Download N2(g)

Q5:
eqn.1
eqn.2
eqn.3
NO (g) ∆H\ = + 90.25 kJmol-1
½ N2(g) + ½ O2(g)
equation of interest:
N2(g) +
Ν2Ο5(g)
2.5 O2(g)
adding (2 x eqn. 1) + eqn. 2 eliminates NO2 to give:
4ΝΟ + 3Ο2 + 4ΝΟ2
4ΝΟ + 3Ο2
4ΝΟ2 + 2Ν2Ο5
2Ν2Ο5
∆H\ = -338.4 kJmol-1
∆H\ = -338.4 kJmol-1
introduce nitrogen, oxygen on left hand side by adding (4 x eqn. 3):
2 N2(g) +
2 O2(g)
4NO (g)
4NO + 3O2 + 2 N2(g) + 2 O2(g)
∆H\ = + 361.0 kJmol-1
2N2O5 + 4NO (g)
∆H\ = + 22.6 kJmol-1
Q5:
eqn.1
eqn.2
eqn.3
NO (g) ∆H\ = + 90.25 kJmol-1
½ N2(g) + ½ O2(g)
equation of interest:
N2(g) +
5O2 + 2 N2(g)
Ν2Ο5(g)
2.5 O2(g)
2N2O5
∆H\ = + 22.6 kJmol-1
∆Hf\ is always expressed per mole of substance so:
2.5O2 + N2(g)
∆Hf\ (Ν2Ο5) = +11.3 kJmol-1
N2O5
∆H\ = + 11.3 kJmol-1
(4 marks)
1
Q1: Calculate the free energy of formation of hydrogen bromide, ∆Gf\(HBr), from
the data:
eqn1
eqn2
eqn3
eqn4
Br2(l)
HBr(g)
Br2(g)
H2(g)
eqn is:
½H2(g) + ½Br2(l)
Br2(g)
H•(g) + Br• (g)
2Br• (g)
2H• (g)
∆G\ = +3.11 kJmol-1
∆G\ = +339.09 kJmol-1
∆G\ = +164.79 kJmol-1
∆G\ = +406.49 kJmol-1
(8 marks)
HBr(g)
∆G\ = ?
Need to eliminate radicals - Divide eqns 3 & 4 by 2 and add together:
eqn3
eqn4
½Br2(g)
½ H2(g)
½Br2(g) + ½ H2(g)
Br• (g)
H• (g)
∆G\ = +82.40 kJmol-1
∆G\ = +203.25 kJmol-1
Br• (g) + H• (g) ∆G\ = +285.65 kJmol-1
Now, reverse equation 2 and add to eliminate radicals and put HBr on RHS…...
Q1: Calculate the free energy of formation of hydrogen bromide, ∆Gf\(HBr), from
the data:
eqn1
eqn2
eqn3
eqn4
Br2(l)
HBr(g)
Br2(g)
H2(g)
eqn is:
½H2(g) + ½Br2(l)
Br2(g)
H•(g) + Br• (g)
2Br• (g)
2H• (g)
∆G\ = +3.11 kJmol-1
∆G\ = +339.09 kJmol-1
∆G\ = +164.79 kJmol-1
∆G\ = +406.49 kJmol-1
(8 marks)
½Br2(g) + ½ H2(g)
HBr(g)
∆G\ = ?
Br• (g) + H• (g) ∆G\ = +285.65 kJmol-1
reverse equation 2 and add to eliminate radicals:
eqn2
H•(g) + Br• (g)
HBr(g)
½Br2(g) + ½ H2(g)
HBr(g)
add half eqn1 to adjust Br to standard liquid state:
½Br2(l)
½Br2(g)
½H2(g) + ½Br2(l)
HBr(g)
∆G\ = -339.09 kJmol-1
∆G\ = -53.44 kJmol-1
∆G\ = +1.56 kJmol-1
∆G\ = -51.9 kJmol-1
2
THERMODYNAMICS
(ch. 6 & 18)
– the study of transformations of energy – especially as heat and work
Used to describe mechanical systems e.g. steam engine
all chemical processes (e.g. combustion, dissolving of a solid, expansion of a
gas) also involve exchange of heat or work.
Terms you’ll use in thermodynamics:
Energy – the capacity to heat or do work
Heat – transfer of energy due to temperature differences
Work – the result of a force acting over a distance work = force x distance
Entropy – a measure of disorder (see Ch. 18.3)
can be transferred in two ways: as heat or work
Energy
cannot be created or destroyed – only converted from one
form to another (Law of conservation of energy)
The System and its Surroundings (p.225)
Universe
We talk in terms of
system and surroundings:
Surroundings assumed to be infinitely large source (or sink) of heat
Chemists usually interested in the system (the reactants and products)
System + surroundings collectively called the universe
3
Can describe systems as:
OPEN (to surroundings)
e.g. open flask, living cells
-volume changes and exchanges
of matter possible
i.e. work can be done
on or by the system
-can exchange
heat with surroundings
CLOSED (no work or matter exchange possible, can exchange heat with
surroundings) e.g. stoppered flask
ISOLATED (no work, matter or heat exchange possible) e.g. a
stoppered, thermally and electrically insulated flask
-ve
+ve
Reactions can either:
negative
positive
i) do work e.g. decomposition of calcium carbonate:
CaCO3 (s)
CO2 (g) + CaO (s)
Large volume of gas produced which must
push back against atmospheric pressure
System loses
internal
energy (U) by
doing work
∆U = -ve
4
(p.229)
System loses internal energy
(∆U = -ve) by doing work
w = -ve (done BY the system)
w = -Pex∆V
- when a gas is compressed e.g.
