Entropy and Spontaneity Section 15.2 (AHL). Introduction Entropy can be regarded as a measure of the...

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