unit 5: thermodynamics ii text: chapter 16 1 st law of thermodynamics: __________________________ q...

UNIT 5: THERMODYNAMICS II Text: Chapter 16 1st Law of Thermodynamics: __________________________ q = heat help us

w = work keep track DE = change in internal energy of the DH = change in enthalpy flow of energy qp(heat content @ constant pressure)

“SPONTANEOUS” describes a _______________________ that occurs without ______________________. (for chemical change, products___________________reactants.

energy in the universe is constant.

chemical or physical change

constant input of energy

are more stable than

more

natural

“NON-SPONTANEOUS” describes a change that requires a ____________________________________. (for chemical change, products___________________reactants.

continuous input of energy in order to occur

are less stable than

reactants

products

heat

products

reactants

needs constant energy input to push it uphill

natural

Thermodynamically favored

Not thermodynamically favored

Spontaneous Processes

• Spontaneous processes are those that can proceed without any outside intervention.

• The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously return to vessel B.

© 2012 Pearson Education, Inc.

Spontaneous Processes• Processes that are spontaneous at one temperature may be

nonspontaneous at other temperatures.• Above 0 C, it is spontaneous for ice to melt.• Below 0 C, the reverse process is spontaneous.


“SPONTANEOUS” processes: 1) have a natural direction examples:

2) depend on the conditions (ie, what’s spontaneous at one temp may be non-spont. at another temp)

examples:

3) we cannot predict “speed” of a process with

thermodynamics only WHETHER the process will occur example:

• copper tarnishing• ice melting at room temp• dye mixing in water• marker drying out without a cap

Cdiamond ----> Cgraphite

spontnon-spont

(spont. but very SLOW process)

water freezing @ 25oC or @ -10oC

WHAT IS SPONTANEOUS IN ONE DIRECTION IS NON-SPONTANEOUS IN THE OPPOSITE DIRECTION!

Processes that “FAVOR” spontaneity:1) exothermic

2) an increase in the DISORDER of the system

Spont: Cu + O2 ---> CuO (tarnish)

Non-Spont: CuO ---> Cu + O2

PEreleaseenergy

before

afterDH = (-)

DS = (+)

2nd Law of Thermodynamics: ___________________________ ____________________________________________________

in any spontaneous process,

there is always an increase in the entropy of the universe

Solutions

Generally, when a solid is dissolved in a solvent, entropy increases.


Entropy Changes

• In general, entropy increases when– Gases are formed from

liquids and solids;– Liquids or solutions are

formed from solids;– The number of gas

molecules increases;– The number of moles

increases.


Standard Entropies

Larger and more complex molecules have greater entropies.


Entropy Changes

Entropy changes for a reaction can be estimated in a manner analogous to that by which H is estimated:

S = nS(products) — mS(reactants)

where n and m are the coefficients in the balanced chemical equation.


So while _______ of the universe is CONSTANT ( ) ________ in universe is INCREASING ( ) DS > 0 means “universe is tending toward greater disorder”

DSuniverse = DSsystem + DSsurroundings

To predict whether a process is SPONTANEOUS,

We must know the sign of DSuniverse

If (+), entropy increases, process is SPONT in direction written If (-), entropy decreases, process is SPONT in opposite direction

energyentropy

DS =(+)

(non-spont)

1st law2nd law

same same(+)(+)(+) (-)(-)(-)

DSsurr = - DH (J)

T (K)by gaining or losing heat

Entropy

• Entropy (S) is a term coined by Rudolph Clausius in the nineteenth century.

• Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered, .

qT


Entropy

• Entropy can be thought of as a measure of the randomness of a system.

• It is related to the various modes of motion in molecules.


Entropy

• Like total energy, E, and enthalpy, H, entropy is a state function.

• Therefore, S = Sfinal Sinitial


Second Law of Thermodynamics

In other words:For reversible processes:

Suniv = Ssystem + Ssurroundings = 0

For irreversible processes:

Suniv = Ssystem + Ssurroundings > 0


Second Law of Thermodynamics

These last truths mean that as a result of all spontaneous processes, the entropy of the universe increases.


