Transcript document

Free Energy and Temperature
•Free energy decreases
(becomes more negative) as
temperature
•At low T, Gm for solid phase is
lower than that of liquid or
vapour, so the solid phase is
prevalent
•As we increase T to Tfus and
higher, the liquid state has a
lower Gm, so it is the phase
that prevails
•As we increase T further to Tb,
the gas phase has the lowest
value of Gm
7.13: Gibbs Free Energy of Reaction
• To determine the spontaneity of a
reaction, we use the change in the
Gibbs Free Energy, G, or the Gibbs
Free Energy of Reaction
G  nG mProducts 

nG
m Reactants
We’ve seen something like
this before somewhere…
Standard Gibbs Free Energy of Formation,
Gf°
 Gf° = The standard Gibbs Free Energy of reaction
per mole for the formation of a compound from its
elements in their most stable form.
• Most stable form?
–
–
–
–
Hydrogen = ?
Oxygen = ?
Iodine = ?
Sodium = ?
 Gf°is for the formation of 1 mole of product
– Different amounts of reactants may be used…Be vigilant!

Gf°: What Does it Mean?
• Compounds with Gf° > 0 are
Thermodynamically Unstable
• Compounds with Gf° < 0 are
Thermodynamically Stable
Gibbs Free Energy and Nonexpansion Work
• we = ‘Extra work’
– Nonexpansion work is any kind of work other than that done
against an opposing pressure
• Stretching a spring, moving a rope, importing a sugar
molecule into a cell are all examples of nonexpansion
work
• All cellular processes are examples of nonexpansion
work
• How are the Gibbs Free Energy and we related?
Gibbs Free Energy and we
 G = we
• If we know the change in free energy,
we know how much nonexpansion work
can be done
• What does this mean?
– Let’s look at the combustion of glucose.
G° and the Combustion of Glucose
C6H12O6 (s) + 6 O2 (g) --> 6 CO2 (g) + 6 H2O (l)
• The G° of the reaction is -2879 kJ
For 1 mole of glucose, we get 2879 kJ of energy
Or
For 180 g of glucose, we get 2879 kJ of energy
• To make one mole of peptide bonds, 17 kJ of work
must be done.
– If we get 2879 kJ of energy from one mole of glucose, we
should be able to make 170 moles of peptide bonds
One molecule of glucose will provide enough
energy to add 170 amino acids to a growing
protein
(in actuality, you can only add 10 amino acids)
The Effect of Temperature on G°
• Remember that H° or S° is the sum of
the individual enthalpies or entropies of the
products minus those of the reactants
• If we change the temperature, both are
affected to the same extent, so the H° and
S° values don’t significantly change
• This is not the case with G°. Why?
G° = H° - TS°
The Effect of Temperature on G°
The Effect of Temperature on G°
The Effect of Temperature on G°
The Effect of Temperature on G°

Thermodynamics Review
Let’s Look at some of the most important
equations we’ve covered over the past
2 chapters…
The First Law of Thermodynamics
• Up until now, we have only considered the changes
in the internal energy of a system as functions of a
single change: either work or heat
• However, these changes rarely occur singly, so we
can describe the change in internal energy as:
U = q + w
(The 1st Law)
• The change in internal energy is dependent upon the
work done by the system and the heat gained or lost
by the system
Heat Capacity
q = CT = mCsT
nCmT
Coffee Cup Calorimeter
System: Solution and chemicals that react
Surroundings: Cup and the world around it!
Assumptions: We use 2 cups to prevent energy
transfer to the surroundings (we assume that it
works as designed)
Expected Changes:
i) As the chemical reaction occurs, the potential
energy in the reactants will be released as heat or
the solution can supply heat to allow formation of a
product with a higher potential energy
ii) The solution will absorb or release energy during
the reaction. We will see this as a temperature
change
qr + qsolution = 0
Constant Pressure Calorimetry: An
Example
We place 0.05g of Mg chips in a coffee cup
calorimeter and add 100 mL of 1.0M HCl, and
observe the temperature increase from 22.21°C
to 24.46°C. What is the ΔH for the reaction?
Mg(s) + 2HCl (aq) --> H2(g) + MgCl2(aq)
Assume: Cp of the solution = 4.20 J/gK
Density of HCl is 1.00 g/mL
Constant Pressure Calorimetry: An
Example
To solve this:
ΔT = (24.46°C – 22.21°C) = (297.61K – 295.36K)=2.25K
Mass of solution =
1g 

