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方程式:
(1) 2NH3(g)+3N2O(g)====4N2(g)+3H2O(l)
△rHm1=﹣1010kj/mol
(2) N2O(g)+3H2(g)====N2H4(l)+H2O(l)
△rHm2=﹣317kj/mol
(3) 2NH3(g)+1/2O2(g) → N2H4(l)+H2O(l)
△rHm3=-143kj/mol
3×② -① 得 9H2 ( g )+4N2( g )→3N2H4 ( l )+2NH3 ( g )
③-④ 得 2NH3 ( g ) →N2H4 ( l )+ H2 ( g )
两式相加得 8H2 ( g )+4N2( g )→4N2H4( l );还要除以4,把N2H4 ( l )系数变为+1.
由上,(3×△rHm2-△rHm1)+ (△rHm3- △rHm4);数据自己代入算,算得的结果再除以4,就是标准摩尔生成焓。