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Chapter 19Spontaneous Change: Entropy and Free Energy
Dr. Peter Warburtonpeterw@mun.cahttp://www.chem.mun.ca/zcourses/1051.php
2
Spontaneous processes
We have a general idea of what we consider spontaneous to mean:
A spontaneous process WILL OCCUR in a system WITHOUT any outside action being performed on the system.
3
Spontaneous processes
Ice melting above zero Celcius is spontatneous
Object falling to earth is
spontaneous
4
Spontaneous processes
Ice will melt above zero Celcius.
We don’t have to DO anything!
Objects will fall to earth.
We don’t have to DO anything!
5
Spontaneous processes
Since we DON’T have to DO anything for these spontaneous processes to occur it APPEARS that an overall energy change
from potential energy
to kinetic energy
IS SPONTANEOUS
6
Non-spontaneous processes
We have a general idea of what we consider non-spontaneous to mean:
A non-spontaneous process WILL NOT OCCUR in a system UNTIL an outside action is performed on the system.
7
Non-spontaneous processes
Ice freezing above zero Celcius is
non-spontatneous
Object rising from earth is
non-spontaneous
8
Non-spontaneous processes
We can make water freeze above zero Celcius by
increasing the pressure.
We can make an object rise from the earth by picking it
up.
9
Non-spontaneous processes
Since we DO have to ACT for these non-spontaneous processes
to occur it APPEARS that an overall energy change from
kinetic energy
to potential energy
IS NON-SPONTANEOUS
10
Chemistry and spontaneity
We know there are chemical processes that are spontaneous because we can put the chemical system together and reactants become products without us having to do anything.
H3O+ (aq) + OH- (aq) 2 H2O (l)
11
Chemistry and non-spontaneity
We know there are chemical processes that are non-spontaneous because we can put the chemical system together and reactants DO NOT become products.
The system we put together stays like it isUNTIL WE CHANGE SOMETHING!
2 H2O (l) 2 H2 (g) + O2 (g)
12
Spontaneous vs. non-spontaneous
It is obvious by the examples we’ve looked at that the opposite of every spontaneous process is a non-spontaneous process.
In chemical systems we’ve seen that if we put a chemical system together a reaction occurs until the system reaches equilibrium.
Whether the forward reaction or the reverse reaction dominates depends on which of the two reactions is spontaneous at those conditions!
13
Equilibrium and spontaneity are
related!
14
Spontaneity and energy
In our examples it APPEARED that the
spontaneous process ALWAYS takes a system to a
lower potential energy.
15
Spontaneity and energy
If this were true all exothermic processes would be spontaneous and all endothermic processes would be non-spontaneous.
THIS ISN’T TRUE! H2O
NH4NO3 (s) NH4+ (aq) + NO3
- (aq)is spontaneous even though
H = +25.7 kJ
16
Recall the First Law
The First Law of thermodynamics stated that the energy of an ISOLATED system is constant.What’s the largest ISOLATED system we can think of?
It’s the UNIVERSE!The energy of the universe is
constant!
17
Recall the First Law
On the universal scale, there is no overall change in energy, and so lower energy CANNOT be the only requirement for spontaneity.
There must be something else as well!
18
Further proof lower energy isn’t enough
If an ideal gas expands into a vacuum at a constant
temperature, then
no work is done
and
no heat is transferred
19
Further proof lower energy isn’t enough
No work done and no heat transferred means
NO OVERALL CHANGE in
energy
of the system
20
Further proof lower energy isn’t enough
This spontaneous process has
no overall change in energy!
21
Entropy
Entropy (from Greek, meaning “in transformation”)
is a thermodynamic property that relates
the distribution of the total energy of the system
to the available energy levels of the particles.
22
Entropy
A general way to envision entropy is
“differing ways to move”Consider mountain climbers on a mountain.
Two factors affect the distribution of mountain climbers on a mountain:
Total energy of all the climbers and how many places can you stop
on the mountain
23
Consider hungry mountain climbers
Hungry mountain climbers have
little total energy amongst themselves to
climb a mountain, so most of them are near the
bottom, while some are distributed on the lower parts of the mountain. Few energy levels
can be reached!
24
Consider well fed mountain climbers
Well-fed mountain climbers have
more total energy amongst themselves to
climb a mountain, so the climbers will be
more spread out on the whole mountain.
More energy levels can be reached!
25
Temperature and total energy
The total energy shared by molecules is related to the temperature.
