6. Thermodynamics II

Thermodynamics II

1 CM1401 CHEMISTRY FOR LIFE SCIENCES

Department of Chemistry

Test Yourself

1) If ΔSsurr = –ΔSsys , the process is at equilibrium. True/False

2) For any process, ΔSsurr and ΔSsys have opposite signs. True/False

3) As long as the disorder of the surroundings is increasing, a process will be spontaneous. True/False


Why Are Some Changes Spontaneous & Some Are Not?

• What is the driving force for spontaneity?

• Examples of spontaneous processes:

a gas expands to fill the vessel, a hot object cools to the temp of its surroundings, iron + water + O2→ rust


Spontaneous Processes

• A process that is spontaneous occurs without outside intervention/without work done

• A non-spontaneous change can be brought about only by doing work

• Driving force for a spontaneous change: tendency of energy & matter to disperse in disorder

40 oC

25 oC


Entropy S

Measure of the disorderly dispersal of energy or matter

• Entropy increases as disorder increases

(order disorder)

(fewer microstates more microstates)

(lower entropy higher entropy)

• Entropy (S) is related to the number of ways that a system can distribute its energy, which in turn is closely related to the freedom of motion of the particles and the number of ways they can be arranged

• S is a state function


• How many ways (microstates) can each arrangement (state) be achieved?

• the number of microstates, the probability of occurrence of the arrangement

Lower entropy

Higher entropy


Ludwig Boltzmann:

S = k In W

k = Boltzmann constant = R/NA = 1.38 x 10-23 J K-1

W = number of microstates

S has units of J K-1


What is the sign of ΔS?

N2 (g) + 3H2 (g) 2NH3 (g)

A phase change

Unfolding of a protein from a compact conformation to a more flexible comformation

Association of 2 reactants to form a more compact structure

ΔS __ 0

ΔS __ 0 in going from solid

to liquid to gas


ΔS __ 0

ΔS __ 0

Second Law of Thermodynamics In any spontaneous process, there is always an increase

in the entropy of the universe

ΔSuniv = ΔSsys + ΔSsurr

We need to consider the entropy of both the system & surroundings when deciding whether or not a process is spontaneous.

ΔSuniv > 0: spontaneous

ΔSuniv < 0: non-spontaneous

ΔSuniv = 0: system is at equilibrium


Entropy Change

Entropy change of a system is equal to the energy transferred as heat to it reversibly divided by the temp at which the transfer takes place.

Reversible transfer of heat (qrev) is smooth, careful, restrained transfer between 2 bodies at the same temp.

What does temp appearing in the denominator signify?

T

qΔS rev


For an isothermal process (constant temperature process) T is in Kelvin

CM1401 CHEMISTRY FOR LIFE SCIENCES 11

T

qΔS rev

)T

Tln(nCΔS

1

2mp,

dqrev = nCp,mdT qrev = ΔHtransition

∆Stransition = ∆Htransition/T

What if T changes from T1 to T2 at constant pressure?

)T

Tln(nCΔS

1

2mp,


If T2 > T1, ΔS > 0

If T2 < T1, ΔS < 0

ΔS & Temperature Change

Calculate ∆S for the process where 5.4 moles of liquid water at 18 ºC is heated to 70 ºC. The entire process takes place at 1 atm. Given: Cp,m of water = 75.3 J K-1 mol-1

13

Constant pressure process

ΔS & Phase Transition

• Melting, freezing, vaporization, condensation are reversible processes taking place at a constant pressure & temp

Fusion and vaporization are endothermic processes, hence ΔS are +ve values. Entropy increases when a substance vaporizes or melts.

T

HΔS

vap

T

HΔS fus


In general, ∆Stransition = ∆Htransition/T

The protein lysozyme, an enzyme that breaks down bacterial cell walls, unfolds at a transition temp of 75.5 oC, and the standard enthalpy of transition is 509 kJ mol-1. Find the ΔSo

trs.

Entropy increases as the lysozyme unravels into a long, flexible chain that can adopt many different conformations, hence dispersal of matter & energy.


∆Stransition = ∆Htransition/T

If you are asked to calculate the change in entropy when 2.00 moles of water is heated from 50 oC to 150 oC, what information do you need?


Entropy Change of Surrounding

T

ΔHΔS

sys

surr

At constant temp & pressure:



Consider the binding of oxidized nicotinamide adenine dinucleotide (NAD+) to enzyme lactate dehydrogenase. It is given that ΔrS

o = -16.8 J K-1 mol-1. Does this mean that the process of NAD+-enzyme complex formation is non-spontaneous?

Given that ΔrHo = -24.2 kJ mol-1, hence at 25 oC,

Hence reaction is spontaneous. The spontaneity is a result of the dispersal of energy that the reaction generates in the surroundings.



We cannot determine the absolute values of H & G, we can only determine ΔH and ΔG.

But we can determine absolute entropy S.


Third Law of Thermodynamics

The entropy of a perfect crystal at 0 K is zero.

A perfect crystal has absolute regular arrangement. Molecular motion virtually ceases. Only 1 way to achieve this perfect order. S = k In W


Starting from S = 0 at T = 0 K, and

we can easily calculate the standard entropy value at a particular temp.

T

HΔS

change phase


)T

Tln(CΔS

1

2p

Calculate So of a substance at 393 K.

So393K = S0K +ΔS

So393K = ΔS =

Cp,solid In(T2/T1) +

Cp,liq In(T3/T2) +

Cp,gas In(T4/T3)

Hfus/Tfus +

Hvap/Tvap +

Equation assumes that the T interval is small enough that the T dependence of Cp can be neglected.

T1 has to be ~0 K, e.g. 0.0001 K


T2 T1 T3 T4

x

ΔrSo = ΣvSo

products – ΣvSoreactants

So : standard entropy; entropy of a substance in its standard state at the temp of interest

v are the stoichiometric coefficients in the chemical equation


ΔrHo = ΣvΔfH

o (products) – ΣvΔfH

o (reactants)

The enzyme carbonic anhydrase catalyses the hydration of CO2 gas in red blood cells:

CO2 (g) + H2O (l) H2CO3 (aq)

Find ΔrSo

at 25oC.

So of CO2 (g) = 213.74 J K-1 mol-1

So of H2O (l) = 69.91 J K-1 mol-1

So of H2CO3 (aq) = 187.4 J K-1 mol-1

ΔrSo = ΣvSo

products – ΣvSoreactants

ΔrSo =

=

Entropy is –ve because a gas is consumed.


Summary

• Spontaneous and Non-spontaneous Processes

• Entropy

• Molecular Interpretation of Entropy

• The Second Law of Thermodynamics

• Entropy Changes

• The Third Law of Thermodynamics

• Standard Entropy


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6. Thermodynamics II