0.050 mol CH2O 500 mL CH2O(g) = H2(g) + CO(g) 0.066 mol/L

0.050 mol CH2O 500 mL CH2O(g) = H2(g) + CO(g) 0.066 mol/L

 

 

A 0.050 mol sample of formaldehyde vapor, CH2O,

was placed in a heated 500 mL vessel and some of it decomposed.

The reaction is: CH2O(g) ⇌ H2(g) + CO(g)

At equilibrium, the CH2O(g) concentration was 0.066 mol/L.

Calculate the value of Kc for this reaction.

 

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초기농도, C = 0.050 mol / 0.500 L = 0.10 M

 

 

 

ICE 도표를 작성하면,

 

............. CH2O(g) . ⇌ . H2(g) . + . CO(g)

초기(M): . 0.10 .............. 0 ............ 0

변화(M): . –x ................ +x ........... +x

평형(M): . 0.10–x ........... x ............ x

 

 

 

0.10–x = 0.066

 

x = 0.10 – 0.066 = 0.034

 

 

 

Kc = [H2] [CO] / [CH2O]

= 0.034^2 / 0.066

= 0.017515

 

 

 

답: Kc = 0.018

 

 

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0.085 mol CH2O 증기를 가열된 500 mL 용기에 넣었다.

평형에서 CH2O의 농도가 0.078 mol/L일 때, Kc는?

 

0.085 mol / 0.500 L = 0.17 M

x = 0.17 – 0.078 = 0.092

Kc = 0.092^2 / 0.078 = 0.11