equilibrium constant

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2012-05-13 7:14 am

26.At 110'C,the equilibrium constant,Kc,for the reaction is 4.7X10^9 mol^-1dm^3.If 0.2 mole COCL2 was placed into 8.0dm^3 reaction vessel at 110'C,calculate the concentration of all substances at equilibrium.

27.At 373K,N2O4 decomposes to NO2 reversibly until an equilibrium is reached.The equilibrium concentration of NO2 is found to be 3.0x10^-3 mol dm^-3 and the equilibrium constant,Kc=0.2 mol dm^-3.Calculate the initial concentration of N2O4

回答 (1)

賣女孩的火柴

2012-05-13 8:34 am

✔ 最佳答案

26.
Suppose the reaction is : COCl2(g) ⇌ CO(g) + Cl2(g)
Initial : [COCl2]o = 0.2/8.0 = 0.025 mol dm⁻³, [CO]o = [Cl2]o = 0 mol dm⁻³
At eqm : [COCl2] = y mol dm⁻³, [CO] = [Cl2] =(0.025 - y) ≈ 0.025 mol dm⁻³

Kc = [CO][Cl2]/[COCl2]
(0.025)²/y = 4.7 x 10⁹
y = 1.33 x 10⁻¹³

Hence, at eqm :
[COCl2] = 1.33 x 10⁻¹³ mol dm⁻³, [CO] = [Cl2] =0.025 mol dm⁻³

27.
N2O4(g) ⇌ 2NO2(g)
Initial : [N2O4]o = y mol dm⁻³, [NO2]o= 0 mol dm⁻³
At eqm : [N2O4] = y - (3.0 x 10^-3)/2 = [y - (1.5 x10^-3)] mol dm⁻³, [NO2] = 3.0 x10^-3 mol dm⁻²

Kc = [NO]²/[N2O4]
(3.0 x 10⁻³)²/[y - (1.5 x 10⁻³)] = 0.2
(0.003)²/(y - 0.0015) = 0.2
0.2y - 0.0003 = 0.000009
0.2y = 0.000309
y = 1.545 x 10^-3

Initial concentration of N2O4 = 1.545 x 10^-3 mol dm⁻³

參考: 賣女孩的火柴

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