chemical thermodynamics the second law of thermodynamics: the is an inherent direction in which any...

CHEMICAL THERMODYNAMICS

The Second Law of Thermodynamics:

The is an inherent direction in which any system not at equilibrium moves

Processes that are spontaneous in one direction are not spontaneous in the reverse direction


Spontaneity, Enthalpy, and Entropy

•Reactions that are exothermic are generally spontaneous, H < 0

•Reactions that are not exothermic may also be spontaneous

H = 0


H > 0


Spontaneity must, therefore, be a function of the degree of randomness in a system.

Spontaneity, Enthalpy, and Entropy

Entropy (S) is a “state function” that describes the randomnessin a system such that, S = S final - S initial


The Second Law of Thermodynamics tells us that

In any spontaneous process, there is always an increase in the entropy of the universe:

S universe = S system + S surroundings > 0


A Molecular Interpretation of Entropy

translational

vibrational

rotational

The Third Law of Thermodynamics: S(0 K) = 0The entropy of the lattice increases with temperature because the number of

possible energy states in which the molecules or atoms are distributed is larger


Calculation of Entropy Changes

The standard entropy (S) is expressed in units of J/mol-K

For any given reaction: aA + bB + … cC + dD + ...

S = [cS(p) + dS (Q) + …] - [ aS(A) + bS(B) + …]

Sample exercise: Calculate the S for the synthesis of ammoniafrom N2 and H2

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

S = [2S(NH3)]- [ S(N2) + S(H2)]

S = [2(192 J/mol-K)]- [ 191.5 J/mol-K+ 3(130.58 J/mol-K)]

S = -198.2 J/K

Ice melting water condensing mixing salt water diffusion butane burning CH4g + 2O2g→ CO2g + 2H2Og


Gibbs Free Energy

Whether a reaction occurs spontaneously is determined by the changes in enthalpy and entropy for that reaction such that:

G = H -TS

The change in free energy is therefore:

G = H - TS•If G is negative, the reaction is spontaneous in the forward direction

•If G is positive, the reaction is not spontaneous in the forward direction Work must be supplied from the surroundings to make it occur.

•If G is zero, the reaction is at equilibrium, there is no driving force

I get it!!


aA + bB + … cC + dD + ...

G = [cGf (p) + dGf (Q) + …] - [aGf (A) + bGf (B) + …]

The standard free energy change ( G) for any reaction:

Calculation of G using standard free energy change of formation


Sample exercise: Using the data from Appendix C, calculate the G using H and S for the following reaction:

BaO(s) + CO2(g) BaCO3(g)

H = [-1216.3 kJ/mol] - [ -553.5 kJ/mol - 393.5 kJ/mol] = -269.3 kJ

G = [-1137.6 kJ/mol] - [ -525.1 kJ/mol - 394.4 kJ/mol] = -218.1 kJ

S = [112.1 J/mol-K] - [ 70.42 J/mol-K - 213.6 J/mol-K] = -171.9 J/mol-K

G = H - TS

G = -269.3 kJ - 298 K (-171.9 J/mol-K)(1 kJ/ 1000 J )

G = -218.07 kJ/mol

Free Energy and the Equilibrium Constant

What happens when you can’t describe G under standard conditions ? G = G ° + RT ln Q (R 8.314 J/K-mol)

When G = 0 then G ° = -RT ln K

G ° is negative: K > 1

G ° is zero: K = 1

G ° is positive: K < 1

Calculate G at 298 K for the following reaction if the reaction mixture consists of 1.0 atm N2, 3.0 atm H2 ,and 1.0 atm NH3 N2(g) + 3H2 2NH3

Q = (1.0)2/ (1.0)(3.0)3 = 3.7 x 10-2

G = G ° + RT ln Q where R 8.314 J/K-mol

G = -33.32 kJ + (8.314x 10-3 kJ/K-mol)(298 K)ln(3.7 x 10-2)

G = -41.49 kJ


Ek= mv2

E = E final - E initial

E = q + w

if V = 0 then, E = q v

if P is constant, H = E + P V or, H = q p

H E

H= H products - H reactants

H fH rxn = n (products) - H f m (reactants)

q = n CT where C is J/mol-C

q = m ST where is C J/g-C

Chapter 5 Hess’s Law

1st Law of Thermodynamics


Chapter 19S = S final - S initial

S universe = S system + S surroundings > 0

For any given reaction: aA + bB + … cC + dD + ...

S = [cS(p) + dS (Q) + …] - [ aS(A) + bS(B) + …] where S= J/mol-K

G = H - TS •If G = spontaneous

•If G =not spontaneous•If G = zero, K = 0

aA + bB + … cC + dD + ...

G = [cGf (p) + dGf (Q) + …] - [aGf (A) + bGf (B) + …]

for any reaction:

G = G ° + RT ln Q where R 8.314 J/K-mol

if G ° = 0 then G ° = RT ln K

G ° is negative: K > 1G ° is zero: K = 0

G ° is positive: K > 1

2nd Law of Thermodynamics

3rd Law of Thermodynamics

chemical thermodynamics the second law of thermodynamics: the is an inherent direction in which any...

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