unit 5: thermodynamics ii text: chapter 16 1 st law of thermodynamics: __________________________ q...
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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
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“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
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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.
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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.
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“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
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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 = (+)
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2nd Law of Thermodynamics: ___________________________ ____________________________________________________
in any spontaneous process,
there is always an increase in the entropy of the universe
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Solutions
Generally, when a solid is dissolved in a solvent, entropy increases.
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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.
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Standard Entropies
Larger and more complex molecules have greater entropies.
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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.
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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
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same same(+)(+)(+) (-)(-)(-)
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DSsurr = - DH (J)
T (K)by gaining or losing heat
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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
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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.
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Entropy
• Like total energy, E, and enthalpy, H, entropy is a state function.
• Therefore, S = Sfinal Sinitial
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Second Law of Thermodynamics
In other words:For reversible processes:
Suniv = Ssystem + Ssurroundings = 0
For irreversible processes:
Suniv = Ssystem + Ssurroundings > 0
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Second Law of Thermodynamics
These last truths mean that as a result of all spontaneous processes, the entropy of the universe increases.
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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.
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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 (+)!!
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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.
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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”
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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
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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
( + )
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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#
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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.
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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.
-
-
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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
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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)
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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)
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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
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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(-)
(+)
(+)
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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
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As TEMPERATURE increases from 0 Kelvin, random vibrational motions increase, thus disorder increases.
(NO motion)
- - - - - - - - - - - - - -
- - - -
- - - -
rotational motion
vibrational motion
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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)
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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
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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
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2SO2(g) + O2(g) -----> 2SO3(g)
DHof :
DHorxn =
-297kJ/mol 0kJ/mol -396kJ/mol
2mol(-396kJ) mol
2mol(-297kJ) mol
=
products - reactants
-792kJ -594kJ
DHorxn = -198kJ to use later
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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
products - reactants
=
to use later
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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
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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
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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
products - reactants
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2mol(-394kJ) mol
+ 4mol(-229kJ) mol
2mol(-163kJ) mol
= -788kJ + -916kJ -326kJ
DGorxn = -1378kJ
-1704kJ + 326kJ
DGorxn = products - reactants
=
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chemical change
chemical change
physical change
absorbed (+) or released(-)
always (-)
/mol
J/g or J/mol
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physical change/mol
J/g or J/mol
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chemical change
/mol
/mol
1 mole dissociating
absorbed (+) or released(-)physical change
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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
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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”
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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 =
products - reactants
-242kJ/mol -286kJ/mol
DHo vap = +44kJ/mol
products - reactants
189J/mol∙K 70J/mol∙K
DSo vap = 119J/mol∙K
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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!
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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)
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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 = (+)
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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
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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
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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
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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 =
products - reactants
-35.4kJ/mol -59.8kJ/mol
DHo vap = +24.4kJ/mol endo
25oC
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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
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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
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DHrxn =
DSrxn =
products - reactants
1mol(-471.5kJ) mol
DHo rxn = +39.8kJ endo
1mol(-511.3kJ) mol
products - reactants
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)