in bicycle pump w = +ve
(work is done ON the system to
increase its internal energy)
∆U = +ve
Or, reactions can:
ii) exchange heat with surroundings e.g. acid-base
neutralisation:
NaOH +
HCl
NaCl
+
H2O
No work is done (negligible volume change)
-but formation of O-H bonds exothermic
-system loses internal energy (∆U = -ve) by losing heat (q = -ve)
5
THE FIRST LAW
Internal energy (U) of system–
a measure of the sum of
kinetic and potential energies
of constituent molecules
U is a state function (p.219) –
value of U independent of the
route by which it was reached
– like altitude on a mountain
n.b. work and heat are not
state functions
internal energy only changes (∆U 0) if heat or work is done on or
by the system
e.g. reaction absorbs 10 KJ heat then ∆U = +10 kJ
reaction does 20 KJ work then ∆U = -20 kJ
∆U = q + w
first law of
thermodynamics
-heat and work are equivalent ways of changing
a system’s internal energy
-i.e. the internal energy of an isolated system is
constant
6
Q: An athlete in gym does 400 kJ work but loses 70 kJ as
heat. What was change in her internal energy?
A:
q = -70 kJ
w = -400 kJ
∆U = q + w = -70 – 400 = -470 kJ
Q: 160 MJ heat were supplied to a balloon its volume increased
from 4.0 x 106 L to 4.7 x 106 L. What was the internal energy
change of the balloon?
A: Use ∆U = q + w and substitute w = - Pex∆V
So:
∆U = q – Pex∆V
Check all units are SI – convert L to m3:
∆V = 0.7 x 106 L = 700 m3
P = 1.0133 x 105 Pa (1 atm)
q = +160 MJ = + 1.6 x 108 J
Pra
c
Qs 6 tise Ex
6
.16
& 6 .1
.18
∆U = + 1.6 x 108 – (1.0133 x 105 x 700) = + 8.9 x 107 J
(internal energy raised by 89 MJ – remaining 71 MJ used to push
back atmosphere)
7
Measuring ∆U by Calorimetry (p. 239)
U is difficult to measure directly but ∆U can be measured.
Since:
∆U = q - Pex∆V
by fixing the volume (making ∆V = 0 i.e. studying reaction in a closed
systems), ∆U = q
simply measure heat change q – use bomb calorimeter
The bomb
calorimeter
(p. 241)
constantvolume
calorimeter
used to
measure heat
changes of
chemical
reactions e.g.
combustion:
8
calorimetry – the science of heat measurement
surround water supplies/absorbs heat of reaction. Temperature
change of water gives q by:
q = mcv∆T
where:
m = mass water
∆T = temperature change of water
cv = heat capacity of water at constant volume – the amount of heat
needed to change water temperature = 4.2 Jg-1K-1
water has high heat capacity compared to other substances
e.g. ethanol cv = 2.4 Jg-1K-1, benzene cv = 1.0 Jg-1K-1
copper cv = 0.4 Jg-1K-1
Q: A 50 g ball of copper at 95 ºC was dropped into a perfectly
lagged beaker of water (100 ml) at 20 ºC. What was the temperature
rise of the water, given that cCu = 0.385 Jg-1K-1 and cH20 = 4.18 Jg1K-1?
A:
q = (mcv∆T)Cu = (mcv∆T)H20
(mcv)Cu(95-T) = (mcv)H20(T-20)
(mcv)Cu(95-T) = (T-20)
(mcv)H20
0.04065(95-T) = (T-20)
rearrange:
T = 23.3 ºC
pra
ctis
Qs e
6.36
,6
& 6 .38
.100
(water absorbs a large amount
of heat in small temperature rise)
9
Enthalpy (p. 232)
changeable volume
Chemistry – most reactions are open systems ⇒ constant pressure
not all energy supplied/released changes U - some exchanged as
work
CaCO3 (s)
CO2 (g)
W = -ve
q = +ve
+ CaO (s)
(endothermic – heat must be supplied). But not all heat raises U of
system – some heat converted to work of pushing back atmosphere
(w = -ve)
e.g. in combustion of phenol, U decreases – heat is evolved but
volume expands (by 3 moles more gas) so work must also be done
(W = -ve)
W = -ve
2C6H5OH(s) + 15O2(g)
12CO2(g) + 6H2O(g) q = -ve
Chemists deal with heat change at constant pressure known as
change of enthalpy (∆H)
At constant pressure:
q = ∆H
- reactions that give out heat are exothermic
e.g. combustions
∆H = -ve
(loss of enthalpy)
sulphur combustion
- reactions that absorb heat from surroundings
are endothermic
∆H = +ve
(gain in enthalpy)
NH4SCN + Ba(OH)2
10
THERMOCHEMISTRY
- the study of
heat exchange in chemical reactions
enthalpy (H) is a state property
so:
enthalpy of a substance is independent of the route
by which it was synthesised – Hess’ Law (p. 246)
This means ∆H is the same for reactants → products regardless of
whether this happens in a single big step or in a series of
smaller steps.
Hess’ Law very convenient for chemists – we can combine
reaction enthalpies of known reactions to calculate ∆H for
unknown (or unfeasible, impossible) reactions e.g.
