ΔG  c GC0  dGD0  aGA0  b GB0
 RT ln[C]c  RT ln[D]d  RT ln[A]a  RT ln[B]b









c
d 
[C] [D]
 ΔG0  RT ln a b
[A] [B]







1501
ΔG  c GC0  dGD0  aGA0  b GB0
 RT ln[C]c  RT ln[D]d  RT ln[A]a  RT ln[B]b









c
d 
[C] [D]
 ΔG0  RT ln a b
[A] [B]







ΔG  ΔG0  RT ln Q
1502
ΔG  c GC0  dGD0  aGA0  b GB0
 RT ln[C]c  RT ln[D]d  RT ln[A]a  RT ln[B]b









c
d 
[C] [D]
 ΔG0  RT ln a b
[A] [B]







ΔG  ΔG0  RT ln Q
where Q is the reaction quotient.
1503
ΔG  c GC0  dGD0  aGA0  b GB0
 RT ln[C]c  RT ln[D]d  RT ln[A]a  RT ln[B]b









c
d 
[C] [D]
 ΔG0  RT ln a b
[A] [B]







ΔG  ΔG0  RT ln Q
where Q is the reaction quotient.
For a reaction system at equilibrium
ΔG  0 and Q = Kc
1504
ΔG  c GC0  dGD0  aGA0  b GB0
 RT ln[C]c  RT ln[D]d  RT ln[A]a  RT ln[B]b









c
d 
[C] [D]
 ΔG0  RT ln a b
[A] [B]







ΔG  ΔG0  RT ln Q
where Q is the reaction quotient.
For a reaction system at equilibrium
ΔG  0 and Q = Kc
and hence
ΔG0   RT ln Kc (a very key result)
1505
For a reaction in which all the species are in the
gas phase,
ΔG0   RT ln Kp
1506
It follows directly from the result
ΔG0   RT ln Kc
1507
It follows directly from the result
ΔG0   RT ln Kc
for ΔG0  0 Kc > 1
1508
It follows directly from the result
ΔG0   RT ln Kc
for ΔG0  0 Kc > 1
for ΔG0  0 Kc < 1
1509
It follows directly from the result
ΔG0   RT ln Kc
for ΔG0  0 Kc > 1
for ΔG0  0 Kc < 1
for ΔG0  0 Kc = 1 In this case the position of the
equilibrium does not favor
either products or reactants
forming.
1510
Example: The standard Gibbs energy for the

reaction ½ N2(g) + 3/2 H2(g)  NH3(g)
is -16.45 kJmol-1. Calculate the equilibrium
constant for the reaction at 25.00 oC. In a certain
experiment the initial concentrations are [H2] =
0.00345 M, [N2] = 0.00956 M, and [NH3] = 0.00678
M. Predict the direction of the reaction.
1511
Example: The standard Gibbs energy for the

reaction ½ N2(g) + 3/2 H2(g)  NH3(g)
is -16.45 kJmol-1. Calculate the equilibrium
constant for the reaction at 25.00 oC. In a certain
experiment the initial concentrations are [H2] =
0.00345 M, [N2] = 0.00956 M, and [NH3] = 0.00678
M. Predict the direction of the reaction.
From ΔG0   RT ln Kc we have
1512
Example: The standard Gibbs energy for the

reaction ½ N2(g) + 3/2 H2(g)  NH3(g)
is -16.45 kJmol-1. Calculate the equilibrium
constant for the reaction at 25.00 oC. In a certain
experiment the initial concentrations are [H2] =
0.00345 M, [N2] = 0.00956 M, and [NH3] = 0.00678
M. Predict the direction of the reaction.
From ΔG0   RT ln Kc we have
0
ΔG
ln Kc  
RT
1513
Example: The standard Gibbs energy for the

reaction ½ N2(g) + 3/2 H2(g)  NH3(g)
is -16.45 kJmol-1. Calculate the equilibrium
constant for the reaction at 25.00 oC. In a certain
experiment the initial concentrations are [H2] =
0.00345 M, [N2] = 0.00956 M, and [NH3] = 0.00678
M. Predict the direction of the reaction.
From ΔG0   RT ln Kc we have
0
ΔG
ln Kc  
RT
3
-1
(

