298 K Kp 0.113 5.40 kJ/mol P_NO2 = 0.122 atm P_N2O4 = 0.453 atm ΔG

298 K Kp 0.113 5.40 kJ/mol P_NO2 = 0.122 atm P_N2O4 = 0.453 atm ΔG

 

 

The equilibrium constant (Kp)

for the following reaction is 0.113 at 298 K,

which corresponds to

a standard free-energy change of 5.40 kJ/mol.

In a certain experiment,

the initial pressures are P_NO2 = 0.122 atm

and P_N2O4 = 0.453 atm.

Calculate ΔG for the reaction at these pressures and

predict the direction of the net reaction at equilibrium,

take the gas constant (R) value as 8.314 J/K•mol.

 

---------------------------------------------------

 

N2O4(g) ⇌ 2NO2(g)

 

Kp = (P_NO2)^2 / (P_N2O4)

 

 

 

Q = (P_NO2)^2 / (P_N2O4)

( 참고: 반응비 https://ywpop.tistory.com/10539 )

 

= (0.122)^2 / (0.453)

= 0.0329

 

 

 

Q < K 이므로,

반응은 정반응(오른쪽) 진행.

 

 

 

ΔG = ΔG° + RTlnQ

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

 

= (5400 J/mol) + (8.314 J/mol•K) (298 K) ln(0.0329)

= (5400) + (8.314) (298) ln(0.0329)

= –3059 J/mol

---> ΔG = –3.06 kJ/mol

 

 

 

ΔG < 0 이므로,

반응은 정반응(오른쪽) 진행.

( 참고 https://ywpop.tistory.com/7438 )

 

 

 

 

[키워드] 자유 에너지 계산 기준문서, 자발적 기준문서, 자발적 방향 기준문서, 자발적 반응 기준문서