N2 + 2O2
2NO2
∆H = ?
(very high temperatures required so impractical to measure in lab)
11
N2 + 2O2
∆H = ?
2NO2
but we know ∆H for each step:
step one:
step two:
N2 + O2
2NO + O2
∆H = +180 kJmol-1
∆H = -112 kJmol-1
2NO
2NO2
adding equations gives desired reaction (cancel 2NO from each side)
N2 + 2NO + 2O2
2NO + 2NO2
likewise, add enthalpy changes:
∆H = +180 + (-112) = +68 kJmol-1 (endothermic)
we can also say that the:
‘heat of combustion of nitrogen’ is +68 kJmol-1
or ‘heat of formation of NO2’ is +34 kJmol-1
For calculations using Hess’ Law remember:
− Reversed (backwards) reactions have the same magnitude
∆H but opposite sign
e.g.
H2 + ½O2
H2O
H2O
H2 + ½O2
∆H = -286 kJmol-1
∆H = +286 kJmol-1
− Enthalpy is an extensive property: heats are cumulative; ∆H
for burning 2 moles is simply double the molar heat of
combustion
H2 + ½O2
2H2 + O2
H2O
2H2O
∆H = -286 kJmol-1
∆H = -572 kJmol-1
12
Enthalpy of Combustion (∆Hc\) – heat change when
1 mole of a substance reacts with oxygen whereby all reactants and
products are in their standard states
\
denotes standard conditions, 1 atm, 298K
- larger molecules generally have larger molar enthalpy of
combustion
- sometimes quantities enthalpy of combustion per mass or per
volume given –
-e.g. high ‘enthalpy density’ required for cars – where a
small tank is required
-values different – e.g. ∆Hc\ for hydrogen is small but very
high per unit mass
for rocket fuels ‘specific enthalpy’ of fuel (energy per mass) much
more important
H2 + ½O2
H2O
∆H = -286 kJmol-1
So ∆Hc\ (H2) = -286 kJmol-1
So we can also say:
∆Hf\(H2O) = -286 kJmol-1
13
Applying Hess’ Law (p. 248-50)
Q: What is the enthalpy change for C(graphite)
given that:
∆Hc(graphite) = -394 kJmol-1
∆Hc(diamond) = -396 kJmol-1
A:
Eqn. 1
Eqn. 2
C(diamond)
CO2 ∆H = -394 kJmol-1
CO2 ∆H = -396 kJmol-1
C(gr) + O2
C(diam) + O2
Need C(diam) on product side of eqn.s so re-write to give:
Eqn. 3
C(diam) + O2
CO2
Add eqns 1 and 3 gives C(graphite)
∆H = +396 kJmol-1
C(diamond)
∆H = -394 + 396 = + 2 kJmol-1
(process weakly endothermic)
pra
ct
Ex. ise
6.9
Applying Hess’ Law (p. 248-50)
Q:
_
eqn. 1
eqn. 2
eqn. 3
A: need HCl on RHS of an equation so re-write eqn 1 in reverse:
eqn 1’
NH4Cl
NH3
+ HCl
∆H = + 176.0 kJmol-1
NH4Cl should now cancel out NH4Cl on RHS of eqn. 3.
Also need to cancel out NH3 so eqn. 2 should be written backwards also:
eqn. 2’
2NH3
N2
+
3H2
∆H = + 92.2 kJmol-1
14
eqn. 1’ NH4Cl
eqn. 2’ 2NH3
eqn. 3
NH3 + HCl
N2 +
3H2
∆H = + 176.0 KJmol-1
∆H = + 92.2 KJmol-1
Now we can cancel NH3 by doubling eqn. 1’ and adding eqn. 2’ to give:
2NH4Cl
2NH3
2NH4Cl
2NH3 + 2HCl
N2 + 3H2
N2 + 3H2 + 2HCl
∆H = (+176 x 2) + 92.2 = +444.2 kJmol-1
now add eqn. 3 to eliminate NH4Cl and cancel:
N2
2NH4Cl
+ 4H2 + Cl2
N2 + 3H2 +
2NH4Cl
H2 + Cl2
2HCl
2HCl
∆H = +444.2 – 628.9
= -184.7 kJmol-1 (exothermic)
– makes sense as hydrogen burns (or explodes) in chlorine gas
pr
Qs 6 actise
.62,
& 6 6.64
.80
Calculating ∆H of Unknown or Implausable Reactions
We can sum ∆Hf\ values of participating compounds/products to work out
∆H for any reaction… (Hess’ Law again!)
Q: what is ∆H for the hydrogenation of acetylene (impossible to perform at
room temperature, pressure )
C2H2 + 2H2
C2H6
∆Hc\ (C2H2) = -1300 kJmol-1
∆Hc\ (C2H6) = -1560 kJmol-1
∆Hf\ (H2O) = -286 kJmol-1
A: first, let’s write out three corresponding balanced equations for these
reactions:
given that:
eqn. 1
eqn. 2
eqn. 3
15
n.b. since combustion equations require two moles hydrocarbon ∆H must be
doubled
-first, ethane must be on RHS so re-write eqn. 2 in reverse:
4CO2 + 6H2O
2C2H6
+
7O2
∆H = +3120 kJmol-1
adding eqn. 1 to eqn. 2’ will eliminate CO2 and give:
2C2H2 + 4H2O
2C2H6 + 2O2
∆H = +3120 –2600
= +520 kJmol-1
-next, introduce hydrogen on LHS and cancel 4H2O, just add 4 x eqn 3…..
eqn. 1
eqn. 2
eqn. 3
to give….
