평형에서의 온도. WO3 + 3H2 → W + 3H2O

평형에서의 온도. WO3 + 3H2 → W + 3H2O

 

 

Tungsten is usually produced by the reduction of WO3 with hydrogen.

WO3(s) + 3H2(g) → W(s) + 3H2O(g)

 

Consider the following data.

.................... WO3(s) ... H2O(g)

Hf° (kJ/mol) ... -839.9 ... -241.8

Gf° (kJ/mol) ... -763.1 ... -228.6

 

a) Calculate ΔS° at 25℃ for the reaction indicated by the equation above.

b) What is the temperature at which G° equals zero for this reaction at 1 atm pressure?

 

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a)

ΔH° = [(0) + 3(-241.8)] - [(-839.9) + 3(0)]

( 식 설명 http://ywpop.tistory.com/3431 )

= +114.5 kJ

= +114.5 kJ/mol (텅스텐 1 mol에 대한 엔탈피이므로,)

 

 

ΔG° = [(0) + 3(-228.6)] - [(-763.1) + 3(0)]

( 식 설명 http://ywpop.tistory.com/7393 )

= +77.3 kJ

= +77.3 kJ/mol (텅스텐 1 mol에 대한 자유에너지이므로,)

 

 

ΔG° = ΔH° - TΔS° 이므로,

( 식 설명 http://ywpop.tistory.com/7438 )

ΔS° = [ΔH° - ΔG°] / T

= [(+114.5) - (+77.3)] / (25+273.15)

= +0.1248 kJ/mol•K

= +124.8 J/mol•K

 

 

 

b)

ΔG° = ΔH° - TΔS° 이고,

평형에서 ΔG° = 0 이므로,

T = ΔH° / ΔS°

= +114.5 / +0.1248

= 917.5 K