Xe(g) + 2F2(g) --> XeF4(s) DH = -251kJ

@ Room Temp 25oC @ 100oC

DSsurr = - DH T

+273 +273

373K298KDSsurr = - DH T

= - (-251kJ)

= - (-251kJ) 298K 373K

= 0.842kJ/K

= 0.673kJ/K DSsurr = 842J/K

DSsurr = 673J/K

For cooler surroundings (lower temps)there is more impact on entropy of surr.

To predict SPONTANEITY of a process, which we have seen is TEMPERATURE dependent, we can use 2 different equations:

1) DSuniverse = DSsystem + DSsurroundings for ALL processes (+) must!

2) DG = DH - TDS for a process @ constant T & P (-) must!

“Going Home To Supper”

When DG is (-) then DSuniverse is (+)!!

Free Energy and Temperature

• There are two parts to the free energy equation: H— the enthalpy term– TS — the entropy term

• The temperature dependence of free energy then comes from theentropy term.


G “Gibbs Free Energy” is defined as maximum energy free to do useful work.

DG = (-) SPONTANEOUS (products more stable than reactants)

DG = (+) NON-SPONTANEOUS

(reactants more stable than products)

DG = 0 EQUILIBRIUM (reactants <----> products)

“product favored”

“reactant favored”

“equally favored”

TDS = DH

We can calculate the TEMPERATURE boundary between

SPONTANEOUS & NON-SPONTANEOUS, when DG = 0:

DG = DH - TDS+TDS

0

+TDS

DSDS

T = DH DS

SIGNS OF DH AND DS are IMPORTANT:

1) when DH = (-) and DS = (+) ex: “exothermic”process that has an entropy increase

is ALWAYS spontaneous!

DG = DH - TDS always ( - ) = ( - ) - ( + )

2) when DH = (+) and DS = (-) ex: “endothermic”process that has an entropy decrease(becomes more ordered) is ALWAYS non-spontaneous!

DG = DH - TDS always ( + ) = ( + ) - ( - )

Ex: combustion( - )

HEATout of

SYSTEM:more disorder

HEATinto

SYSTEM:more order

( + )

3) when DH = (-) and DS = (-) (both negative) reaction is spontaneous at LOW temperatures.

DG = DH - TDSthen ( - ) = ( - ) - lowT( - ) larger neg# smaller pos# 4) when DH = (+) and DS = (+) (both positive) reaction is spontaneous at HIGH temperatures.

DG = DH - TDSthen ( - ) = ( + ) - highT( + )

smaller pos# larger neg#

Which are “SPONTANEOUS”? DH = 25kJ DG = DH - TDS

DS = 5.0J/K T = 27oC

+273300.K

--> J

= (25,000J)-(300.K)(5.0J) K= (25000J)-(1,500J)

DG

= 23,500J

= +23.5kJ

Since is DG =(+), then non-spontaneous.

Which are “SPONTANEOUS”? DH = -3.0kJ DG = DH - TDS

DS = 45.0J/K T = 25oC

+273

298K

--> J

= (-3,000J)-(298K)(45.0J) K= (-3000J)-(13,400J)

DG

= -16,400J

= -16.4kJ

Since is DG =(-), then spontaneous.

-

-

Now see other handout with 4 different scenarios: At what Temperature will this be SPONTANEOUS?

DH = -18kJ T = DH DS = -60.J/K DS

= -18,000J-60.J/K

--> J

T = 300K-

Test it: Pick an easy # like 1 to substitute for T in Going Home To Supper

DG = DH - TDS

= -18,000J (1K)(-60.J) K

-

-18,000J= +60.J

DG = -17,940J

Since SPONT at 1 K (which is below 300K), then

Ans: SPONT below 300K

SPONT

RUBBERBAND

stretching is non-spont

DG = DH - TDS

DS must be ____

+- ?