2C2H2 +
4H2
2C2H6
∆H = +520 + (4 x –286)
= -624 kJmol-1
divide through by 2 - to simplify to target equation:
C2H2 + 2H2
C2H6
∆H = -624/2
= -312 kJmol-1
The best way to understand how to apply Hess’ Law
is practise! (Examples in Chang and similar textbooks!)
pra
ct
Q 6 ise
.80
&6
.112
eqn. 1
eqn. 2
eqn. 3
16
T H E R M O D Y N A M I C S - 2nd LAW
(Ch. 18)
– the study of transformations of energy – especially as heat and
work
Terms you’ll use in thermodynamics:
Energy – the capacity to heat or do work
Heat – transfer of energy due to temperature differences
Work – the result of a force acting over a distance
Entropy – a measure of disorder
SPONTANEITY and ENTROPY (ch .18)
First law of thermodynamics accounts for energy as it is transferred
in chemical reactions:
But:
It gives us no clue as to why reactions have tendency to occur in a
given direction – i.e. the spontaneity of processes
Spontaneous reaction – occurs without outside influence
But: spontaneity tells us nothing about rate of reaction e.g.
i) gradual dissolving of limestone statue
ii) explosive decomposition of nitroglycerine
are both spontaneous processes – occur at very different rate!
17
Everyday examples of spontaneity (p. 784):
•ball rolling down a hill
•wood burning
•iron rusting
•gas spreading out evenly
(e.g. smoke filling a room)
•heat spreading from hot objects to cooler ones
(try to imagine these processes happening in
reverse – absurd!)…..but why?
-Need to consider extra property – ENTROPY – a measure
of disorder
The driving force behind a spontaneous process is an
increase in entropy of the universe
Second Law of Thermodynamics
Don’t forget - universe includes system and surroundings –
sometimes either system or surroundings can become more
ordered but, overall, entropy increases (∆Suniv = +ve)
e.g. unevenly distributed gas always spreads out to an even
density (p.785):
But
why?
18
Microstates (p. 787)
e.g. consider 4 gas molecules
in same apparatus (p. 787):
-only one way to arrange 4:0
arrangement (1/11 chance)
-four ways to arrange 3:1
arrangement (4/11 chance)
-six ways to arrange 2:2
arrangement (6/11 chance)
2:2 arrangement has highest
positional probability of
occurring
……..nature always tends towards state that has highest probability of
occurring
19
S = klnW
Where k = Boltzmann constant
= R/NA = 1.38 x 10-23 J/K
W = spatial disorder, number of
ways the particles can be arranged
e.g. in tetrahedral chloroform molecule
CHCl3, the molecule can be oriented
four ways
So there are W = 4NA ways per mole
of arranging the molecules
so S = kln(4NA) = (R/NA)ln(4NA) = R ln4 = +11.5 JK-1
(measured value = +10.1 suggests almost random orientation)
The Third Law (p. 787)
S\ values are positive - all substances have some disorder because
i) molecules are free to move e.g. in fluids.
Recall: Ssolid < Sliquid << Sgas
ii) spatial irregularities e.g. a gap in crystal
lattice, poorly aligned molecules
iii) vibrations of molecules – increases with
temperature
Only a perfect crystal at absolute
zero (0K) has zero entropy
(Third Law of Thermodynamics)
Real solids have irregularities (S > 0 even at 0K)
20
positional entropy changes (S) in different phases:
many more positional possibilities for particles in gas than
solid, liquid
Ssolid < Sliquid
<<
Sgas
Liquid resembles solid but some positional freedom causes
slightly increased entropy
Standard Entropies (p. 789)
For most chemical reactions we
consider ∆H or ∆G
(absolute values of H and G are
impossible to measure –
arbitrarily fixed as zero for
elements)
But we can determine absolute
(standard) entropy values (S\)
– entropy of substance under
standard conditions
S\ jumps instantly upwards at
melting and boiling points
21
Comparing Absolute Entropies
for all substances:
absolute entropy S\ always
increases with temperature
Note:
-high value for sucrose due to large
number of bonds per mole
-low value for diamond– highly
positional ordered and rigidity
means minimal thermal disorder
Entropy Changes in Reactions
Can estimate sign of ∆S in chemical reactions that involve change of
state because
S\solid < S\liquid
<<
S\gas
e.g.
CaCO3(s)
CO2(g) + CaO(s)
\
expect ∆S = + ve (gas produced from a solid)
2SO2(g) + O2(g)
2SO3(g)
\
expect ∆S = – ve (three moles gas become two - smaller volume)
Mg(s) + 2HCl(aq)
MgCl2(aq) + H2(g)
expect ∆S\ = + ve (gas produced from a solution)
pra
Ex 1 ctise
8
Qs 1 .3 &
8.14
22
Entropy Changes in Reactions
Q 18.57 (Chang)
A:
(a)
(b)
(c)
(d)
∆S\ > 0
∆S\ < 0
∆S\ > 0
∆S\ > 0
pra
Ex 1 ctise
8.
