Hydrogen iodide Kc 0.0156 400℃ 0.550 mol HI 2.00 L

Hydrogen iodide Kc 0.0156 400℃ 0.550 mol HI 2.00 L

 

 

Hydrogen iodide decomposes according to the equation 2HI(g) ⇌ H2(g) + I2(g), for which Kc = 0.0156 at 400℃. Suppose 0.550 mol HI is injected into a 2.00 L reaction vessel at 400℃. What is the concentration of HI at equilibrium?

 

 

 

[HI] = 0.550 mol / 2.00 L

= 0.275 mol/L

= 0.275 M

 

 

 

ICE 도표를 작성하면,

............. 2HI(g) ⇌ H2(g) + I2(g)

초기(M): 0.275 ... 0 ... 0

변화(M): –2x ... +x ... +x

평형(M): (0.275 – 2x) ... x ... x

( 참고 https://ywpop.tistory.com/7136 )

( 2NO(g) = N2(g) + O2(g) ICE table 참고하세요. )

 

 

 

Kc = [H2] [I2] / [HI]^2

= (x) (x) / (0.275 – 2x)^2

= x^2 / (0.275 – 2x)^2

= [x / (0.275 – 2x)]^2

= 0.0156

 

x / (0.275 – 2x) = 0.0156^(1/2) = 0.125

 

x = 0.125(0.275 – 2x)

 

x = 0.125(0.275) – 2(0.125)x

 

x + 0.250x = 0.125(0.275)

 

1.250x = 0.125(0.275)

 

x = 0.125(0.275) / 1.250

= 0.0275

 

 

 

평형에서 [HI] = 0.275 – 2x

= 0.275 – 2(0.0275)

= 0.220 M

 

 

 

답: 0.220 M

 

 

 

 

[ 관련 글 https://ywpop.tistory.com/21575 ]

H2(g) + I2(g) ⇌ 2HI(g) is 57 at 700 K

 

 

 

[키워드] 2HI(g) ⇌ H2(g) + I2(g) 평형 기준, 2HI(g) ⇌ H2(g) + I2(g) 평형 사전