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Entropy and Spontaneity
•Section 15.2 (AHL)
Introduction
Entropy can be regarded as a measure of the
disorder or dispersal of energy in a system
It measures the randomness or disorderness
Is given the symbol “S”
Compare a solid and a gas at the same
temperature; the gas has greater entropy because
its particles are moving rapidly in all directions
Effect of Temperature Change When the temperature is increased,
disorder and hence entropy
increases
The reverse takes place when the
temperature is lowered
Effect of Change of State
The disorder of particles increases
from solid to liquid to gas (of the
same substances), increasing
entropy
Effect of a Change of the Number of Particles
If the number of particles increases,
disorder and hence entropy increases
Ex. N2O4(g) → 2NO2(g) (1 mole of gas to 2
moles, so entropy increases)
Conversely, entropy decreases when the
number of particles decreases
Effect of Mixing of Particles Mixing of particles increases entropy
because it increases disorder
This is easily seen when one
substance is dissolved in another as
the particles are free to move
around randomly
Predicting the Sign of a Change in Entropy
If the products are more disordered than the
reactants, then the entropy change of the
system is positive, +ΔS
If the products are less disordered than the
reactants, then the entropy change is negative,
-ΔS
Some Examples Reaction or change
Melting
Boiling
Condensing
Freezing
Crystallization from a
solution
Chemical reaction: solid or
liquid from gas
Entropy change
Increase
Large increase
Large decrease
Decrease
Decrease
Large decrease
Calculating Entropy Changes
The units of entropy and entropy change are J K-1 mol-
1
Entropy values are absolute values and can be measured
experimentally
Standard entropy change = sum of entropies of products –
sum of entropies of reactants
ΔSθ = ΣSθ[products] – ΣSθ
[reactants]
Entropy values are in the data booklet
Example Problem
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Sθ[methane] = 186 J K-1 mol-1
Sθ[oxygen] = 205 J K-1 mol-1
Sθ[carbon dioxide] = 214 J K-1 mol-1
Sθ[water] = 70 J K-1 mol-1
ΔSθ = 214 + (2x 70) – (186 + 2 x 205)
ΔSθ = -242 J K-1 mol-1
Spontaneity Spontaneous process: has a natural
tendency to occur and involves an increase in the entropy of the universe (at the expense of energy available to do useful work).
Non spontaneous processes result in a decrease in the entropy of the universe
Standard Ambient conditions (SATP): 100 kPa of pressure and 298 K
Spontaneous processes may occur very quickly or very slowly
Spontaneity Determination Gibbs free energy: the energy
available to do work Spontaneity is determined by the
sign of the Gibbs free energy change, ΔGθ (AKA free energy)
Gibbs equation: ΔGθ = ΔHθ – TΔSθ
T = standard temperature in K ΔHθ = enthalpy change
More on Spontaneity and Gibbs Free energy
ΔSθ = entropy change ΔGθ is measured in kJ mol-1
- ΔGθ means the reaction or process is spontaneous
+ΔGθ means the reaction or process is non-spontaneous
ΔGθ = 0, the reaction or process is at equilibrium
Example Problem #1 Calculate the Gibbs free energy
change for the following reaction under standard conditions
N2(g) + 3H2(g) → 2NH3(g)
ΔHθ = -95.4 kJ ΔSθ = -198.3 J K-1mol-1 (needs to be
in kJ) T = 298 K
Problem #1 (Continued)
ΔGθ = ΔHθ – TΔSθ ΔGθ = -95.4 – (298) (-0.1983) ΔGθ = -36.3 kJ mol-1
-ΔGθ means the reaction is spontaneous at this temperature
Part 2: calculate the temperature at which the reaction ceases to occur spontaneously
Part 2
Need to find when ΔGθ =0 0 =ΔHθ – TΔSθ
TΔSθ = ΔHθ T = ΔHθ / ΔSθ
T = -95.4
-0.1983 T = 481 K = 208 °C
Gibbs Free Energy Change of Formation
This is the free energy change that occurs when 1 mol of a compound formed from its elements in their standard states under standard conditions
Each compound has a Gibbs free energy change of formation, ΔGθ
f
Some values are in the data booklet (Table 12)
More The ΔGθ
f of elements is zero Gibbs free energy change for a
reaction = [sum of Gibbs free energy of formation of the products] – [sum of Gibbs free energy of formation of the reactants]
ΔGθ = ΣGθf(products) - ΣGθ
f(reactants)
Problem # 2 Use the following Gibbs free
energy changes of formation to calculate the free energy change for the decomposition of MgCO3
MgCO3(s) → MgO(s) + CO2(g)
ΔGθ (MgCO3(s)) = -1012 kJ mol-1
ΔGθ (MgO(s)) = -569 kJ mol-1
Problem # 2 Continued ΔGθ (CO2(g))
= -394 kJ mol-1
ΔGθ = [-394 + (-569)] - (-1012)
ΔGθ = -963 + 1012
ΔGθ = + 49 kJ mol-1 + ΔGθ indicates that the reaction is
non-spontaneous under standard conditions
Spontaneity and Temperature
Positive entropy changes are favorable and negative enthalpy changes are favorable and drive the reaction forward
Negative entropy changes are unfavorable and positive enthalpy changes are unfavorable and drive the reaction backward
See the following chart to summarize the effect of temperature on spontaneity
Think of the equation: ΔGθ = ΔHθ – TΔSθ, as this will guide you to determine the impact of each variable. Remember for a reaction to be spontaneous, : ΔGθ has to be NEGATIVE (-)
Problem # 3Determine ΔG for the following reaction:
CaCO3(s) → CaO(s) + CO2(g) (at 500K)
The required data are: (BE CAREFUL OF
UNITS!)
First Calculate ΔH
ΔH = ΣΔHf(products) - ΣΔHf(reactants)
ΔH = [(-636) + (-394)] - (-1207)
ΔH = +177 kJ mol-1 = 177000 J
mol-1
Calculate ΔS
ΔS = ΣΔS(products) - ΣΔS(reactants)
ΔS = (40 + 214) – 93
ΔS = 161 J K-1mol-1
Calculate ΔG
ΔG = ΔHθ – TΔSθ
ΔG = 177000 – (500)(161)
ΔG = 96500 J mol-1 or 96.5 kJ mol-1