Chapter 16

Chapter 16

• Spontaneous processes– Occur without assistance– Fast or slow

– Thermodynamics predicts if reactions will occur but only focuses on the initial and final states

• Entropy (S)– Randomness of a system

S(gas) >> S(liq.) > S(solid)

• 1st law of thermodynamics– Energy of the universe is conserved

• 2nd law of thermodynamics– Entropy of the universe increases for any

spontaneous process Suniv = Ssys + Ssurr

Suniv = + spont. Suniv = - spont. In opposite direction

• Effect of temperature– H2O (l) H2O (g) more entropy as gas Ssys = + Ssurr = - Ssurr depends on the H and T

Ssurr = - H / T T must be in Kelvin

• Free Energy (G)

G = H - TS all are system

G = - spont. = + rev. spont. = 0 equil.

G / T = - H / T + S

- G / T = Ssurr + S = Suniv

– So G has to be neg. to give a pos. Suniv

– From the equation we can predict when reactions will be spontaneous given H and S

G = H - TS

H S G

+ + - at high T

+ - +

- - - at low T

- + -

• Entropy in chemical reactions– We can predict entropy for reactions based on

the states of the substances and/or the amount of gas particles present.

– Na2CO3(s) Na2O(s) + CO2(g) + S– 2H2(g) + O2(g) 2H2O(g) - S

• 3rd law of thermodynamics– Entropy of a perfect crystal at 0K is zero

• We can calculate S by using the following equationSo =npSoprod - nrSoreact

Unlike H all chemicals will have a value

Pg 800

• Free energy and chemical reactions– Standard Free Energy Change Go

Go = Ho - TSo

What’s the free energy for an exothermic reaction that releases 45 kJ of energy and has an increase in entropy of 250 J at 300K?

Go = -45000 J – (300K)(250J) = -120000 J

– “Hess’s Law” method -- add reactions together to get an overall reaction and add the Go to get an overall Go

– Calculate the Go for the following reaction

– C diamond + O2 (g) CO2 (g) Go = -397 kJ– CO2 (g) C graphite + O2 (g) Go = 394 kJ

– C diamond C graphite

Go = -397 kJ + 394 kJ = -3 kJ

– Standard free energy of formation Gof

Gof for elements is zero

Go = np Gof prod - nr Gof react

2CH3OH (g) + 3O2 (g) 2CO2 (g) + 4H2O (g)Go = ?

Go = 2n(-394kJ/n)+4n(-229kJ/n)-3n(0)-2n(-163kJ/n)

Go = -1378 kJ

• Dependence on pressure– Since pressure effects entropy it also effects

free energy– G = Go + RT ln(P) R = 8.3145 J/nK

G = Go + RT ln(Q) Q = (Ppa / Prb)

• Meaning of Free Energy G tells us which side of a reaction is favored

but it doesn’t tell us that the reaction will go to completion.

– The lowest G is at the equilibrium point

• Free energy and Equilibrium G = Go + RT ln(Q)– At equilibrium G = 0 and Q = K

– 0 = Go + RT ln(K) Go = - RT ln(K)

Go = 0 then K = 1 and no shift will occur Go = - then K > 1 and the reaction shifts right Go = + then K < 1 and the reaction shifts left

Free energy and work

wmax = G