K = 6.40×10^(-7) 0.250 mol CO2(g) in 1.000 L container

K = 6.40×10^(-7) 0.250 mol CO2(g) in 1.000 L container

 

 

Carbon monoxide gas is used for synthesis of organic chemicals.

At 2000℃, the value of the equilibrium constant is

6.40×10^(-7) for the decomposition of carbon dioxide gas, CO2(g),

into carbon monoxide, CO(g), and oxygen O2(g).

Calculate the concentrations of all entities at equilibrium

if 0.250 mol of carbon dioxide is placed in 1.000 L container.

 

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[CO2] = 0.250 mol / 1.000 L = 0.250 M

 

 

 

ICE 도표를 작성하면,

............. 2CO2(g) . → . 2CO(g) + O2(g)

초기(M): 0.250 ............ 0 ............ 0

변화(M): –2x ............... +2x ......... +x

평형(M): 0.250–2x ...... 2x ........... x

 

 

 

K = [CO]^2 [O2] / [CO2]^2

 

6.40×10^(-7) = (2x)^2 (x) / (0.250–2x)^2

 

 

 

초기농도 / K > 100 이므로,

0.250–2x ≒ 0.250 이라 근사처리하면,

6.40×10^(-7) = 4x^3 / (0.250)^2

 

x = [(6.40×10^(-7)) (0.250)^2 / 4]^(1/3)

= 0.00215 M

 

 

 

답:

[CO2] = 0.250 – 2x

= 0.250 – 2(0.00215) = 0.246 M

 

[CO] = 2x = 2(0.00215) = 0.00430 M

 

[O2] = 0.00215 M

 

 

 

 

[키워드] 2CO2(g) → 2CO(g) + O2(g) ICE table, 근사 처리 기준문서, 근사 처리 방법 기준문서, 근사 처리 사전, 근사 처리 방법 사전