(-)

smaller(-)# greater(+)#

(+) - T=

feels warm so exo

(-)

(+)

(-)

so stretched rubber band has less disorder (less entropy)

contracting is spont“natural”

RUBBERBAND

DG = DH - TDS

DS must be ____

+- ?

(+)

smaller(+)# greater(-)#

(-) - T=

feels cold so endo

(+)

(-)

(+)

so relaxed rubber band has more disorder (more entropy)

CHEMICAL REACTIONS WITH GASES: The relative # of moles of gaseous reactants vs. products dominates “positional entropy” 1) N2(g) + 3H2(g) ---> 2NH3(g)

DS = ___ meaning________________ 2) 4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O(g)

DS = ___ meaning_________________

2mol4mol(-) less disorder

9mol 10mol(+) more disorder

3) CaCO3(s) ---> CaO(s) + CO2(g)

DS = _____ 4) C(s) + 2H2(g) ---> CH4(g)

DS = _____ 5) F2(g) ---> 2F(g)

DS = _____

0mol 1mol

2mol

1mol 2mol

1mol(-)

(+)

(+)

3rd LAW OF THERMODYNAMICS: _________________ ___________________________________________________ meaning that its internal arrangement is “absolutely regular” with “motionless” particles so NO “DISORDER”: S = 0 at zero Kelvin

the entropy of

a perfect crystal at zero Kelvin is zero

As TEMPERATURE increases from 0 Kelvin, random vibrational motions increase, thus disorder increases.

(NO motion)

- - - - - - - - - - - - - -

- - - -

- - - -

rotational motion

vibrational motion

So = “standard molar entropy” or “absolute entropy”

units are _________

always______________________ for an element _________________________________ values _________________________________________

DSoreaction = SnpSo(products) - SnrSo(reactants)

J/mol•K

(+) value (never zero)

So ≠ 0 as for DHof & DGo

f

of So are at Room Temp (298K) & 1atm

(use chart)

Problem: Calculate DSorxn at 25oC for the reaction:

2NiS(s) + 3O2(g) ---> 2SO2 (g) + 2NiO(s)

So ______ _______ _______ _____ DSo

rxn =

53J/mol•K 205J/mol•K 248J/mol•K 38J/mol•K

2mol(248J) mol∙K

+ 2mol(38J) mol∙K

2mol(53J) + mol∙K

3mol(205J) mol∙K

= 496J/K + 76J/K 106J/K + 615J/K

DSorxn = -149J/K

572J/K 721J/K

products - reactants

DGo “standard free energy change” is defined as the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states. (298K and 1 atm) There are 3 ways to solve for DGo

:

1) Given: “Heats of Formation” DHof & Std. Entropy Values So

Find : DHo and DSo

Then substitute answers: DG = DH - TDS

2SO2(g) + O2(g) -----> 2SO3(g)

DHof :

DHorxn =

-297kJ/mol 0kJ/mol -396kJ/mol

2mol(-396kJ) mol

2mol(-297kJ) mol

=


-792kJ -594kJ

DHorxn = -198kJ to use later

2SO2(g) + O2(g) -----> 2SO3(g)

So:

DSo =

248J/mol·K 205J/mol·K 257J/mol·K

2mol(257J) mol∙K

2mol(248J) + mol∙K

1mol(205J) mol∙K

= 514J/K 496J/K + 205J/K

DSorxn = -187J/K

514J/K 701J/K


=

to use later

Now substitute DG o = DHo - TDSo

for DHorxn and DSo

T = 298K (std.state)= (-198kJ)-(298K)(-0.187kJ) K= -198kJ + 55.7kJ

DG = -142kJ -142.3

2) Given DGo for related reactions,

we can use Hess’s Law procedures since free energy G is a state function like enthalpy &

entropy: Find DGo for the following reaction:

2CO(g) + O2(g) -----> 2CO2(g)

Given: 2CH4(g) + 3O2(g) ----> 2CO(g) + 4H2O(g) DGo = -1088 kJ

CH4(g) + 2O2(g) ----> CO2(g) + 2H2O(g) DGo = -801 kJ

flip

2

2CO(g) + 4H2O(g) ----> 2CH4(g) + 3O2(g) DGo = +1088 kJ

2CH4(g) + 4O2(g) ----> 2CO2(g) + 4H2O(g)

2( )

DGo = -1602 kJ

2CO(g) + O2(g) ----> 2CO2(g) DGo = -514 kJ

3) Given DGof for reactants and products.

