298 K ΔG° –4.5 kcal/mol Keq ΔH° –3.2 kcal/mol ΔS°

298 K ΔG° –4.5 kcal/mol Keq ΔH° –3.2 kcal/mol ΔS°

 

 

298 K에서 ΔG°가 –4.5 kcal/mol이면 Keq는 얼마인가?

만약 ΔH° = –3.2 kcal/mol이면 ΔS°는 얼마인가?

 

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(–4.5 kcal/mol) (4.184 kJ / 1 kcal)

= (–4.5) (4.184)

= –18.828 kJ/mol

= –18828 J/mol = ΔG°

 

 

 

ΔG° = –RT lnK

( 참고 https://ywpop.tistory.com/10336 )

 

K = e^[–ΔG° / RT]

= e^[–(–18828 J/mol) / ((8.314 J/mol•K) (298 K))]

= e^[–(–18828) / ((8.314) (298))]

= 1997

= 2.0×10^3

 

 

 

ΔG° = ΔH° – TΔS°

( 참고 https://ywpop.tistory.com/7438 )

 

ΔS° = (ΔH° – ΔG°) / T

= [(–3.2) – (–4.5)] / 298

= 0.00436 kcal/mol•K

= 4.4×10^(-3) kcal/mol•K

= 4.4 cal/mol•K

 

 

 

 

[키워드] ΔG° = –RT lnK 기준문서, ΔG° = ΔH° – TΔS° 기준문서, ΔG = –RT lnK 기준문서, ΔG = ΔH – TΔS 기준문서