A given number of molecules at a low temperature (less total energy) have less “differing ways to move” than the same number of molecules at a
high temperature (more total energy).
26
Higher mountain means more places to stop
If a well-fed mountain climber tries to climb Signal Hill, they will most likely reach the
top. They have only a few places to stop (levels) because Signal Hill is a small
mountain.The same well-fed mountain climber on
Mount Everest has a greater number of places to stop (levels) because it is a
larger mountain.
27
Volume and energy levels
The number of levels of energy distribution of molecules is related to
the volume.A given number of molecules in a
small volume have less “differing ways to move” than the same
number of molecules in a larger volume.
28
Entropy
The greater the number of “differing ways to move” molecules can take
amongst the available energy levels of a system
of a given state (defined by temperature and volume, and number of molecules),
the greater the entropy of the system.
29
Expansion into vacuum
A gas expands into a vacuum because the
increased volume allows for a greater
number of “differing ways to move” for
the molecules, even if the temperature is
the same.
30
Expansion into vacuum
That is, the entropy increases when the gas is allowed to expand into
a vacuum.
Entropy increase plays a role in spontaneity!
31
Entropy is a state function
Entropy, S, is a state function like enthalpy or internal energy.
The entropy of a system DEPENDS ONLY on the current state
(n, T, V, etc.) of the system, and NOT how the system GOT TO BE
in that state.
32
Boltzmann equation and entropy
More available energy levels when the size of a box increases – like expanding a gas into a vacuum
– ENTROPY INCREASES!
More energy levels are accessible when the temperature increases – ENTROPY INCREASES!
33
Change in entropy is a state function
Because entropy is a state function, then change in entropy S is ALSO
a state function.
The difference in entropy between two states ONLY depends on the
entropy of the initial and final states, and NOT the path taken to get there.
34
Hess’s Law
Recall Hess’s Law – as long as we get from the same initial state to the same
final state then H will be the same regardless of the steps we add together.
Change in entropy S will work exactly the same way! As long as we get from the same initial state to the same final state then S will be the same regardless of
the steps we add together.
35
Boltzmann equation and entropy
n, T, V help define the number of states (number of available energy levels) the system
can have. The many “different ways to move” of
molecules in a particular state are called microstates.
Hopefully it makes sense that more total states should automatically mean more total
microstates.
The number of microstates is often symbolized by W.
36
Playing cards
Say we have a deck of 52 playing cards.
Choosing one playing card is a state.
If we choose the first card out of the pack, there are 52 microstates for this first state.
The second card (second state) we choose has 51 microstates, and so on.
37
Overall there are
W = 52! 8 x 1067
possible distributions
(total microstates) for
52 playing cards!
Playing cards
38
If we flip 52 coins (a coin is one state), with two possible microstates
(heads or tails) each, there are W = 252 = 4.5 x 1015
possible distributions. (total microstates)
A deck of 52 playing cards has greater entropy than 52 coins!
Playing cards and coin flips
39
Boltzmann equation and entropy
Ludwig Boltzmann formulated the relationship between the number of
microstates (W) and the entropy (S).
S = k ln W
The constant k is the Boltzmann constant which has a value equal to the gas constant R divided by Avagadro’s
number NA
40
Boltzmann equation and entropy
S = k ln W
where k = R / NA
k = (8.3145 JK-1mol-1) / (6.022 x 1023 mol-1)
k = 1.381 x 10-23 JK-1
We can see the units for entropy will be Joules per Kelvin (JK-1)
41
Measuring entropy change
From the units for entropy (JK-1)
we get an idea of how we might measure entropy change S
It must involve some sort of energy change relative to the
temperature change!
42
Measuring entropy change
S = qrev / T
The change in entropy is the heat involved in a
reversible process at a constant temperature.
43
Heat IS NOT a state function
Since heat IS NOT a state function we need a reversible process to make it ACT LIKE a
state function.
44
Reversible processes
In a reversible process a change in one direction is exactly equal and opposite
to the change we see if we do the change in the reverse direction.
In reality it is impossible to make a reversible process without making an
infinite number of infinitesimally small changes.
45
Reversible processes
We can however imagine the process is done
reversibly and calculate the heat involved in it, so
we can calculate the reversible entropy
change that could be involved in a process.
Srev = qrev / T
46
Endothermic increases in entropy
In these three processes the
molecules gain greater “differing ability to
move.”
The molecules occupy more available
microstates at the given temperature, and so the entropy increases in all three processes!