DGof “standard free energy of formation” is defined as

the change in free energy that accompanies the formation of 1 mole of a substance from its constituent elements with all reactants and products in their standard states. (most stable state @ 298K, 1 atm). Using formula: DGo

rxn = SnpDGof - Snr DGo

f

2CH3OH(g) + 3O2(g) ---> 2CO2(g) + 4H2O(g)

DGof:

DGorxn =

-163kJ/mol 0kJ/mol -394kJ/mol -229kJ/mol


2mol(-394kJ) mol

+ 4mol(-229kJ) mol

2mol(-163kJ) mol

= -788kJ + -916kJ -326kJ

DGorxn = -1378kJ

-1704kJ + 326kJ

DGorxn = products - reactants

=

chemical change

chemical change

physical change

absorbed (+) or released(-)

always (-)

/mol

J/g or J/mol

physical change/mol

J/g or J/mol

chemical change

/mol

/mol

1 mole dissociating

absorbed (+) or released(-)physical change

State Function Name Meaning Signs For Elements

H

enthalpy heat content

(-) exo, heatreleased by system(+) endo, heatabsorbed by system

DHof = 0

kJ/mol

S

entropy disorder (+) more disorder(-) less disorder

So > 0 J/Kmol

G Gibbs free energy

spontaneity(products favored over reactants)

(-) spontaneous(o)at equilibrium(+) non-spontan

DGof = 0

kJ/mol

Note npHof products - nrHo

f reactants] can be used to calculate the

“enthalpy change” during phase change, such as the “heat of vaporization” of a substance. Standard “Heat of Vaporization” Ho

vap when XY(l) ------> XY(g) reactants products And npS

o products - nrS

o reactants] can be used to calculate the

“entropy change” during phase change, such as the “entropy of vaporization” of a substance. Standard “Entropy Change” So

vap when XY(l) ------> XY(g) reactants products And npS

o products - nrS

o reactants] can be used to calculate the

“entropy change” during phase change, such as the “entropy of vaporization” of a substance. Standard “Entropy Change” So

vap when XY(l) ------> XY(g) reactants products

At Room Temp 25oC

At Room Temp 25oC

“evaporating”

Ex: calculate “standard heat of vaporization” Hovap when H2O(l) ----> H2O(g)

IF For H2O(l) Hof = -286kJ/mol and For H2O(g) Ho

f = -242kJ/mol Ho

vap = Ex: calculate “entropy change” So

vap when H2O(l) ----> H2O(g)

IF For H2O(l) So = 70.J/mol K and For H2O(g) Ho

f = 189J/mol K So

vap = Ex: calculate “entropy change” DSo

vap when H2O(l) ----> H2O(g)

IF For H2O(l) So = 70.J/mol K and For H2O(g) So = 189J/mol K

DSo

vap =


-242kJ/mol -286kJ/mol

DHo vap = +44kJ/mol


189J/mol∙K 70J/mol∙K

DSo vap = 119J/mol∙K

Calculating Gibbs Free Energy using enthalpy and entropy changes:

DG = DH - TDS “going home to supper”

kJ kJ K(J/K) careful with units!!! (-) = (-)exo - (+) SPONT at all temps! (+) = (+)endo - (-) NON-SPONT at all temps!