Q18 3 &
.60
recall: we met entropy when discussing solutions:
ordered solid dissolving in solvent ⇒ big increase in
positional entropy (many more possibilities for mixed
condition than separate solvent and solute): ∆Ssys = +ve
– entropy driving force often outweighs endothermicity
of solution e.g. dissolution of NaNO3
23
Quantifying the Second Law (p. 790)
A process is spontaneous if there is an accompanying increase in the
entropy of the universe (∆Suniv = +ve)
∆Suniv = ∆Ssys + ∆Ssurr
e.g. tree growth is spontaneous – living cells highly ordered so ∆Ssys
= -ve. Large energy input (sunlight) required so ∆Ssurr is larger and
+ve so overall:
∆Suniv = +ve i.e. the tree grows!
Temperature and Entropy (p. 793-4)
- consider the boiling of water:
H2O(l)
H2O(g)
∆Suniv = ∆Ssys + ∆Ssurr
∆Ssys = +ve due to increased positional entropy of vapour - favours
spontaneity
24
Temperature and Entropy (p. 793-4)
∆Ssurr depends on heat flow in and out of system. Vapourisation is
endothermic (∆Hvap = +ve) – loss of heat from surround causes it to
become less random, more ordered i.e. ∆Ssurr = -ve
which parameter controls ∆Suniv?
The overall sign of ∆Suniv depends on relative magnitude of ∆Ssys and
∆Ssurr i.e. depends on temperature
-entropy change of system similar at all temperatures
-but entropy change of surround depends on how ordered it is already
i.e.
∆Ssurr = – ∆Hsys
T
-eqn. tells us: why exothermicity is driving force for spontaneity
-eqn. tells us: exothermicity is stronger driving force at lower
temperatures (small T causes large, positive ∆Ssurr)
25
∆Ssurr = – ∆Hsys,rev
T
-e.g. at low temperatures injection of 1
kJ of heat into surround causes greater
entropy increase to surround than
would 1 kJ of heat at higher
temperature
-analogy: stone thrown into pond
causes larger ripple than stone into sea
Since:
∆S\univ = ∆S\sys + ∆S\surr
∆S\univ = ∆S\sys – ∆H\sys/T
(spontaneity defined in
terms of system alone - useful)
so for boiling water:
high T makes ∆S\sys controlling so ∆S\univ = +ve (spontaneously boils)
low T makes ∆S\surr controlling so ∆S\univ = -ve (boiling nonspontaneous)
\
\
\
intermediate temperature ∆S sys = –∆H sys/T so ∆S univ = 0 (boiling
and condensation equally favoured)
T = 373K – the boiling point of water
In words: ‘Above 373K boiling is spontaneous because then the entropy
loss of surroundings becomes smaller than the entropy gain of the
molecules’
26
Free Energy (G) (p. 796)
Have seen that:
∆S\univ = ∆S\sys + ∆S\surr
∆S\univ = ∆S\sys – ∆H\sys/T (defined in
terms of system alone)
To express two entropy changes in terms of just energy of system,
we introduce property called Gibbs’ Free Energy (G):
G is defined so that G = H – TS:
Mult by –T:
-T∆S\univ = -T∆S\sys + ∆H\sys
Substitute H – TS = G
so:
∆G\ = ∆H\ – T∆S\
Gibbs-Helmholtz Equation
To check how Gibbs-Helmholtz equation relates to spontaneity,
divide by -T:
−∆G\ = ∆S\ – ∆H\
T
T
looks familiar?
∆S\ – ∆H\ = ∆S\univ
T
so:
∆G\ = –T∆S\univ
(constant temperature and pressure)
∆G\ is a direct expression of spontaneity,
of opposite sign to ∆Suniv
i.e. a process is spontaneous in the direction
in which free energy decreases (∆G\ = -ve)
J.W. Gibbs
27
Applying Gibbs-Helmholtz : Phase Changes
- consider the melting of ice (∆H\ = 6030 Jmol-1, ∆S\ = +22.1 JK1mol-1) at three different temperatures: -10°C, 0°C, +10°C
two opposing factors:
i) entropy increase of system
and
ii) entropy decrease of surroundings (due to endothermicity)
∆H\ = +6030 Jmol-1
at -10°C
∆H is bigger than T∆S so ∆G = +ve
(melting not spontaneous)
at +10°C
T∆S is bigger than ∆H so ∆G = -ve
(spontaneously melts)
at 0°C
T∆S is equal to ∆H so ∆G = 0
(system in equilibrium)
n.b. ∆G values agree with ∆Suniv (opposite signs)
28
Applying Gibbs-Helmholtz : Phase Changes
Often ∆S and ∆H counteract i.e. they have same sign – temperature
decides driving force:
Cl2(l)
Q: For the melting of solid chlorine Cl2(s)
(∆H\ = +6.41 KJmol-1, ∆S\ = +37.3 JK-1mol-1).
What’s chlorine’s freezing point?
A: Use Gibbs-Helmholtz: ∆G\ = ∆H\ – T∆S\
Melting point is an equilibrium so: ∆G\ = 0 so ∆H\ = T∆S\
T = ∆H\ /∆S\ = 6410/37.3 = 171.8
Remember:
K
(-101.3 °C)
- use units consistent (best to stick to J throughout)
- temp in Kelvin!
Applying Gibbs-Helmholtz : Phase Changes
Q: Predict the boiling point of liquid sodium, given that the
entropy of Na(l) changes by 84.8 J/Kmol when it vapourises at 1
atm and the enthalpy of vapourisation is +98 KJ/mol.