47
Generally entropy increases when…
…we go from solid to liquid.…we go from solid or liquid to gas.…we increase the amount of gas in a reaction.…we increase the temperature.…we allow gas to expand against a vacuum.…we mix gases, liquids, or otherwise make solutions of most types.
48
Problem
Predict whether entropy increases, decreases, or we’re uncertain for the following processes or reactions:
(g) Cl (g) H (aq)OH 2(l) OH 2 (aq) Cl 2 d)
(s) Ag 2 (s) ZnO(s) OAg 2 (s) Zn c)
(g) O (l) Hg 2 (s) HgO 2 b)
(g) OH 2 (s) S 3 (g) SO (g) SH 2 a)
22-iselectrolys
2-
2
2
222
Answers: a) decreases b) increases c) uncertain d) increases
49
Evaluating entropy and entropy changes
Phase transitions – In phase transitions the heat change does occur reversibly, so we can use the formula
Srev = qrev / Tto calculate the entropy change. In this case the heat is the enthalpy of the phase transition and the temperature is the transition temperature
S = Htr / Ttr
50
Evaluating entropy and entropy changes
Phase transitions – For water going from ice (solid) to liquid, Hfus = 6.02 kJmol-1 at the melting point (transition temp.) of 273.15 K (0 C)
Sfus = Hfus / Tmp
Sfus = 6.02 kJmol-1 / 273.15 K
Sfus = 22.0 JK-1mol-1
51
Evaluating entropy and entropy changes
Phase transitions – For water going from liquid to gas, Hvap = 40.7 kJmol-1 at the boiling point (transition temp.) of 373.15 K (100 C)
Svap = Hvap / Tbp
Svap = 40.7 kJmol-1 / 373.15 K
Svap = 109 JK-1mol-1
52
Problem
What is the standard molar entropy of vapourisation Svap for CCl2F2 if its boiling point is -29.79 C and Hvap = 20.2 kJmol-1?
Answer: Svap = 83.0 JK-1mol-1
53
Problem
The entropy change for the transition from solid rhombic sulphur to solid monoclinic sulphur at 95.5 C is Str = 1.09 JK-1mol-1. What is the standard molar enthalpy change Htr for this transition?
Answer: Htr = 402 Jmol-1
54
Absolute entropies
Say we imagine a system of molecules that has no total energy. At this zero-point energy there can ONLY be ONE
possible distribution of microstates, as no molecule has the energy to occupy a
higher energy level.
The entropy CAN NEVER get smaller than its value in this situation, so we define
the entropy S of this situation as ZERO.
55
Absolute entropies
This kind of imagining is the Third Law of Thermodynamics which states that
The entropy of a pure perfect crystal at 0K is zero.
At conditions other than at absolute zero, our entropy is that of the perfect system (zero)
PLUS any entropy changes that come changing temperature and/or volume.
These are absolute entropies!
56
Methyl chloride entropy as a function
of temperature
57
Standard molar entropies
One mole of a substance in its standard state will have an absolute entropy that we often call the standard molar entropy S.
These are usually tabulated at 298.15 K
In a chemical process we can then use these standard molar entropies to calculate the entropy change in the process.
58
Standard molar entropies
S = [pS(products) - rS(reactants)]
Hopefully this looks somewhat familiar!
We have seen a special treatment of Hess’s Law in Chem 1050 where
H = [pHf(products) - rHf(reactants)]
We can do something similar with ANY thermodynamic property that IS A STATE
FUNCTION!
59
Standard molar entropies
S = [pS(products) –
rS(reactants)]
H = [pHf(products) –
rHf(reactants)]
Enthalpies of formation ARE NOT absolute!
60
Problem
Use the data given to calculate the standard molar entropy change for the synthesis of ammonia from its elements.
N2 (g) + 3 H2 (g) 2 NH3 (g)
S298 for N2 = 191.6 JK-1mol-1
S298 for H2 = 130.7 JK-1mol-1
S298 for NH3 = 192.5 JK-1mol-1Answer: -198.7 JK-1 mol-1 (per mole of rxn)
61
Problem
N2O3 is an unstable oxide that readily decomposes. The decomposition of 1.00 mol N2O3 to nitrogen monoxide and nitrogen dioxide at 25 C is accompanied by the entropy change S = 138.5 JK-1mol-1. What is the standard molar entropy of N2O3 (g) at 25 C?
