Boudouard 2CO(g) ⇌ C(s) + CO2(g) 5.07 MPa 1.00 mol% CO

Boudouard 2CO(g) ⇌ C(s) + CO2(g) 5.07 MPa 1.00 mol% CO

 

 

Consider the Boudouard reaction:

2CO(g) ⇌ C(s) + CO2(g)

Thermodynamic data on these gases are given in Appendix C.

You may assume that ΔH° and ΔS° do not vary with temperature.

a) At what temperature will an equilibrium mixture of

101.3 kPa total pressure contain 1.00 mol% CO?

b) At what temperature will an equilibrium mixture of

5.07 MPa total pressure contain 1.00 mol% CO?

 

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▶ Thermodynamic data on these gases

> CO(g): H = –110.5, G = –137.2, S = 197.9

> C(s): H = 0, G = 0, S = 5.69

> CO2(g): H = –393.5, G = –394.4, S = 213.6

 

 

 

ΔH° = [(0) + (–393.5)] – [2(–110.5)]

= –172.5 kJ/mol

( 참고 https://ywpop.tistory.com/3431 )

 

 

 

ΔS° = [(5.69) + (213.6)] – [2(197.9)]

= –176.51 J/mol•K

= –0.17651 kJ/mol•K

( 참고 https://ywpop.tistory.com/7404 )

 

 

 

> Kp = P_CO2 / (P_CO)^2

 

> ΔG° = –RT lnK

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

 

> ΔG° = ΔH° – TΔS°

( 참고 https://ywpop.tistory.com/7438 )

 

---> –RT lnK = ΔH° – TΔS°

 

 

 

1.00 mol% CO means

P_CO = 0.01P_tot

( 참고 https://ywpop.tistory.com/48 )

 

---> P_CO2 = 0.99P_tot

 

 

 

b)

> P_CO = 0.01 × 5070 kPa = 50.7 kPa

> P_CO2 = 0.99 × 5070 kPa = 5019.3 kPa

 

 

 

Kp = P_CO2 / (P_CO)^2

= 5019.3 / (50.7)^2

= 1.95266

 

 

 

–RT lnK = ΔH° – TΔS°

–(8.314/1000) (T) ln(1.95266) = (–172.5) – T(–0.17651)

( R = 8.314 J/mol•K )

 

–0.005563667T = (–172.5) + 0.17651T

 

(–0.005563667 – 0.17651)T = (–172.5)

 

T = (–172.5) / (–0.005563667 – 0.17651)

= 947 K

 

 

 

 

[ 관련 예제 https://ywpop.tistory.com/20153 ] a) 101.3 kPa