A process at constant T and P can be described as spontaneous if ?G < 0 and nons

Question

A process at constant T and P can be described as spontaneous if ?G < 0 and nonspontaneous if ?G > 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 ?H and ?S [assumed independent of temperature and equal to ?H° and ?S°, respectively] and then use the definition of ?G.)

2 SnO2(s) ? 2 SnO(s) + O2(g)

?Hf° (kJ mol-1) -580.7 -285.0 S° (J K-1 mol-1) 52.3 56.5 205.1

?H° = kJ ?S° = J K-1 The calculated value of T at which ?G° = 0 is K. Reaction is spontaneous .

Explanation / Answer

Given:

Hof(SnO2(s)) = -580.7 KJ/mol

Hof(SnO(s)) = -285.0 KJ/mol

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

Balanced chemical equation is:

2 SnO2(s) ---> 2 SnO(s) + O2(g)

?Ho rxn = 2*Hof(SnO(s)) + 1*Hof(O2(g)) - 2*Hof( SnO2(s))

?Ho rxn = 2*(-285.0) + 1*(0.0) - 2*(-580.7)

?Ho rxn = 591.4 KJ

Given:

Sof(SnO2(s)) = 52.3 J/mol.K

Sof(SnO(s)) = 56.5 J/mol.K

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

Balanced chemical equation is:

2 SnO2(s) ---> 2 SnO(s) + O2(g)

?So rxn = 2*Sof(SnO(s)) + 1*Sof(O2(g)) - 2*Sof( SnO2(s))

?So rxn = 2*(56.5) + 1*(205.1) - 2*(52.3)

?So rxn = 213.5 J/K

?Ho = 591.4 KJ/mol

?So = 213.5 J/mol.K

= 0.2135 KJ/mol.K

use:

?Go = ?Ho - T*?So

0.0 = 591.4 - T * 0.2135

T = 2770 K

Answer: 2770 K

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.