Question

1 For the process 2F(g)→F2(g),

Select one:

a. ΔH is + and ΔS is + for the reaction.ΔH is + and ΔS is + for the reaction.

b. ΔH is + and ΔS= 0 for the reaction.ΔH is + and ΔS= 0 for the reaction.

c. ΔH is – and ΔS is – for the reaction.ΔH is – and ΔS is – for the reaction.

d. ΔH is – and ΔS is + for the reaction.ΔH is – and ΔS is + for the reaction. e. ΔH is + and ΔS is – for the reaction.ΔH is + and ΔS is – for the reaction.

2 For a certain reaction, ΔH∘= +13.3  kJ and ΔS∘= −233J/K. If n=2, calculate For a certain reaction, ΔH∘= +13.3  kJ and ΔS∘= −233J/K. If n=2, calculate E°cell for the reaction at 25°C. (F = 96485 C/mol)

3 For the reaction  ΔH°=+61.14 kJ and  ΔS° = +132 J/K at 25°C.. At what minimum temperature will the reaction be spontaneous?

Answer 1

Q1.

Note that: there will be a Bond formation for 2F ---> F-F). Therefore, this is typically an exothermic reaction.

dH is therefore, NEGATIVE, since it releases heat

dS --> this can be considered NEGATIVE as well, since this decreases amount of chaos, F-F is ordered, whereas 2 F molecules are less ordered, so entropy decreases

best answer is

c. ΔH is – and ΔS is – for the reaction

Q2.

Apply:

dG = -n*F*E°cell

dG = dH - T*dS

substitute data

dH - T*dS =  -n*F*E°cell

(13.3*10^3) - (25+273)*(-233) = -2*96485*E°cell

Solve for E° cell

82734 = -2*96485*E°cell

E°cell = 82734 /(-2*96485) = -0.4287 V

E°cell = -0.4287 V

Q3.

For:

dG = dH - T*dS ... this is the free energy

in order to be spontaneous, dG < 0

so:

dG = dH - T*dS

dH - T*dS <0

dH < T*dS

(61.14*10^3) < T*(132)

(61.14*10^3) /132 < T

463.181 < T

T > 463.181 K

T must be higher 463.181 K