100ml x
  0.05g  100.05 g
ml 

Now, let’s calculate qsolution:
qsolution = mCmΔT = (100.05g)(4.20 J/gK)(2.25K)
= 945.5 J
Now, let’s calculate qr:
qr = -qsolution = -945.5 J
Enthalpy
• In a constant volume system in which no work is done
(neither expansion nor non-expansion), we can rearrange
the first law to:
U = q + w
(but w=0)
U = q
• Most systems are constant pressure systems which can
expand and contract
• When a chemical reaction takes place in such a system, if
gas is evolved, it has to push against the atmosphere in
order to leave the liquid or solid phase
– Just because there’s no piston, it doesn’t mean that no work is
done!
Enthalpy
• Let’s look at an example:
• If we supply 100J of heat to a system at constant pressure
and it does 20J of work during expansion, the U of the
system is +80J (w=-20J)
– We can’t lose energy like this
• Enthalpy, H, is a state function that we use to track energy
changes at constant pressure
H=U + PV
• The change in enthalpy of a system (H) is equal to the heat
released or absorbed at constant pressure
Enthalpy
• Another way to define enthalpy is at constant
pressure:
H = q
Enthalpy is a tricky thing to grasp, but we can look at it this way:
• Enthalpy is the macroscopic energy change (in the form of
heat) that accompanies changes at the atomic level (bond
formation or breaking)
• Enthalpy has the same sign convention as work, q and U
– If energy is released as heat during a chemical reaction
the enthalpy has a ‘-’ sign
– If energy is absorbed as heat from the surrounding during
a reaction, the enthalpy has a ‘+’ sign
The 2nd Law of Thermodynamics
The entropy of an isolated system
increases in the course of any
spontaneous change
• We can summarize this law
mathematically as:
q
S 
T
Entropy Change as a Function of
Temperature at Constant Volume
T2 
S = C ln 
T1 
• If T2 > T1, then the logarithm is ‘+’ and entropy
increases
– Makes
 sense since we are raising the temperature and
thermal motion will increase
• The greater (higher) the heat capacity, the higher the
entropy change
Entropy Change as a Function of
Changing Volume
• We can use a similar logic to derive the change in
entropy when the volume changes:
V2 
S = nR ln  
V1 
• When V2 > V1, the entropy increases
• Note: Units
 are still J/K

Entropy Change as a Function of
Pressure
• Remember Boyle’s Law?
• We can substitute this relationship into
the equation for entropy change as a
function of volume to get:
P1 
S = nR ln  
P2 
Entropy decreases
for a samples that
has been
compressed
isothermally (P1>P2)

Boiling Water and Entropy
Let’s get 3 facts straight:
1. At a transition temperature (Tf or Tb), the temperature remains
constant until the phase change is complete
2. At the transition temperature, the transfer of heat is reversible
3. Because we are at constant pressure, the heat supplied is equal to the
enthalpy
Water Boiling and Entropy
S =
H vap
Tb
(at the boiling temperature)
• We use the ‘ º ’ superscript to denote
the standard
entropy
or
the
entropy
at
1

bar of pressure
H vap
S =
Tb
Ice Melting and Entropy
• We use the same logic to determine the
entropy of fusion, Sfus
Sfus

H fus
=
Tb
Sfus

H fus
=
Tb
The Boltzmann Formula
Where:
k = Boltzmann’s constant
S = k lnW
= 1.381 x10-23 J/K
W=# of ways atoms or molecules in
the system can be arranged and still
give the same total energy
• W is a reflection of the ensemble, the
collection of molecules in the system
• This entropy value is called the
statistical entropy
Calculating the Entropy of a Reaction
• Sometimes we can’t always use our
judgement and we need to calculate the
entropy
• In order to do this, we need the
standard molar entropies of the
products and the reactants as well as
the number of moles of each
S =

nS
 m (products) -

nS
 m (reactants)

The Surroundings
STot = SSystem + SSurr
• If STot is positive, the
reaction is spontaneous
• If the Ssystem is negative,
the reaction will still be
spontaneous if SSurr is that
much more positive
Gibbs Free Energy
G = H - TS
G = -TS
(at constant T)
(at constant T and P)
• A negative value of G indicates that a reaction will
spontaneously occur
• Large negative H values (like we’d have in a
combustion reaction) would probably give you a large
negative G
• If TS is large and positive, the value of G may be
large and negative