A: first, let’s list what we know:
∆H\ = +98 kJmol-1, ∆S\ = +84.8 JK-1mol-1
Next, use Gibbs-Helmholtz: ∆G\ = ∆H\ – T∆S\
Boiling point is an equilibrium so: ∆G\ = 0
so ∆H\ = T∆S\
∆H\
/∆S\
T =
= 98 x 103 / 84.8
= 1156 K (883 °C)
pra
ctise
Q
18.6
0
29
try
at hom
e
Q:18.59 (from Chang)
+
A: 625 K (352 °C)
Q: Estimate the enthalpy of vapourisation of benzene, given that it
the liquid hydrocarbon boils at 80 °C and its entropy changes by 85
J/Kmol during vapourisation.
A: + 30 kJ/mol
entropy is a state function so we can calculate ∆S\ values for
reactions by subtracting S\ values of reactants from S\ values of
products
e.g. for:
2H2 + O2
2H2O(l)
entropy values are:
S\(H2) = 130.7 JK-1mol-1
S\(O2) = 205 JK-1mol-1
S\(H2O, l) = 70 JK-1mol-1
So ∆S\ = (2 x 70) – 205 – (2 x 130.7) = -327 JK-1
highly entropically unfavourable – makes sense:
– 3 moles gas becomes 2 moles liquid
S is a state property
(route to final value
is not important)
30
-Yet we know H2 combustion to be spontaneous so must be increase
in entropy of surroundings. Let’s check in two ways:
i) by comparing ∆S\sys (= -327 JK-1) with ∆S\surr
- ∆H\ = 2 x –286 kJmol-1
-so ∆S\surr = - ∆H\/T = +1920 JK-1 (easily outweighs entropy loss
of molecules)
ii) by using Gibbs-Helmholtz: ∆G\ = ∆H\ – T∆S\
= - (572 x 103) – (298 x -327)
∆G\ = - 474 kJmol-1 (confirms very spont.)
Finally lets confirm our two ‘checks’ agree with each other –
remember:
∆G\ = –T∆S\univ = –298 x (1920 – 327) = -474 x 103 Jmol-1
Applying Gibbs-Helmholtz : Chemical Reactions (p. 800-802)
Q: Given the data:
S\(MgCO3) = 66 JK-1mol-1 ∆Hf\ (MgCO3) = -1096 kJmol-1
S\(MgO) = 27 JK-1mol-1
∆Hf\ (MgO) = -602 kJmol-1
S\(CO2) = 214 JK-1mol-1
∆Hf\ (CO2) = -394 kJmol-1
Calculate ∆G\,, ∆H\ and ∆S\ for the endothermic decomposition
MgCO3
heat
MgO + CO2
Is MgCO3 stable at room temperature?
What’s the lowest temperature MgCO3 decomposes?
31
Q: Given the data:
S\(MgCO3) = 66 JK-1mol-1 ∆Hf\ (MgCO3) = -1096 kJmol-1
S\(MgO) = 27 JK-1mol-1
∆Hf\ (MgO) = -602 kJmol-1
\
-1
-1
S (CO2) = 214 JK mol
∆Hf\ (CO2) = -394 kJmol-1
Calculate ∆G\,, ∆H\ and ∆S\ for the endothermic decomposition
MgCO3
heat
MgO + CO2
Is MgCO3 stable at room temperature?
What’s the lowest temperature MgCO3 decomposes?
A: ∆S\ is found by subtracting entropy of reactant from entropy of
products:
∆S\ = (27 + 214) – 66
= +175 JK-1mol-1
makes sense? Yes large S\ of CO2 makes reaction entropically
favoured.
Q: Given the data:
S\(MgCO3) = 66 JK-1mol-1 ∆Hf\ (MgCO3) = -1096 kJmol-1
S\(MgO) = 27 JK-1mol-1
∆Hf\ (MgO) = -602 kJmol-1
S\(CO2) = 214 JK-1mol-1
∆Hf\ (CO2) = -394 kJmol-1
Calculate ∆G\,, ∆H\ and ∆S\ for the endothermic decomposition
MgCO3
heat
MgO + CO2
Is MgCO3 stable at room temperature?
What’s the lowest temperature MgCO3 decomposes?
∆H\ can be found by similarly summing heats of formation:
∆H\ = (-394 – 602) – (-1096) = +100 kJmol-1
makes sense? Yes - endothermicity agrees well with needing to
heat reaction
32
Q: Given the data:
S\(MgCO3) = 66 JK-1mol-1 ∆Hf\ (MgCO3) = -1096 kJmol-1
S\(MgO) = 27 JK-1mol-1
∆Hf\ (MgO) = -602 kJmol-1
\
-1
-1
S (CO2) = 214 JK mol
∆Hf\ (CO2) = -394 kJmol-1
Calculate ∆G\,, ∆H\ and ∆S\ for the endothermic decomposition
MgCO3
heat
MgO + CO2
Is MgCO3 stable at room temperature?
What’s the lowest temperature MgCO3 decomposes?
Find ∆G\ by using Gibbs-Helmholtz: ∆G\ = ∆H\ – T∆S\
= (100 x 103) – (298 x 175) = + 48 kJmol-1
makes sense? Yes: decomposition non-spontaneous at room temperature
– MgCO3 stable
- agrees well with needing to heat reaction and MgCO3 as component of
many rocks.
