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Lecture Notes 4-DELAAT (1).pdf
Lecture Notes 4-DELAAT (1).pdf

Lecture Notes 4-DELAAT (1).pdf

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University of Guelph

Chemistry

CHEM 1050

Richard Delaat

Winter

Description

THERMO - PART 4
SPONTANEITY and GIBBS FREE ENERGY
Spontaneous processes tend to minimize energy and maximize disorder. NOTES:
Some processes are " Enthalpy Driven" with S = _______
(e.g. a ball rolling downhill)
Some processes are "Entropy Driven" with H = _______
(e.g. two inert gases mixing).
Therefore, for a chemical reaction:
If H Rxnis NEGATIVE and S Rxnis POSITIVE, then the rxn will be
_______________ (in the forward direction) as both enthalpy and entropy
_________________ the reaction.
If H Rxnis POSITIVE and S Rxnis NEGATIVE, then the rxn will be
_________________ as both Enthalpy and Entropy _______________ the
reaction (reverse reaction will be spontaneous).
What about the cases where
(a) H and S are BOTH positive, OR
rxn rxn
(b) H and S are BOTH negative
rxn rxn
How do we decide if a reaction is “spontaneous” or not?
This balance between energy and entropy can be handled by a new term, the
__________________, and it gets us away from having to worry about S Surroundings
and hence S Universe
We know that: S UniverseS Surroundings SSystem (1)
We also know that at constant T and P:
S SurroundingsqSurroundings = (2)
(this is a key relationship!)
Therefore, substituting (2) into equation (1):
S = _______________ + S
Universe System
Multiply both sides by –T:
S UniverseH System– TS System
***** We define G = H - TS = –TS Universe,
where G is called Gibbs Free Energy ****
42 NOTES:
Significance of the Sign of G
A reaction at constant temperature and pressure proceeds spontaneously when
G rxn< 0, i.e. for a spontaneous process, rxnwill be NEGATIVE.
If G rxnis N EGATIVE, the rxn is _____________________________.
(i.e., rxn proceeds in the forward direction from reactants to products).
If G rxnis ZERO, the reaction is _______________________________.
If G rxnis POSITIVE, the reaction is NOT
__________________________________________________________.
From here on we will use G which includes both energy anndentropy. For a reaction to
proceed spontaneously, G must be negative. (Recall that the 2 Law of Thermodynamics
says that for a spontaneous process, S > 0.) This means that the direction of spontaneous
change it toward lower free energy (i.e., free energy minimization).
The equation: G = H – TS can have 4 possibilities
Enthalpy Change Entropy Change Spontaneous Reaction?
Endothermic (+) Increase (+) If TS > H
1.
(H > 0) (S > 0) (______________ driven)
Exothermic (–) Increase (+)
2. (H < 0) (S > 0) _______, G < 0
Endothermic (+) Decrease (–) _________, G > 0
3. (H > 0) (S < 0) (But rxn is ______________________.)
4. Exothermic (–) Decrease (–) If TS < H
(H < 0) (S < 0) (_____________ driven)
43 NOTES:
Note: For 2 and 3, spontaneity depends on size of H, S and T.
Some examples: H in kJ, S in J/K)
1. H 2(g)+ Br 2(l) 2HBr (g; H –73 & S +114
2. NH 3(g) + HCl (g) NH Cl 4 (s; H –176 & S –285
3. NH 4l (s) NH 3(g)+ HCl (g; H +176 & S +285
4. 3O 2(g) 2O 3(g; H +285 & S –137
For each example, state when, if ever, the reaction will be spontaneous:
Example 1:
Example 2:
Example 3:
Example 4:
For the examples above, we can plot graphs of G vs. temperature:
44 So, for some reactions you might need to check out the possible competing H and NOTES:
S requirements (i.e., is the reaction enthalpy or entropy driven).
Standard Molar Free Energies of Formation, G f
Definition: G fs the G for the reaction in which 1 MOLE of a substance in its
STANDARD STATE is FORMED from its ELEMENTS in their STANDARD
STATES.
G m,fof any pure element (in its standard state) = _______ (just like
______________) Refer to Appendix C for G valuesf
G m,fis a measure of stability with respect to decomposition of a compound into
its elements under standard conditions.
If Gm,fis negative at a particular temperature, the elements that make up
the compound form the compound spontaneously (i.e., the compound is
__________ stable than the elements).
If Gm,fis positive at a particular temperature, the compound decomposes
spontaneously into its elements (i.e., the compound is _____________ stable
than the elements).
Recall Standard States:
pure liquid; pure solid (most stable form at 1 atm and specified T)
gases (1 atm)
solutes (1M concentration)
Question: The standard free energy of formation (G ) for CS (gis +66.85 kJ/mol.
f 2
What does this say about the molecule CS 2g?
G is a STATE FUNCTION (independent of path)
We can get G rxnEITHER from:
a) Tables of G m,fdata (at a specific temperature, e.g., 25C)
i.e., Use:G Rxn = ________________________
NOTE: This equation can ONLY be used for 25C when
using G values given for 25C.
f
OR
b) H anf S datm (at any temperature).
45 i.e., For ANY temperature we useG Rxn = __________________ NOTES:
This equation assumes that both H and S are independent of temperature
(this is OK at our level).