S298 for NO (g) = 210.8 JK-1mol-1
S298 for NO2 (g) = 240.1 JK-1mol-1
Answer: 312.4 JK-1 mol-1
62
The second law of thermodynamics
We’ve seen that entropy MUST play a role in spontaneity,
because the total energy of the universe doesn’t change.
We could say that an entropy increase leads to spontaneity,
but we have to be careful.
63
The second law of thermodynamics
Ice freezing below 0 Celcius is spontaneous, but the entropy of the
water decreases in the process!
Sfreeze = -Hfus / Tmp
Since Hfus and Tmp are +ve,
then Sfreeze is –ve!-ve since “freezing” is the reverse of “fusion” (like we do in Hess’s Law)
64
The second law of thermodynamics
The water is ONLY the system.
The rest of the universe (the surroundings) must experience an
opposite heat change as it takes the heat the freezing water gave off (a +ve H for
the surroundings), which means
the entropy of the REST OF THE UNIVERSE INCREASES in the process
of water freezing!
65
The second law of thermodynamics
The total entropy change in any process is the entropy change for the system
PLUSthe entropy change for the surroundings
Suniverse = Stotal = Ssys + Ssurr
Now we can connect entropy and spontaneity!
66
The second law of thermodynamics
In any spontaneous process
the entropy of the universe INCREASES.
Suniverse = Ssys + Ssurr > 0
This is the Second Law of Thermodynamics!
67
Water freezing
So while water freezing below zero Celcius decreases the entropy of the
system, the heat given off to the surroundings increases the entropy of the surroundings to a greater extent.
The total entropy change of the universe is positive and the process of
water freezing below 0 Celcius is spontaneous!
68
Free energy
We’ve seen entropy increases when molecules
have more ways to distribute themselves amongst the
energy levels.
69
Free energy
However, some of the energy a molecule uses to put itself at a higher energy level CAN NO LONGER be used to to do work
because doing work would put the molecule back at a lower energy level,
which would automatically decrease entropy.
The energy is NOT free (or available) to be used!
70
S = Hrev / TReversible since the rest of the universe is SO BIG
then
TS = H
Free energy
Suniverse = Ssys + Ssurr
TSuniverse = TSsys + TSsurr
TSuniverse = TSsys + Hsurr
71
Free energy
TSuniverse = TSsys + Hsurr
TSuniverse = TSsys - Hsys
Since Hsys = -Hsurr
72
Free energy
TSuniverse = TSsys – Hsys-TSuniverse = Hsys - TSsys
G = Hsys – TSsys
G is the free energy(Gibbs free energy)
73
Free energy
G = -TSuniv
For a spontaneous process Suniv > 0, which means for a spontaneous
process G < 0!
74
Free energy
G < 0 is spontaneous
G > 0 is non-spontaneous
G = 0 is at equilibrium
75
76
Problem
Predict the spontaneity at low and high temperatures for:
N2 (g) + 3 H2 (g) 2 NH3 (g) H = -92.22 kJ
2 C (s) + 2 H2 (g) C2H4 (g) H = 52.26 kJ
77
Problem
N2 (g) + 3 H2 (g) 2 NH3 (g) H = -92.22 kJ
Spontaneous @ low T and nonspontaneous @ high T
2 C (s) + 2 H2 (g) C2H4 (g) H = 52.26 kJ
Nonspontaneous @ all T
78
Standard free energy change G
Just like we can have a standard enthalpy change H for chemicals, we can also define the standard free energy change…
G = H - TS
79
Standard free energy of formation Gf
Standard enthalpies of formation Hf of elements in their standard states are zero:
Hrxn = [p Hf(products) –
r Hf(reactants)]Standard free energies of formation Gf of elements in their standard states are zero:
G = [p Gf(products) –
r Gf(reactants)]
80
Problem
Calculate G at 298.15 K for the reaction
4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s)by using the two following sets of data. Compare your answers.a) H = -1648 kJ mol-1
and S = -549.3 JK-1 mol-1
b) Gf (Fe) = 0 kJmol-1
Gf (O2) = 0 kJmol-1
Gf (Fe2O3) = -742.2 kJmol-1
81
Problem answer
a) G = -1484 kJmol-1
b) G = -1484.4 kJmol-1
The answers are the same because free energy is a state function.
