 # CHEMICAL THERMODYNAMICS

## Presentation on theme: "CHEMICAL THERMODYNAMICS"— Presentation transcript:

CHEMICAL THERMODYNAMICS
AP Chemistry Chapter 19 Notes CHEMICAL THERMODYNAMICS

1st Law of Thermodynamics Energy in the universe is conserved
REVIEW 1st Law of Thermodynamics Energy in the universe is conserved

value depends on how process takes place (i.e. q, w)
REVIEW Path Function value depends on how process takes place (i.e. q, w)

heat q = mCpT

REVIEW State Function value independent of path - is a defined reference or zero point (i.e. H)

Hf0 of elements in natural form at 250C
Enthalpy - H zero point Hf0 of elements in natural form at 250C

Enthalpy - H types: Hrxn = nHf0(prod) - nHf0(react)

Enthalpy - H types: Hfus , Hvap

a measure of randomness or disorder of a system
ENTROPY “S” a measure of randomness or disorder of a system

Entropy is NOT conserved. The universe seeks maximum disorder.

2nd Law of Thermodynamics for a spontaneous process, entropy increases

Entropy - state function
zero reference: S = 0 for a perfect crystal at absolute zero

Ssolid < Sliquid << Sgas

S0 = standard entropy of elements & cmpds at P = 1 atm; T = 250C units = J/K mol [Appendix L, text]

Calculate S using Hess’ Law S(rxn) = nS0(prod) - nS0(react)

Example: Ca(s) + C(gr) + 3/2 O2  CaCO3(s)

Suniv = Ssys + Ssurr if Suniv > 0 process spontaneous

Suniv = Ssys + Ssurr if Suniv < 0 process ?

Suniv = Ssys + Ssurr if Suniv = 0 process ?

to relate S and H consider

where the water is the system & everything else is the surroundings
H2O(s)  H2O(l) where the water is the system & everything else is the surroundings entropy of water increases, therefore Ssys > 0 the vaporization of water is an endothermic process, therefore heat is flowing from the surroundings to the system & the random motion of the atoms in the surroundings decrease. thus, Ssurr < 0 so, Suniv = Ssys(+) + Ssurr(-) ; which predominates? we know that above 1000C, water spontaneously evaporates and that below 1000C, water spontaneously condenses, thus TEMPERATURE must have an effect on relative importance of Ssys and Ssurr .

Temperature Dependence S = J/mol (1/T) Ssurr = -H/T
introduces central idea that the entropy changes in the surroundings are primarily determined by heat flow. exothermic process increases the entropy of the surroundings, thus is an important driving force for spontaneity. Remember that a system tends to undergo changes that lower its energy. The SIGN on Ssurr depends on the direction of the heat flow. The MAGNITUDE of Ssurr depends on the temperataure.

Ssys Ssurr Suniv spon

Ssys Ssurr Suniv spon

Ssys Ssurr Suniv spon

Ssys Ssurr Suniv spon + + + Y - - - N + - ? D - +
It will be spontaneous if Ssys has a larger magnitude larger than Ssurr.

Ssys Ssurr Suniv spon + + + Y - - - N + - ? D - + ? D
It will be spontaneous if Ssurr has a larger magnitude larger than Ssys.

For a phase change:

GIBB’S FREE ENERGY G

most abstract of thermodynamic state functions

w w = reversible work w2 P V

G = w + PV Definition: w: reversible work PV: pressure/volume work
isothermal, reversible path

G = w + PV + VP at constant P P = 0 so VP = 0 G = w + PV

G = w + PV + VP at constant P G = w + PV at constant V V = 0 so PV = 0 G = w = useful work

must measure G over a process
G cannot be measured must measure G over a process

ZERO REFERENCE G = 0 for elements in stable form under Standard Thermodynamic Conditions T = 25oC ; P = 1 atm

Gf0 = standard Free Energy of Formation from the elements
Appendix L, text

G follows Hess’ Law: G0 (rxn) = nGf0(p) - nGf0(r)

Summary of Laws of Thermodynamics Zeroth Law: Heat Gain = Heat Loss

Summary of Laws of Thermodynamics First Law: Law of Conservation of Energy

Summary of Laws of Thermodynamics Second Law: Defines Entropy

Summary of Laws of Thermodynamics Third Law: Defines Absolute Zero

GIBB’S HELMHOLTZ EQUATION

combine G H T

G = -aT + H G, H are state functions, thus “a” must be a state function G = -S(T) + H

Gibb’s Helmholtz Equation G = H - TS

Units on the State Functions

thus, a process is spontaneous
if and only if G is negative

enthalpy (minimum energy)
Spontaneity controlled by enthalpy (minimum energy)

enthalpy entropy (maximum disorder)
Sponaneity controlled by enthalpy entropy (maximum disorder)

Sponaneity controlled by enthalpy entropy both

Predict Spontaneity IF H(-) and S(+) G = -H - T(+S) G < 0, => spontaneous

Predict Spontaneity IF H(+) and S(-) G = +H - T(-S) G > 0, => NOT spontan

Summary of Spontaneity
H S G Spont. yes no or ? or ?

