5.63 g C5H6O3(g) 2.50 L 200.℃ 1.63 atm

5.63 g C5H6O3(g) 2.50 L 200.℃ 1.63 atm

 

 

Consider the decomposition of the compound C5H6O3 as follows:

C5H6O3(g) ⇌ C2H6(g) + 3CO(g)

 

When a 5.63 g sample of pure C5H6O3(g)

was sealed into an otherwise empty 2.50 L flask and heated to 200°C,

the pressure in the flask gradually rose to 1.63 atm

and remained at that value.

Calculate K for this reaction.

 

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C5H6O3의 몰질량 = 114.10 g/mol 이므로,

5.63 g / (114.10 g/mol) = 0.0493427 mol C5H6O3

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

PV = nRT 로부터,

( 참고 https://ywpop.tistory.com/3097 )

 

초기 C5H6O3의 압력을 계산하면,

P_C5H6O3 = nRT / V

= [(0.0493427) (0.08206) (273.15+200.)] / (2.50)

= 0.766 atm

 

 

 

ICE 도표를 작성하면,

 

 

 

 

평형에서,

P_tot = 1.63 atm = (0.766 atm – x) + x + 3x

 

x = (1.63 – 0.766) / 3 = 0.288 atm

 

 

 

> P_C5H6O3 = 0.766 – 0.288 = 0.478 atm

> P_C2H6 = 0.288 atm

> P_CO = 3(0.288) = 0.864 atm

 

 

 

K_p = [(P_C2H6) (P_CO)^3] / (P_C5H6O3)

= [(0.288) (0.864)^3] / (0.478)

= 0.389

 

 

 

Kc = Kp / (RT)^Δn 이므로,

( 참고 https://ywpop.tistory.com/6523 )

 

Δn = (1 + 3) – (1) = 3

 

K = Kp / (RT)^Δn

= 0.389 / (0.08206 × (273.15+200.))^3

= 6.65×10^(-6)

 

 

 

답: 6.65×10^(-6)

 

 

 

 

[키워드] C5H6O3 화합물의 분해