82
Free energy and equilibrium
We’ve already seen that
G < 0 is spontaneous G > 0 is non-spontaneous
One process in a written reaction is spontaneous while the reverse is not, as long as G ≠ 0. Therefore
G = 0 is at equilibrium
83
Water and steam
H2O (l, 1 atm) H2O (g, 1 atm)
G373.15 K = 0 kJmol-1
The system is at equilibrium at 1 atm (standard
conditions) and at the boiling point temperature!
84
Water and steam
H2O (l, 1 atm) H2O (g, 1 atm)
G298.15 K = 8.590 kJmol-1
The system is not at equilibrium at 1 atm
(standard conditions) and at the room temperature!
85
Water and steam
H2O (l, 1 atm) H2O (g, 1 atm)
G298.15 K = 8.590 kJmol-1
The forward process is non-spontaneous (G > 0) so the
reverse process is spontaneous and condensation occurs.
86
Water and steam
H2O (l, 0.03126 atm)
H2O (g, 0.03126 atm) G298.15 K = 0 kJ
The system is at equilibrium at 0.03126 atm (non-standard
conditions) and at the room temperature!
87
Water and steam
H2O (l, 0.03126 atm)
H2O (g, 0.03126 atm) G298.15 K = 0 kJmol-1
Water CAN evaporate at room temperature,
just not to give an equilibrium pressure of 1 atm!
88
Non-standard conditions
As we’ve seen with the previous water example,
our interest in an equilibrium system is often at non-standard conditions,
so knowing G is usually not as useful as
knowingG.
89
Non-standard conditions
For an ideal gas H does not change if pressure changes, so at all non-
standard conditions H = H.
For an ideal gas S does change if pressure changes (expansion into vacuum shows us this!), so at all
non-standard conditions S ≠ S.
90
Non-standard free energy
Because of these facts, the non-standard free energy change is
G = H - TSBut the standard free energy change
is
G = H - TS
91
Non-standard free energy
The difference between standard and non-standard free energy is totally
due to the difference in entropy change between the standard and
non-standard conditions
G - G = - T(S-S)G = G + T(S-S)
92
Boltzmann distribution
By the Boltzmann distribution
S = R ln W for one mole of particles
G = G + T(R ln W - R ln W)G = G + T(R ln W / W)
We are comparing a real system with a standard one. We are
dealing with activities!
93
Boltzmann distribution
G = G + RT ln Qeq
In the comparison, we are looking at a reaction quotient!
In other words, we are looking at how our non-standard system is
different from the system we have at standard conditions!
94
Boltzmann distribution
G = G + RT ln Qeq
If our system
is at equilibrium
then Qeq = Keq
and G = 0
95
Boltzmann distribution
G = G + RT ln Keq = 0
which means
G = -RT ln Keq
96
Equilibrium constant
The thermodynamic equilibrium constant for a reaction is directly
related to the standard free energy change!
G = -RT ln Keq
Keq = e-G/RT
97
Equilibrium constant
G 0 Keq 1
G >> 0 then Keq is very small
G << 0 then Keq is very large
98
Equilibrium constant at 298 K
G Keq Meaning
Keq = e-G/RT
99
Predicting reaction direction
G < 0 means the forward reaction is spontaneous at the given non-standard conditions
G < 0 means the forward reaction is spontaneous at standard conditions and so K > 1
100
Predicting reaction direction
G = 0 means the system is at equilibrium at the given non-standard conditions
G = 0 means the system is at equilibrium at standard conditions and so Keq = 1 which only occurs at one specific T!
101
Predicting reaction direction
G > 0 means the forward reaction is non-spontaneous at the given non-standard conditions
G > 0 means the forward reaction is non-spontaneous at standard conditions and so K < 1
102
Predicting reaction direction
G = G
ONLY at standard conditions!
103
Thermodynamic equilibrium constant Keq
The thermodynamic equilibrium constant Keq is expressed in terms of activities, which
are unitless quantities.
Activities relate properties like concentration or pressure compared to a standard property
value, like 1 M for concentration or 1 bar for pressure.
104
Thermodynamic equilibrium constant Keq
a A + b B c C + d D
b
Ba
A
dD
cC
eqaa
aaK
where ax = [X] / c0 (c0 is a standard concentration of 1 M)
or ax = Px / P0 (P0 is a standard pressure of 1 atm)
Note that ax = 1 for pure solids and liquids
105
Free energy and equilibrium constants
G = -RT ln Keq
IS ALWAYS TRUEG = -RT ln Kc
andG = -RT ln Kp
DO NOT have to be true!
