standard entropy change CaF2(s) → CaF2(aq) at 298 K

standard entropy change CaF2(s) → CaF2(aq) at 298 K

 

 

Calculate the standard entropy change

for the dissolution of CaF2 in water:

CaF2(s) → CaF2(aq)

at 298 K, if

S° of CaF2(s) = 68.87 J/K•mol

S° of CaF2(aq) = –80.8 J/K•mol

 

ΔH°_f of CaF2(s) = –1219.6 kJ/mol

ΔH°_f of CaF2(aq) = –1208.09 kJ/mol

 

 

Using the given data, what is the standard free-energy change?

 

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ΔS°_rxn = [생성물들의 S°의 합] – [반응물들의 S°의 합]

( 참고 https://ywpop.tistory.com/7404 )

 

= (–80.8) – (68.87)

= –149.67 J/K

 

 

 

 

ΔH°_rxn = [생성물들의 ΔH°_f 합] – [반응물들의 ΔH°_f 합]

( 참고 https://ywpop.tistory.com/3431 )

 

= (–1208.09) – (–1219.6)

= 11.51 kJ

 

 

 

 

ΔG° = ΔH° – TΔS°

( 참고 https://ywpop.tistory.com/7438 )

 

= (11510 J) – (298 K) (–149.67 J/K)

= 56112 J

= 56.11 kJ

 

또는

반응물, 생성물이 1 mol이므로,

= 56.11 kJ/mol

 

 

 

 

[키워드] ΔG = ΔH – TΔS 기준, G = H – TS 기준