16.45
x
10
J
mol
)
ln Kc  
(8.314 JK-1mol-1)(298.15K)
1514
Hence
ln Kc = 6.636
Kc  e6.636
1515
Hence
ln Kc = 6.636
Kc  e6.636
therefore
Kc = 762
1516
Hence
ln Kc = 6.636
Kc  e6.636
therefore
Kc = 762
(Note almost one significant digit is lost in
evaluating the exponential.)
1517
Hence
ln Kc = 6.636
Kc  e6.636
therefore
Kc = 762
(Note almost one significant digit is lost in
evaluating the exponential.)
To predict the direction of the reaction, use
ΔG  ΔG0  RT ln Q
1518
Now
Q 
[NH3]
[N2]1/2[H2]3/2
1519
Now
Q 

[NH3]
[N2]1/2[H2]3/2
0.00678
(0.00956)1/2(0.00345)3/2
1520
Now
Q 
[NH3]
[N2]1/2[H2]3/2

0.00678
(0.00956)1/2(0.00345)3/2

0.00678
(0.0978)(0.000203)
= 342
1521
Now
Q 
[NH3]
[N2]1/2[H2]3/2

0.00678
(0.00956)1/2(0.00345)3/2

0.00678
(0.0978)(0.000203)
= 342
Since Q < Kc the reaction will move in the direction
to produce more NH3.
1522
Employing
ΔG  ΔG0  RT ln Q
1523
Employing
ΔG  ΔG0  RT ln Q
ΔG  16.45kJmol-1  (8.314 JK-1mol-1)(298.15K)ln (342)
1524
Employing
ΔG  ΔG0  RT ln Q
ΔG  16.45kJmol-1  (8.314 JK-1mol-1)(298.15K)ln (342)
= – 16.45 kJ mol-1 + 14.46 kJ mol-1
= – 1.99 kJ mol-1
1525
Employing
ΔG  ΔG0  RT ln Q
ΔG  16.45kJmol-1  (8.314 JK-1mol-1)(298.15K)ln (342)
= – 16.45 kJ mol-1 + 14.46 kJ mol-1
= – 1.99 kJ mol-1
Since ΔG < 0, the reaction takes place in the
forward direction.
1526
Are diamonds forever?
1527
Are diamonds forever?
In chemical jargon, is the transition
Cdiamond
Cgraphite
spontaneous at room temperature?
1528
Are diamonds forever?
In chemical jargon, is the transition
Cdiamond
Cgraphite
spontaneous at room temperature?
Tabulated data: S0(diamond) = 2.44 J K-1 mol-1
S0(graphite) = 5.69 J K-1 mol-1
1529
Are diamonds forever?
In chemical jargon, is the transition
Cdiamond
Cgraphite
spontaneous at room temperature?
Tabulated data: S0(diamond) = 2.44 J K-1 mol-1
S0(graphite) = 5.69 J K-1 mol-1
ΔH0 for the reaction Cgraphite
Cdiamond = 1.90 kJ
1530
For the reaction
Cdiamond
Cgraphite
1531
For the reaction
Cdiamond
Cgraphite
ΔS0 = 1 mol x 5.69 J K-1 mol-1 – 1 mol x 2.44 J K-1 mol-1
= 3.25 J K-1
1532
For the reaction
Cdiamond
Cgraphite
ΔS0 = 1 mol x 5.69 J K-1 mol-1 – 1 mol x 2.44 J K-1 mol-1
= 3.25 J K-1
Using
ΔG0  ΔH0 - TΔS0
= - 1.90 x 103J - 298 K (3.25 J K-1)
1533
For the reaction
Cdiamond
Cgraphite
ΔS0 = 1 mol x 5.69 J K-1 mol-1 – 1 mol x 2.44 J K-1 mol-1
= 3.25 J K-1
Using
ΔG0  ΔH0 - TΔS0
= - 1.90 x 103J - 298 K (3.25 J K-1)
= - 1.90 x 103J - 969 J
= -2.87 kJ
1534
For the reaction
Cdiamond
Cgraphite
ΔS0 = 1 mol x 5.69 J K-1 mol-1 – 1 mol x 2.44 J K-1 mol-1
= 3.25 J K-1
Using
ΔG0  ΔH0 - TΔS0
= - 1.90 x 103J - 298 K (3.25 J K-1)
= - 1.90 x 103J - 969 J
= -2.87 kJ
Conclusion: The reaction Cdiamond
Cgraphite is
spontaneous!
1535
Electrochemistry
1536
Electrochemistry
Two Key Ideas in this section:
1537
Electrochemistry
Two Key Ideas in this section:
(1) Use spontaneous redox chemistry to generate
electricity.
1538
Electrochemistry
Two Key Ideas in this section:
(1) Use spontaneous redox chemistry to generate
electricity.
(2) Use electric current to drive non-spontaneous
redox reactions.
1539
Oxidation-reduction reactions
1540
Oxidation-reduction reactions