106
Problem
Write thermodynamic equilibrium constant expressions for each of the following reactions and relate them to Kc and Kp where appropriate:a) Si (s) + 2 Cl2 (g) SiCl4 (g)
b) Cl2 (g) + H2O (l) HOCl (aq) + H+ (aq) + Cl- (aq)
107
Problem answer
a) Si (s) + 2 Cl2 (g) SiCl4 (g)
b) Cl2 (g) + H2O (l) HOCl (aq) + H+ (aq) + Cl- (aq)
2222
-
Cl
-
Cl
-
OHCl
ClHHOCleq P
]][Cl[HOCl][H
1P
P
c][Cl
c][H
c[HOCl]
aa
aaaK
p2
Cl
SiCl
2
Cl
SiCl
2ClSi
SiCleq K
P
P
PP
1
PP
aa
aK
2
4
2
4
2
4
108
Problem
Use the given data to determine if the following reaction is spontaneous at standard conditions at 298.15 K:
N2O4 (g) 2 NO2 (g)
Gf (N2O4) = 97.89 kJmol-1
Gf (NO2) = 51.31 kJmol-1
Answer: G = 4.73 kJmol-1, not spontaneous at standard conditions.
109
Problem
Based on the problem of the previous slide, determine which direction the reaction will go in if 0.5 bar of each gas is placed in an evacuated container:
N2O4 (g) 2 NO2 (g)
Answer: Since G = 4.73 kJmol-1, then Keq = 0.15 = Kp. Since Qp = 0.5 the reaction should go from right to left.
110
Problem
Determine the equilibrium constant at 298.15 K for the following reaction using the given data:
AgI (s) Ag+ (aq) + I- (aq)
Gf (AgI) = -66.19 kJmol-1
Gf (Ag+) = 77.11 kJmol-1
Gf (I-) = -51.57 kJmol-1
111
Problem answer
Since G = 91.73 kJmol-1, then Keq = 8.5 x 10-17 = Kc = Ksp.If we compare to the Ksp value in Table 18.1 (8.5 x 10-17) we see we are definitely in the right ballpark!
112
G and Keq are functions of T
We’ve already seen that equilibrium constants change with temperature.
Why?
Keq = e-G/RT
and
G = H - TS
113
G and Keq are functions of T
G = -RT ln Keq = H - TS
ln Keq= -(H/RT) + (S/R)
114
Problem
At what temperature will the following reaction have Kp = 1.50 x 102?
2 NO (g) + O2 (g) 2 NO2 (g)
H = -114.1 kJmol-1
and S = -146.5 J K-1mol-1
Answer: T = 606 K
115
G and Keq are functions of T
G = -RT ln Keq = H - TS
ln Keq= -(H/RT) + (S/R)
If H and S are constant over a temperature range
then
116
G and Keq are functions of T
ln K1= -(H/RT1) + (S/R)minus
ln K2= -(H/RT2) + (S/R)
ln K1 - ln K2 =
-(H/RT1) - (-H/RT2)
117
G and Keq are functions of T
ln K1 - ln K2 =
[-H/R] [(1T1) - (1T2)]OR
ln [K1/K2] =
[-H/R] [(1T1) - (1T2)]
118
van’t Hoff equation
The van’t Hoff equation relates equilibrium constants to
temperatures
ln [K1/K2] = [-H/R] [(1T1) - (1T2)]It looks very similar to the Arrhenius
equation and so a plot of ln K versus 1/T should give a straight line
with a slope of [-H/R]
119
Plot of ln Kp versus 1/T for the
reaction.
slope = -H / R, so for this reaction
H = -180 kJmol-1
2 SO2 (g) + O2 (g) 2 SO3 (g)
120
Problem
The following equilibrium constant data have been determined for the reaction
H2 (g) + I2 (g) 2 HI (g)
Kp = 50.0 @ 448 CKp = 66.9 @ 350 C
Estimate H for the reaction.
Answer: H = -11.1 kJmol-1.
121
Coupled reactions
If we have a non-spontaneous (G > 0) reaction with a product that appears as a reactant in a different reaction that is spontaneous (G < 0) then we can couple the two reactions (do them in the same container at the same time) to drive the non-spontaneous reaction!
122
Coupled reactions
This should make sense because if we have two equilibria in one container, one with a small K (like Ksp) and one with a large K (like Kf for complex formation), then the coupled reaction that is the sum of the two reactions has a K value that is larger than Ksp!
We have made the non-spontaneous reaction occur!
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