Example of calculating G Rxn from H Rxnand S Rxn values
Calculate the standard free energy change for the formation of methane at 298 K
given H fnd S mor reactants and products.
Equation: C(s) + 2H2(g) CH 4g)
Find H Rxn:
H Rxn=
H Rxn=
Find S Rxn
S Rxn=
S Rxn=
Find G Rxn:
G Rxn= H Rxn- T SRxn
G Rxn=
G Rxn= G f - 50.7 kJ
Reaction goes at 298 K
Note: At 1000K,
G Rxn= –74.8 kJ – {1000 K × –80.8 J/K × 1 kJ/ 1000 J}
46 G Rxn = G f + 6 kJ NOTES:
Reaction is entropically unfavoured at high T.
Standard Molar Free Energies of Formation
of Aqueous Ions G (aq) f
Recall that for Hfof aqueous ions, it was not possible to separate the cation and
anion contributions. We got around this by defining H (H ,aq) = _____.
f
Similarly, S (H , aq) = _____ and G (H , aq) = ____.
m m,f
Concentration and Free Energy
G Rxncan be obtained from:
a) G vflues (e.g., at 25C) and
G Rxn= nG m,f(Products) - nG m,f(Reactants)
b) H afd S vmlues (at any temperature) and G Rxn = H rxn– TS rxn
We have a problem however. We RARELY have standard conditions (1 atm, 1M).
These conditions are unusual and difficult to set up (except for changes of state at 1 atm
or 1 M). Much more useful if we have an expression that we can use with any
concentrations. The value of G (NOT G) is much more useful and is easily
obtained.
For any reactions, Reactants Products, the sign of GRxntells us if we are going
forward, backward or at equilibrium. Complete the following table on your own:
G Reaction (, or )
Rxn
negative
zero
positive
Hence G Rxnmust depend on concentrations of reactants and products.
You already know this!
Think about the REACTION QUOTIENT, Q (material from CHEM*1040)
Remember, for a reaction with any concentrations or pressures values:
aA + bB cC + dD
Q C QP=
47 NOTES:
Units: molarity (solutions) atm (gases)
Note: Pure solids and liquids 1 within these expressions.
This expression contains ANY values, not just those at equilibrium.
If Q = K, then we are at ________________________.
If Q < K, the reaction wants to go ________, until Q = K (at equilibrium).
i.e, This means that in the expression, the top portion is too small and the bottom
portion is too big. To get Q to equal K, the reaction __________________ [C] and
[D] and ___________________ [A] and [B] We would say the reaction is
____________________.
If Q > K,the top of the expression has values that are too large and the bottom has
values that are too small. We must drive the reaction ________________ to
_____________ C & D, and __________ A & B.
i.e. if Q > K the reaction goes L R until Q = K.
This rxn is _________________ as written, but _________________ in the reverse.
How are G rxn and Q/K related?
We can write:G rxn= RT ln(Q/K) = 2.303 RT log(Q/K)
Q/K ln(Q/K) G rxn Direction Spontaneity?
< 1 – – L R
> 1 L R
1 L R
Spontaneous reactions ALWAYS proceed so as to move
TOWARDS a STATE OF EQUILIBRIUM
i.e. Q K i.e. Q/K 1
Thus G rxnvia the value of Q) depends on concentrations of Reactants and
Products. We need to think carefully about what we mean brxnnd Gºrxn
Using logarithm rules (division within a logarithm is the same as subtracting the
logarithm of the top from the bottom):
48 NOTES:
G rxn = RT ln(Q/K) = RT lnQ – RT lnK
which gives us:
G rxn=
Grxnis the "standard conditions value" (i.e. value appropriate to unit
pressures and concentrations).
RT lnQ term is the "correction factor" to account for non–unit concentrations
or pressures (i.e., non-standard conditions).
N.B. G rxncalculated in this way reflects the ____________________ of a reaction
under a particular set of conditions of concentrations or pressures – and as reactants
are consumed, G rxnwill change and eventually go to 0 as equilibrium is achieved.
G rxnis then a quantitative expression of ______________________.
*******************************************************************
Example: Reaction: N 2(g)+ 3H 2(g) 2NH 3(g)
At 298K,the partial pressures in a reaction mixture areN2 = 0.25 atm, PH2 = 0.55
atm and P NH3 = 0.95 atm. Detemine G .rxn
Problem Solving Strategy:
Step 1: Determine the value of G rxnusing tables of Gm,fdata and
G Rxn = nG m,fProducts) - nG m,fReactants).
Step 2: Determine Q based on the conditions given in the question.
Step 3: Solve for G rxn
Step 1: G Rxn= nG (frods) - nG (Rfacts) =
G Rxn = -33.28 kJ
Step 2: Q = [Prod]/[Reactants] =
Q = 21.7
Step 3: G rxn= G rxn+ RT ln Q
G = -25.7 kJ
rxn
49 NOTES:
G is negative. Therefore, the reaction is __________________ in the forward
rxn
direction. An increase in 3H (g) will continue until Q becomes equal to K (the
equilibrium constant) and hence G will become zero (i.e., at equilibrium).
rxn
To see the relationship between Q and Greview the following set of
rxn
experiments where the partial pressures for the reactants and products are measured
for the following reaction2 N (g) +23H (g) 23H (g) where
G = -33.28 kJ + RT ln {P 2/ P × P 3}
rxn NH3 N2 H2
Note: The equilibrium constant, K, for this reaction at 298K is 6.8×10 .
Expt. PN2 PH2 PNH3 Q RT lnQ

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