Kc 730℃ 2.18×10^6 H2(g) + Br2(g) ⇌ 2HBr(g) 12.0 L HBr 3.20 mol

Kc 730℃ 2.18×10^6 H2(g) + Br2(g) ⇌ 2HBr(g) 12.0 L HBr 3.20 mol

 

 

The equilibrium constant Kc for the reaction

H2(g) + Br2(g) ⇌ 2HBr(g)

is 2.18×10^6 at 730°C.

Starting with 3.20 moles of HBr in a 12.0 L reaction vessel,

calculate the concentrations of H2, Br2, and HBr at equilibrium.

 

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반응 전 HBr의 몰농도를 계산하면,

[HBr] = 3.20 mol / 12.0 L = 0.267 M

 

 

 

ICE 도표를 작성하면,

 

............. H2(g) + Br2(g) ⇌ 2HBr(g)

초기(M) . 0 ......... 0 ........... 0.267

변화(M) . +x ....... +x .......... –2x

평형(M) . x ......... x ............ 0.267–2x

 

 

 

 

 

Kc = [HBr]^2 / [H2] [Br2]

 

2.18×10^6 = (0.267–2x)^2 / (x) (x)

 

2.18×10^6 = (0.267–2x)^2 / (x)^2

 

2.18×10^6 = [(0.267–2x) / (x)]^2

 

(2.18×10^6)^(1/2) = (0.267–2x) / (x)

 

1476.48x = 0.267–2x

 

1478.48x = 0.267

 

x = 0.267 / 1478.48

= 0.00018059 M

≒ 0.000181 M

 

 

 

답:

[H2] = [Br2] = x = 1.81×10^(-4) M

 

[HBr] = 0.267 – 2(0.000181)

= 0.266638 ≒ 0.267 M