反应N2+3H2=2NH3在密闭容器中进行达到平衡，平衡时c(N2)=3.5mol/L,c(H2)=1MOL/L,c(NH3)=5MOL/L
（1）设N2与H2发开始浓度分别为x、y，转化为n，则 N2（g）+3H2（g）?2NH3（g），开始 x y 0转化 n 3n 2n平衡3.5 1 52n=5，解得n=2.5，所以x=3.5+2.5=6mol/L，y=1+3×2.5=8.8mol/L，答：N2和H2的起始浓度分别为6mol?L-1、8.5mol?L-1；（2）N2的转化率为2.5mol/L6mol/L×1005=41.67%，答：N2的转化率为41.67%．
反应N2+3H2=2NH3在密闭容器中进行达到平衡，平衡时c(N2)=3.5mol/L,c(H2)=11MOL/L,c(NH3)=5MOL/L.求： 1.N2和H2的起始浓度。 用等效平衡算 开始浓度:c(N2)=6MOL/L c(N2)=18.5MOL/L N2+3H2=2NH3 能反应出c(NH3)=5MOL/L 那么原来就需要2.5MOL/L的N2 加上平衡时的3.5MOL/L就是c开始(N2)=6MOL/L H2同理 原来需要7.5MOL/L 家上平衡的11就是c开始(H2)=18.5MOL/L 2.N2的转化率 41.67％(5/12) N2开始的时候的浓度是6MOL/L反应转化了的浓度2.5MOL/L 那么2.5/6*100％就是转化率 就大约是41.67％ 3.平衡时压强为开始时压强得？% 由气体对气壁的压强定义 条件相同时(温度压强)体积相同的容器的气体压强比等于两容器中的气体物质的量比 反应开始时有气体重物质的量18.5+6=24.5MOL 反应后3.5+11+5=19.5MOL 那么19.5/24.5*100％就是所求大约等于79.59％ 4.平衡时NH3占总体积的体积分数 由阿佛加得罗定理 同条件下(温度 压强)的物质的量比等于气体的体积比 那么反应后一共有气体19.5MOL NH3有5MOL 那么NH3的体积分数就等于n(NH3)/n(总)=5/19.5*100％≈25.64％ 明白了么? 祝进步 歆蟹
1.初始c(N2)=6mol/L 【计算 5/2+3.5】 c(H2)=8.5mol/L 【计算（5/2）*3+1】2. （6-3.5）/6=41.67%
3.（3.5+1+5）/（6+8.5）=65.52%
4 5/（5+1+3.5）=52.63%
达到化学平衡时生成的氨气所需要消耗的H2和N2的量分别为:
体积恒定,物质的量直接转化为浓度计算
c(N2)=5/2=2.5mol/L
c(H2)=(3*5)/2=7.5mol/L
所以初始浓度为:
c(N2)=3.5+2.5=6mol/L
c(H2)=11+7.5=18.5mol/L
N2的转化率=消耗的N2/初始的N2=2.5/6=41.67%
由于体积恒定,压强比等于体积比,
P平/P初=V平/V初=(3.5+11+5)/(6+18.5)=75.6%
NH3占总体积的体积分数为:
5/(3.5+11+5)=25.6%
K=4*4/(3*9*9*9)=0.007316 C(N2)=3 4/2=5mol/L C(H2)=9 4/2*3=15mol/L
JFG290
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