Chapters 17,18,19

W

# of micro states

Calculate ΔG^0 for the following reactions: (a) 2Mg(s) + O2 (g) ---> 2MgO(s) (b) 2SO2 (g) + O2 (g) ---> 2SO3(g) (c) 2C2H6(g) + 7O2(g) ---> 4CO2(g) + 6H2O (l)

(a) -1139 kJ/mol (b) -140.0 kJ/mol (c) -2935 kJ/mol

Using the data in Appendix 3, calculate the standard entropy changes for the following reactions at 25C: (a) H2(g) + CuO(s) ---> Cu(s) + H2O(g) (b) 2Al(s) + 3ZnO(s) ---> Al2O3(s) + 3Zn(s) (c) CH4(g) +2O2(g) ---> CO2(g) + 2H2O(l)

(a) 47.5 J/K-mol (b) -12.5 J/K-mol (c) -242.8 J/K-mol

Find the temperatures at which reactions with the following ΔH and ΔS values would become spontaneous: (a) ΔH = -126 kJ/mol, ΔS = 84 J/K-mol (b) ΔH = -11.7 kJ/mol, ΔS = -105 J/K-mol

(a) at all temperatures (b) below 111K

State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and explain your predictions: (a) PCl3(l) + Cl2(g) ---> PCl5(s) (b) 2HgO(s) ---> 2Hg(l) + O2(g) (c) H2(g) ---> 2H(g) (d) U(s) + 3F2(g) ---> UF6(s)

(a) ΔS < 0 (b) ΔS > 0 (c) ΔS > 0 (d) ΔS <0

The equilibrium constant Kp for the reaction: H2 (g) + CO2(g) <---> H2O(g) + CO(g) is 4.40 at 2000K. (a) Calculate ΔG^0 for the reaction (b) Calculate ΔG for the reaction when partial pressures are P(H2) = 0.25 atm, P(CO2) = 0.78 atm , P(H2O) = 0.66 atm, P (CO) = 1.20 atm

-24.6 kJ/mol

The equilibrium constant Kp for the reaction: CO(g) + Cl2(g) <------> CoCl2(g) is 5.62 x 10^35 at 25C. Calculate ΔGf^0 for COCl2 and 25C

-341 kJ/mol

Standard Entropy

1 atm 298 K (25C)

Boltzmann Constant

1.38x10^23 J/K

Calculate ΔG and Kp for the following equilibrium reaction at 25C: 2H2O(g) <----> 2H2(g) + O2(g)

4.572 x 10^2 kJ/mol

For the auto ionization of water at 25C: H2O(l) <------> H+ (aq) + OH- (aq) Kw is 1.0 x 10^-14 ; what is ΔG for process?

8.0 x 10^1 kJ/mol

how to calculate E^0

E^0cell= E^0 cathode - E^0anode

S

Entropy

Third Law of Thermodynamics

Entropy of a perfect crystalline substance is zero at absolute zero of temperature

S=klnW

Entropy=(1.38x10^-23 J/K)lnW

E^0=( RT / nF) lnK

F= 95600

Gibbs Free Energy Change in Gibbs Free Energy

G= H- TS ΔG=ΔH-TΔS

ΔG>0

The reaction is non-spontaneous. The reaction is spontaneous in the opposite direction

ΔG<0

The reaction is spontaneous in the forward direction

ΔG=0

The system is at equilibrium. There is no net change

EMF (electromotive force)

a measure of voltage

Arrange the following substances (1 mole of each) in order of increasing entropy at 25C: (a) Ne (g) (b) SO2 (g) (c) Na (s) (d) NaCl (s) (e) H2 (g)

c d e a b

anode

electrode at which oxidation occurs

cathode

electrode at which reduction occurs

what does II mean?

salt bridge

E^0

standard reduction potential; voltage associate with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm

Electrochemistry

the branch of chemistry that deals with the interconversion of electrical energy and chemical energy

Second Law of Thermodynamics

the entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process

Galvanic Cell

the experimental apparatus for generating electricity through the use of a spontaneous reaction

cell voltage or cell potential

voltage across the electrodes of a galvanic cell

How do we calculate the standard free energy of a reaction?

ΔG(rxn)= [ΔGmoles of products] - [ΔGmoles of reactants]

Relationship between ΔG and ΔG^0

ΔG= ΔG^0 + RT ln Q

ΔG at equilibrium

ΔG^0 = -RT ln K

The standard enthalpy of formation and the standard entropy of gaseous benzene are 92.3 kJ/mol and 269.2 J/K-mol. Calculate ΔH and ΔS and ΔG for the process at 25C: C6H6(l) ----> C6H6(g)

ΔH = 33.89 kJ/mol ΔS = 96.4 J/mol ΔG = 5.2 kJ/mol

How do we calculate the standard enthalpy of a reaction?

ΔS(rxn)= [cS(C) + dS(D)] - [aS(A) + bS(B)] or ΔS(rxn)= { (sum of products) - { (sum of reactants)

Relationship between ΔS(surr) and ΔH(sys)

ΔS(surr)=-ΔH(sys) / T

Consider the following reaction at 298K: 2H2(g) + O2(g) ---> 2H2O(l) ΔH^0= -571.6 kJ/mol Calculate ΔS(sys), ΔS(surr), and ΔS(univ) for reaction

ΔS(sys) = -327 J/K-mol ΔS(surr) = 1918 J/K-mol ΔS(univ) = 1591 J/K-mol

Equilibrium Process

ΔS(univ)=ΔS(sys)+ΔS(surr)=0

Spontaneous Process

ΔS(univ)=ΔS(sys)+ΔS(surr)>0