Uses of the Gibb’s Helmholtz Equation

1. Find the molar entropy of formation for ammonia.

2. Elemental boron, in thin fibers, can be made from a boron halide:
BCl3(g) + 3/2 H2(g) ->B(s) + 3HCl(g)

Calculate: H0, S0 and G0. Spontaneous? Driving force?

3. Using thermodynamic information, determine the boiling point of bromine.

Thermodynamic Definition of Equilibrium Geq = 0

by definition G = H - TS & H = E + PV

thus, G = E + PV - TS take derivative of both sides
dG = dE + PdV + VdP - TdS - SdT

for a reversible process
TdS = q

derivative used for state function while
partial derivative used for path function

if the only work is PV work of expansion PdV = w

First Law of Thermodynamics E = q - w dE = q - w = 0

thus q = w or TdS = PdV

by substitution dG = 0 + PdV + VdP - PdV - SdT or dG = VdP - SdT

let’s assume we have a gaseous system at equilibrium, therefore, examine Kp at that constant temperature

at constant T SdT = 0 thus dG = VdP

but

if “condition 2” is at standard thermodynamic conditions, then G2 = G0 and P2 = 1 atm

thus

determine G for aA + bB  cC + dD where all are gases

G = Gprod - Greact = cGC + dGD - aGA - bGB

but and likewise for the others

and

but

Thus, DG = DG0 + (RT) ln Q

But aA + bB  cC + dD G = 0

thus or in general

THERMODYNAMICS & EQUILIBRIUM

information, determine
3. Using thermodynamic information, determine the boiling point of bromine.

Thermodynamics and Keq
FACT: Product-favored systems have Keq > 1.

Thermodynamics and Keq
Therefore, both ∆G˚rxn and Keq are related to reaction favorability.

Thermodynamics and Keq
Keq is related to reaction favorability and thus to ∆Gorxn. The larger the value of K the more negative the value of ∆Gorxn

Thermodynamics and Keq
∆Gorxn= - RT lnK where R = J/K•mol

Thermodynamics and Keq
∆Gorxn = - RT lnK Calculate K for the reaction N2O4 2 NO2 ∆Gorxn = +4.8 kJ K = 0.14 When ∆G0rxn > 0, then K < 1

∆G, ∆G˚, and Keq ∆G is change in free energy at non-standard conditions. ∆G is related to ∆G˚ ∆G = ∆G˚ + RT ln Q where Q = reaction quotient

∆G, ∆G˚, and Keq When Q < K or Q > K, reaction is spontaneous.
When Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln K

Product favored reaction
∆G, ∆G˚, and Keq Product favored reaction –∆Go and K > 1 In this case ∆Grxn is < ∆Gorxn , so state with both reactants and products present is MORE STABLE than complete conversion.

∆G, ∆G˚, and Keq Product-favored reaction. 2 NO2  N2O4
∆Gorxn = – 4.8 kJ Here ∆Grxn is less than ∆Gorxn , so the state with both reactants and products present is more stable than complete conversion.

Thermodynamics and Keq Overview
∆Gorxn is the change in free energy when reactants convert COMPLETELY to products.

Keq is related to reaction favorability.
When ∆Gorxn < 0, reaction moves energetically “downhill”

4. For the following reaction, calculate the temperature at which the reactants are favored.

THERMODYNAMICS OF CHEMICAL REACTIONS

5. How much useful work can be obtained from an engine fueled with 75
5. How much useful work can be obtained from an engine fueled with 75.0 L of hydrogen at 10 C at 25 atm?

reacting silver with water.
6. The reaction to split water into hydrogen and oxygen can be promoted by first reacting silver with water. 2 Ag(s) + H2O(g) Ag2O(s) + H2(g) Ag2O(s)  2 Ag(s) + 1/2 O2(g)

Calculate H0, S0 and G0 for each reaction.

Combine the reactions and calculate H0 and G0 for the combination.
Is the combination spontaneous?

At what temperature does the second reaction become spontaneous?

7. The conversion of coal into hydrogen for fuel is:
C(s) + H2O(g)  CO(g) + H2(g)

Calculate G0 and Kp at 250C. Is the reaction spontaneous?

At what temperature does the reaction become spontaneous?
Calculate the temperature at which K = 1.0 x 10-4.

8. The generation of nitric acid in the upper atmosphere might destroy the ozone layer by: NO(g) + O3(g)  NO2(g) + O2(g)

Calculate G0 (reaction)
and K at 250C.

Similar presentations