Question

积为2.0L恒容的密闭容器中充入0.80mol和1.60mol,反应在673K、30MPa下达到平衡时,NH3的体积分数

Answer 1

假定N2和H2分别为0.80 mol,1.60mol

N2 + 3H2 = 2NH3

0.80 1.60 0

0.80-x 1.60-3x 2x (平衡时)

平衡时NH3的体积分数(与物质的量成正比) = 2x/(0.80 - x + 1.60 - 3x + 2x) = x/(1.2 - x) = 20%

x = 0.20

氮气的转化率是:0.20/0.80 = 25%

平衡时:[N2] = (0.80 - 0.20)/2 = 0.30 mol/L

[H2] = (1.60 - 3*0.20)/2 = 0.50 mol/L

[NH3] = 2*0.20/2 = 0.20 mol/L

K = [NH3]²/([N2]*[H2]³) = 0.20²/(0.30*0.5³) = 1.07