Q: Given the data:
S\(MgCO3) = 66 JK-1mol-1 ∆Hf\ (MgCO3) = -1096 kJmol-1
S\(MgO) = 27 JK-1mol-1
∆Hf\ (MgO) = -602 kJmol-1
S\(CO2) = 214 JK-1mol-1
∆Hf\ (CO2) = -394 kJmol-1
Calculate ∆G\,, ∆H\ and ∆S\ for the endothermic decomposition
MgCO3
heat
MgO + CO2
Is MgCO3 stable at room temperature?
What’s the lowest temperature MgCO3 decomposes?
∆H\ = T∆S\ at equilibrium so
T = ∆H\ /∆S\ = 100 x 103/175 = 570 K
So above 570K (297 °C) T∆S\ outweighs endothermicity and
MgCO3 decomposes spontaneously.
33
Applying Gibbs-Helmholtz : Chemical Reactions (p. 800-802)
Q18.52 (Chang)
A: Need to know ∆S° and ∆H° for the reaction:
∆H° = (1)(−110.5 kJ/mol) + (1)(0)] − [(1)(−241.8 kJ/mol) + (1)(0)]
∆H° = +131.3 kJ/mol (endothermic)
∆S° = S°(CO) + S°(H2) − [S°(H2O) + S°(C)]
∆S° = [(1)(197.9 J/K⋅mol) + (1)(131.0 J/K⋅mol)] − [(1)(188.7
J/K⋅mol) + (1)(5.69 J/K⋅mol)]
∆S° = +134.5 J/K⋅mol (favourable)
Setting ∆G° = 0
0 = ∆H° − T∆S°
T =
∆H °
=
∆S °
1000 J
1 kJ
= 976 K = 703°C
134.5 J/K ⋅ mol
131.3 kJ/mol ×
⇒ the temperature must be greater than 703°C for the reaction to be
spontaneous.
Free Energies of Formation & Thermodynamic Stability
As a short-cut to calculating reaction spontaneity we simply sum
free energies of formation (∆Gf\) of reagents and products.
Free energies of formation are measured with respect to chemical
elements (taken to have G = 0)
Like ∆Hf\, ∆Gf\ is the change in free energy that occurs as a
compound forms from elements in standard states
e.g. ethane
2C(gr) + 3H2 (g)
C2H6(g)
∆G\ = - –33 kJmol-1
∆Gf\(C2H6) = is –33 kJmol-1. Value is -ve so ethane is
thermodynamically stable with respect to its elements
34
6C(gr) + 3H2 (g)
C6H6(l)
∆Gf\ = +124 kJmol-1
benzene is thermodynamically unstable with respect to its
elements
i.e. it would be impossible to make benzene directly from carbon
and hydrogen
-doesn’t mean benzene will decompose – huge kinetic barrier to
decomposition means benzene won’t decompose
n.b. spontaneity tells us nothing about rate of process
activation energies (kinetics) tells us about reaction rate
-when ∆Gf\ is negative elements have tendency to form compounds
i.e compound is thermodynamically stable
-when ∆Gf\ is positive compound has tendency to decompose into
elements
i.e compound is thermodynamically unstable
∆Gf\ (and not ∆Hf\) is a true measure of compound’s
thermodynamic stability
35
Q: Work out the free energy of formation (∆Gf\) of hydrogen iodide
in the equation:
H2 + I2
2HI
from the data:
S\(HI) = 206.6 JK-1mol-1
S\(H2) = 130.7 JK-1mol-1
\
-1
-1
S (I2) = 116.1 JK mol
∆Hf\ (HI) = +26.5 KJmol-1
Α: ∆Sf\ = S\prod – S\react
∆Sf\ = (2 x 206.6) – (130.7 + 116.1) = +166.4 JK-1mol-1
For ∆Gf\ use Gibbs-Helmholtz:
∆Gf\ = ∆Hf\ – T∆Sf\
= (2 x 26.5 x 103) – (298 x 166.4) = +3.4 KJmol-1
but eqn features two moles HI so ∆Gf\ = +1.7 KJmol-1
HI does not form spontaneously – heating will
be required to allow T∆Sf\ to dominate
Pra
18.1 ctise Q
s
8&
18.7
2
Q: Do the following compounds become more or less stable with
respect to their elements as the temperature is raised? Use the GibbsHelmholtz equation.
a) CS2
b) Hg2Cl2
c) NO
Α: a) CS2
calculate ∆Sf\ to see how T∆Sf\ affects ∆Gf\.
C(s) + 2S(s)
CS2(l)
∆Sf\ = S\prod - S\react
∆Sf\ = 151 - [5.7 + (2 x 32.5)] = +80.3 JK-1mol-1
Using Gibbs-Helmholtz:
∆Gf\ = ∆Hf\ - T∆Sf\
∆Sf\ = positive means T∆Sf\ (and hence ∆Gf\) becomes more
negative with increasing T
⇒ stability of CS2 increases with T
36
Q: Do the following compounds become more or less stable with
respect to their elements as the temperature is raised? Use the GibbsHelmholtz equation.
a) CS2
b) Hg2Cl2
c) NO
Α: b) Hg2Cl2
calculate ∆Sf\ to see how T∆Sf\ affects ∆Gf\.