Oxidation number: A number equal to the charge
an atom would have if its shared electrons were
held completely by the atom that attracts them
more strongly.
1541
Oxidation-reduction reactions
Oxidation number: A number equal to the charge
an atom would have if its shared electrons were
held completely by the atom that attracts them
more strongly.
Need to be able to distinguish reactions where
there is a change in oxidation number (these are
redox reactions) from reactions where there is no
change in oxidation number.
1542
1543
Example:
Na2SO4(aq) + BaCl2(aq)
2 NaCl(aq) + BaSO4(s)
1544
Example:
+
2-
2+
-
Na2SO4(aq) + BaCl2(aq)
+
-
2+
2-
2 NaCl(aq) + BaSO4(s)
1545
Example:
+
2-
2+
-
Na2SO4(aq) + BaCl2(aq)
+
-
2+
2-
2 NaCl(aq) + BaSO4(s)
There is no change in oxidation number, there is no
electron transfer taking place in this reaction.
Example: 2 Na(s) + Cl2(g)
2 NaCl(s)
1546
Example:
+
2-
2+
-
Na2SO4(aq) + BaCl2(aq)
+
-
2+
2-
2 NaCl(aq) + BaSO4(s)
There is no change in oxidation number, there is no
electron transfer taking place in this reaction.
0
0
Example: 2 Na(s) + Cl2(g)
+
-
2 NaCl(s)
1547
Example:
+
2-
2+
-
Na2SO4(aq) + BaCl2(aq)
+
-
2+
2-
2 NaCl(aq) + BaSO4(s)
There is no change in oxidation number, there is no
electron transfer taking place in this reaction.
0
0
+
-
Example: 2 Na(s) + Cl2(g)
2 NaCl(s)
In this example, there is a change in oxidation
number, so electron transfer is taking place.
1548
Half-Equations:
1549
Half-Equations:
Na
Na+ + e-
(1)
1550
Half-Equations:
Na
2e- + Cl2
Na+ + e-
(1)
2 Cl-
(2)
1551
Half-Equations:
0
Na
0
2e- + Cl2
+
Na+ + e-
(1)
-
2 Cl-
(2)
1552
Half-Equations:
0
Na
0
2e- + Cl2
+
Na+ + e-
(1)
-
2 Cl-
(2)
These are called the half-equations.
1553
Half-Equations:
0
Na
0
2e- + Cl2
+
Na+ + e-
(1)
-
2 Cl-
(2)
These are called the half-equations. Multiply
equation (1) by 2 and add equation (2) leads to the
overall balanced equation.
2 Na + Cl2
2 NaCl
1554
Oxidation: A process in which electrons are lost.
Na
Na+ + e-
1555
Oxidation: A process in which electrons are lost.
Na
Na+ + eReduction: A process in which electrons are gained.
2e- + Cl2
2 Cl-
1556
Oxidation: A process in which electrons are lost.
Na
Na+ + eReduction: A process in which electrons are gained.
2e- + Cl2
2 ClOxidizing Agent: A substance that gains electrons
(in the reduction step). It undergoes a decrease in
oxidation number. Example: Cl2 in the above eq.
1557
Oxidation: A process in which electrons are lost.
Na
Na+ + eReduction: A process in which electrons are gained.
2e- + Cl2
2 ClOxidizing Agent: A substance that gains electrons
(in the reduction step). It undergoes a decrease in
oxidation number. Example: Cl2 in the above eq.
Reducing Agent: A substance that donates electrons
in a redox reaction (specifically in the oxidation
step). It undergoes an increase in oxidation
number. Example: Na in the above oxidation.
1558
Redox reaction: A reaction that can be split apart
into an oxidation step and a reduction step.
1559
Redox reaction: A reaction that can be split apart
into an oxidation step and a reduction step.
Example:
2 Na + Cl2
2 NaCl
1560