# Lecture%205%20Free%20energy ```Free Energy
We have established that entropy increases with
increasing temperature.
In an endothermic reaction, the decrease in enthalpy will
cause the entropy of the surroundings to…………….. decrease.
An endothermic reaction can only take place if the
increase in entropy within the reaction mixture is great
enough to overcome the unfavourable enthalpy change.
If we want to predict whether a reaction will take place
spontaneously, two factors must be considered.
•
•
∆H
∆S
A negative value of ∆H favours reaction.
An increase in entropy (positive value for
∆S favours reaction.)
Since the units of entropy are JK-1, the
entropy value must be considered in
conjunction with the entropy value, so
T∆S is used.
The combination of enthalpy changes and entropy changes is
known as the free energy change and is given the symbol ∆G.
∆G = ΔH - T∆S
For a reaction to be feasible, ∆G must be negative.
In using the equation
∆G = ΔH - T∆S
units must be considered.
ΔH has the units kJmol-1
ΔS has the units JK-1
In calculations, entropy values should be converted to kJ K-1
and temperature should be expressed in K (Kelvin).
Like other thermodynamic values, the
standard free energy change ΔG&ordm;
can be calculated.
ΔG&ordm; = ∑G&ordm;f (products) - ∑G&ordm;f (reactants)
Example:
Calculate the standard free energy change for
the reaction given that ΔG&ordm;f (NO2) = 52kJmol-1
and ΔG&ordm;f(N2O4) = 98kJmol-1
2NO2
N2O4
ΔG&ordm; = ∑G&ordm;f (products) - ∑G&ordm;f (reactants)
= 98 - (2 x 52)
= -6kJmol-1
Since ΔG&ordm; is negative, the reaction is feasible.
Summary
•
•
•
•
•
•
Exothermic reactions are more likely to take
place spontaneously than endothermic reactions.
Entropy (S) is the degree of disorder in a system.
The Third Law of Thermodynamics states that for
a spontaneous process, the total entropy
always increases.
ΔG = ΔH – TΔS
For a reaction to be feasible, ΔG must be less
than zero.
The value of ΔG only indicates the favoured
direction in a reaction – it provides no information
Example 2:
Using the data given below, calculate ΔG&ordm; at 298 K
and 1 atmosphere pressure for the reaction:
H2(g) + &frac12;O2(g)
H2O(l)
ΔH&ordm; (kJ mol-1)
ΔS&ordm; (J K-1mol-1)
H2(g)
0
131
O2(g)
0
205
H2O(l)
-286
70
H2(g) + &frac12;O2(g)
H2O(l)
ΔH&ordm; (kJ mol-1)
ΔS&ordm; (J K-1mol-1)
H2(g)
0
131
O2(g)
0
205
H2O(l)
-286
70
ΔH&ordm; = ∑ΔH&ordm;f (products) - ∑ΔH&ordm;f (reactants)
= -286 -0
= -286 kJ mol-1
ΔS&ordm; = ∑S&ordm;products-∑S&ordm;reactants
= 70 – (131 + (0.5 x 205))
= -163.5 J K mol-1
ΔG&ordm; = ΔH&ordm; - TΔS&ordm;
= –286 - (298 x (-0.1635))
= -286 + 48.7
= -237.3 kJ mol-1
The standard free energy for the reaction is -237.3 kJ mol-1.
This indicates that the reaction is feasible (i.e. that the products
are favoured thermodynamically.
It does not tell you anything about the rate of reaction.
At room temperature, virtually no reaction occurs between H2 and
O2 until energy is supplied to overcome the activation energy
barrier.
Predicting the sign of ΔG and reaction feasibility
ΔH
ΔS
negative
positive
negative
negative
positive
positive
positive
negative
Feasible?
ΔG
Yes
negative
Only if TΔS
is smaller
than ΔH
Only if TΔS
is greater
than ΔH
No
negative
negative
positive
Complete tutorial questions – page 52 questions 5-7
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