
ΔG = ΔH - T ΔS
Predict spontaneity:
ΔH
ΔS
+
+
+
-
-
+
-
-
ΔG
Spontaneous
GIBBS FREE ENERGY
In reality, spontaneity is determined by Gibbs Free Energy (ΔG) and
depends on:
• Enthalpy (kJ mol-1)
ΔG>0 non-spontaneous
• Entropy (kJ K-1 mol-1)
ΔG<0 spontaneous
• Temperature (K)
ΔG = ΔH - T ΔS
Predict spontaneity:
ΔH
ΔS
ΔG
Spontaneous
+
+
Depends on T
With high T
+
-
+
No
-
+
-
Yes
-
-
Depends on T
With low T
GIBBS FREE ENERGY EXAMPLE
Calculate ΔG for the following reaction at 25°C. Will the reaction occur (be
spontaneous)? How do you know?
NH3(g) + HCl(g) → NH4Cl(s)
Also given for this reaction:
ΔH = -176.0 kJ·mol-1
ΔS = -284.8 J·K-1·mol-1
GIBBS FREE ENERGY EXAMPLE
Solution
We will calculate ΔG using the formula
ΔG = ΔH - TΔS
but first we need to convert units for ΔS and temperature to Kelvin:
ΔS = -284.8 J·K-1 mol-1 = -0.2848 kJ·K-1mol-1
K = 273 + °C = 273 + 25 = 298 K
Now we can solve our equation:
Since ΔG < 0 the reaction will be spontaneous.
ΔG = -176.0 - (298)(-0.2848)
ΔG = -176.0 - (-84.9)
ΔG = -91.1 kJ mol-1
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