Question

A process at constant T and P can be described as spontaneous if ΔG < 0 and nonspontaneous if ΔG > 0. Over what range of temperatures is the following process spontaneous? Assume that gases are at a pressure of 1 atm. (Hint: Use the data below to calculate ΔH and ΔS [assumed independent of temperature and equal to ΔH° and ΔS°, respectively] and then use the definition of ΔG.)

	3 PbO2(s)	→	Pb3O4(s)	+	O2(g)
ΔHf° (kJ mol-1)	-277.0		-718.4
S° (J K-1 mol-1)	69.0		211.3		205.1

ΔH° =  kJ

ΔS° =  J K^-1

The calculated value of T at which ΔG° = 0 is  K.

Reaction is spontaneous _________above this temperaturebelow this temperatureat all temperaturesat no temperature.

Answer 1

1)

Given:

Hof(PbO2(s)) = -277.0 KJ/mol

Hof(Pb3O4(s)) = -718.4 KJ/mol

Hof(O2(g)) = 0.0 KJ/mol

Balanced chemical equation is:

3 PbO2(s) ---> Pb3O4(s) + O2(g)

ΔHo rxn = 1*Hof(Pb3O4(s)) + 1*Hof(O2(g)) - 3*Hof( PbO2(s))

ΔHo rxn = 1*(-718.4) + 1*(0.0) - 3*(-277.0)

ΔHo rxn = 112.6 KJ

Answer: 112.6 KJ

2)

Given:

Sof(PbO2(s)) = 69.0 J/mol.K

Sof(Pb3O4(s)) = 211.3 J/mol.K

Sof(O2(g)) = 205.1 J/mol.K

Balanced chemical equation is:

3 PbO2(s) ---> Pb3O4(s) + O2(g)

ΔSo rxn = 1*Sof(Pb3O4(s)) + 1*Sof(O2(g)) - 3*Sof( PbO2(s))

ΔSo rxn = 1*(211.3) + 1*(205.1) - 3*(69.0)

ΔSo rxn = 209.4 J/K

Answer: 209.4 J/K

3)

ΔGo = 0.0 KJ/mol

ΔHo = 112.6 KJ/mol

ΔSo = 209.4 J/mol.K

= 0.2094 KJ/mol.K

use:

ΔGo = ΔHo - T*ΔSo

0.0 = 112.6 - T *0.2094

T = 538 K

Answer: 538 K

4)

As T increases above 538 K, ΔGo becomes negative.

So, the reaction become spontaneous

Answer: Above