2Hg(l) + Cl2(g)
Hg2Cl2(s)
∆Sf\ = S\prod - S\react
∆Sf\ = 196.2 - [223 + (77.4 x 2)] = -182 JK-1mol-1
Using Gibbs-Helmholtz:
∆Gf\ = ∆Hf\ - T∆Sf\
∆Sf\ = negative means T∆Sf\ (and hence ∆Gf\) becomes more
positive with increasing T
⇒ stability of Hg2Cl2 decreases with T (rapidly)
Q: Do the following compounds become more or less stable with
respect to their elements as the temperature is raised? Use the GibbsHelmholtz equation.
a) CS2
b) Hg2Cl2
c) NO
Α: c) NO
calculate ∆Sf\ to see how T∆Sf\ affects ∆Gf\.
0.5 N2(g) + 0.5 O2(g)
NO(g)
∆Sf\ = S\prod - S\react
∆Sf\ = 210.6 - 0.5[191.5 + 205] = +12.4 JK-1mol-1
Using Gibbs-Helmholtz:
∆Gf\ = ∆Hf\ - T∆Sf\
∆Sf\ = positive means T∆Sf\ (and hence ∆Gf\) becomes more
negative with increasing T
⇒ stability of Hg2Cl2 increases with T (gradually)
37
∆G at arbitary concentrations (Ch. 18.6)
So far we’ve considered reaction free energies only as ∆G\ i.e.
reactions with stoichiometric (exact reacting ratios) of reactants are
present – tendency of pure reactants to form products expressed
as ∆G\
∆G is different – its the reaction free energy at a non-standard
composition
∆G
-changes throughout the course of the reaction as
the composition of the reaction mixture changes.
-expresses the spontaneity of the forward reaction
at a given moment in the reaction
∆G is related to ∆G\ by:
∆G = ∆G\ + RTlnQ
where Q is called the reaction quotient and expresses the
composition of the reaction mixture:
For a gas-phase reaction: aA + bB
Q=
cC + dD
PCc PDd
PAa PBb
n.b. Q can be calculated using partial pressures (P) or molar
concentrations (a)
38
Q: Calculate ∆G at 298K for the reaction
CO(g) + 2H2(g)
CH3OH(l)
in which CO at 5 atm pressure and hydrogen (3 atm) are converted to liquid
methanol, given that:
∆Gf\ (MeOH) = -166 kJmol-1
∆Gf\ (CO) = -137 kJmol-1
A: First we need to calculate ∆G\ = -166 - (-137) = -29 kJmol-1
Now calculate Q:
Q = 1/(PCO2 x PH22)
= 1/(5 x 32) = 0.0222
Solids and liquid concentrations appear as ‘unity’ i.e. = 1 in quotient
calculations
∆G = ∆G\ + RTlnQ
= -(29 x 103) + (8.314 x 298)ln 0.0222 = -38 kJmol-1
⇒ reaction more favourable at these high pressures than at 1 atm
∆G and Equilibrium (p. 804)
∆G\ is a measure of the initial spontaneity of pure reagents to form
products – i.e. when no product is present
∆G changes throughout the course of the reaction (as Q changes)
since:
∆G = ∆G\ + RTlnQ
….until equilibrium is reached (reaction mixture with the lowest free
energy)
39
If the reaction starts with pure B, the backwards reaction (∆G\ = +ve) also
proceeds to the same equilibrium mixture.
- the value of reaction quotient Q has a special name when ∆G = 0:
the equilibrium constant (K)
since: ∆G = ∆G\ + RTlnQ ……
…..and at equilibrium ∆G = 0 and Q = K we can write:
∆G\ = -RTlnK
e.g. if the reaction
13
87
2A
3B
goes 87% to completion by equilibrium, then:
K =
[B]3
[A]2
= 872/132
= 3895 (unitless)
so:
∆G\ = -RTlnK
= -8.314 x 298 x ln(3895)
= -2 x 104 Jmol-1 (-20 kJmol-1)
Read Ex. 18.6
& try
Q 18.28
40
∆Gf\ (NO2) = +51.3 kJmol-1
∆Gf\ (N2O4) = +97.9 kJmol-1
What is the room temperature eq. constant for the reaction
Q: Given the data:
N2O4 (g)
2NO2 (g) ?
Calculate the equilibrium constant to see if the reaction nears
completion.
A: First calculate ∆G\ = (2 x 51.3) – 97.9 = +4.8 kJmol-1
(non-spontaneous)
Find K from ∆G\ = -RTlnK
so:
lnK = −∆G\/RT = -4800/(8.314 x 298) = -1.91
K = e-1.91 = 0.148
p
Qs 1 ractis
8.30 e
&18
.62
K < 1 so equilibrium lies towards
reactants rather than products
Q:18.28 (Chang)
A: a) Using ∆G° = −RTln Kp
try
at hom
e
∆G° = −(8.314 J/mol⋅K)(2000 K) ln (4.40)
= −2.46 × 104 J/mol = −24.6 kJ/mol
+
b) Under non-standard-state conditions, ∆G is related to the reaction
quotient Q by the following equation: ∆G = ∆G° + RTln Qp
We are using Qp in the equation because this is a gas-phase reaction.
Step 1: Calculate Qp:
Step 2:
∆G
∆G
∆G
∆G
=
=
=
=
Qp =
PH 2O ⋅ PCO
PH 2 ⋅ PCO2
=
(0.66)(1.20)
= 4.1
(0.25)(0.78)
∆G° + RTln Qp
−2.46 × 104 J/mol + (8.314 J/mol⋅K)(2000 K) ln (4.1)
(−2.46 × 104 J/mol) + (2.35 × 104 J/mol)
−1.10 × 103 J/mol = −1.10 kJ/mol
41
42