The greater the magnitude of the negative DG for a process, the more thermodynamically favorable it is. The only time that there is ”temperature dependence” is when you have “like signs” for DH

and DS . Substitute and solve. When DG = 0 (both products and reactants are favored equally) then we can calculate the boundary temp between SPONT and NON-SPONT: T = DH

DS

Ex: Normal Melting Pt. S L Boiling Pt. L G(1 atm)

Processes that result in “ENTROPY” changes:

1) phase changes sol--->liq--->gas DS = (+)

2) temp changes T incr, motion incr DS = (+)

3) volume changes (due to Dpressure) P decr, V incr DS = (+)

4) mixing (incr V) DS = (+) or separating (decr V) DS = (-) 5) # atoms/molecule # incr DS = (+) (more rotations & vibrations possible)

6) # moles of gases # moles incr, V incr DS = (+)

Which of the following is SPONTANEOUS at LOW temps only?

a) NH4Br(s) + 188kJ ----> NH3(g) + Br2(l)

b) NH3(g) + HCl(g) ----> NH4Cl(s) + 176kJ

c) 2H2O2(l) ----> 2H2O(l) + O2(g) + 196kJ

d) both a and be) both a and c

endo DH = (+)

exo DH = (-)

exo DH = (-)

DS = (+)

DS = (-)

DS = (+)

Decomposition

SynthesisMnO2

Decomposition

Now if B is spontaneous at all Temps below 347oC,

find DSorxn

We know DH = -176kJ

DG = DH - TDS

And 347oC is the “boundary” Temp when DG = 0 then

0 = DH - TDS

TDS = DH

+TDS +TDS

T T

DS = DH T

DS = DH T

= -176kJ

+273620.K

620.K

-0.28387

DS = -0.284kJ/K

DS = -284J/K

The “enthalpy of vaporization” DHvap is a function of temperature DHvap kJ/mol Temp oC for WATER:45.054 0 43.990 25 43.350 4042.482 6041.585 8040.657 100

Note: as temp increases, the energy required to overcome the inter-molecular forces in the liquid decreases

DHovap = (Standard Heat of

Vaporization) at Rm Temp!!!!

DHvap = Heat of Vaporization at Normal Boiling Point

1) DHovap called “standard heat of vaporization” means @ what

temp? ____

So it is NOT the same as DHvap which is at Normal BPt Problem: Find DHo

vap when bromomethane vaporizes (evaporates at Rm Temp)

CH3Br(l) -------> CH3Br(g) (at 25oC!!)

DHof CH3Br(l) = -59.8kJ/mol DHo

f CH3Br(g) = -35.4kJ/mol

DHovap =


-35.4kJ/mol -59.8kJ/mol

DHo vap = +24.4kJ/mol endo

25oC

2) Ethanol boils @ 78.4oC (N. BPt) and its DHvap= 38.56kJ/mol while its DHo

vap= 42.32kJ/mol

Find the change in entropy during the boiling of ethanol @ 1 atm.Formula?? Hint: C2H5OH(l) <-----> C2H5OH(g) both favored so DG = 0

T = DH DSDS = DH T= 38.56kJ/mol

78.4oC+273

351.4

351K

0.109686DS =

= 0.110kJ/mol·K

110J/mol·K

DHvap

L<-->G

1) Estimate the BPt of SnCl4 at 1 atm. If

SnCl4(l) : DHof = -511.3 kJ/mol

So = 258.6 J/mol·K

SnCl4(g) : DHo

f = -471.5 kJ/mol So

= 366 J/mol·K Equation: SnCl4(l) <-------> SnCl4(g)

Since reactants and products are equally favored at the N BPt (1atm) then DG = 0 (equilibrium) so we can use T = DH

DS

Normal BPt

DHrxn =

DSrxn =


1mol(-471.5kJ) mol

DHo rxn = +39.8kJ endo

1mol(-511.3kJ) mol


1mol(366J) mol·K

1mol(258.6J) mol·K

DSrxn =

T = DH DS

107J/K 107.4

T = 39.8kJ

0.107kJ/K

371.96…

T = 372K -273

T = 99oC (BPt)

unit 5: thermodynamics ii text: chapter 16 1 st law of thermodynamics: __________________________ q...

Documents

natural slide

universe slide

favored slide

o c slide

whats spontaneous

constant pressure spontaneous

pearson education

chemical change