發(fā)射嫦娥二號的火箭究竟需要灌注多少燃料？2N2H4(g)＋2NO2(g)=3N2(g)＋4H2O(g)ΔH＝？

第三節(jié)化學(xué)反應(yīng)熱的計算

不管化學(xué)反應(yīng)是一步或分幾步完成，其反應(yīng)熱是相同的。即：化學(xué)反應(yīng)的反應(yīng)熱只與反應(yīng)的始態(tài)（各反應(yīng)物）和終態(tài)（各生成物）有關(guān)，而與具體反應(yīng)的途徑無關(guān)。蓋斯定律則有ΔH＝ΔH1＋ΔH2

＝ΔH3＋ΔH4＋ΔH5如何理解蓋斯定律？化學(xué)反應(yīng)的反應(yīng)熱相當(dāng)于山的高度，與登山途徑無關(guān)!S(始態(tài)）L（終態(tài)）ΔH1<0ΔH2>0ΔH1＋ΔH2=0H2O(g)==H2(g)＋?O2(g)

ΔH＝＋242kJ·mol－1H2(g)＋?O2(g)==H2O(g)ΔH＝-242kJ·mol－1已知：①C(s)+O2(g)=CO2(g)ΔH1=-393.5kJ/mol②CO(g)+1/2O2(g)=CO2(g)ΔH2=-283.0kJ/mol計算：③C(s)+1/2O2(g)=CO(g)ΔH3=?小組討論

ΔH1C

CO2

ΔH3

ΔH2

CO

由蓋斯定律得ΔH1＝ΔH2＋ΔH3即：ΔH3＝ΔH1—ΔH2（虛擬路徑）CO(g)+?O2(g)C(s)+O2(g)CO2(g)ΔH1=-393.5kJ/molΔH3ΔH2=-283.0kJ/mol由蓋斯定律得ΔH3=ΔH1-ΔH2

=-393.5+283.0）=-110.5kJ/molC(s)+1/2O2(g)=CO(g)ΔH3=?①C(s)+O2(g)=CO2(g)ΔH1=-393.5kJ/mol②CO(g)+1/2O2(g)=CO2(g)ΔH2=-283.0kJ/mol③C(s)+1/2O2(g)=CO(g)ΔH3

因①=②+③

則ΔH1=ΔH2+ΔH3ΔH3=ΔH1-ΔH2=-393.5-（-283.0）=-110.5kJ/mol應(yīng)用了什么規(guī)律？疊加法C(s)+CO2(g)=2CO(g)ΔH4=?=-110.5kJ/mol=ΔH3-ΔH2=2ΔH3-ΔH1=ΔH1-2ΔH2=+172.5kJ/mol例1：已知：①P4(s，白磷)＋5O2=P4O10(s)ΔH1＝－2983.2kJ·mol－1②P(s，紅磷)＋5/4O2(g)=1/4P4O10(s)ΔH2＝－738.5kJ·mol－1寫出白磷轉(zhuǎn)化為紅磷的熱化學(xué)方程式用一用蓋斯定律①P4(s，白磷)＋5O2=P4O10(s)ΔH1＝－2983.2kJ·mol－1②P(s，紅磷)＋5/4O2(g)=1/4P4O10(s)ΔH2＝－738.5kJ·mol－1③P4(s，白磷)=4P(s，紅磷)ΔH＝？因為=①－②×4則ΔH＝ΔH1－ΔH2×4＝－2983.2－（－738.5）×4＝－29.2kJ·mol－1例2：嫦娥二號，用N2H4（肼）在NO2中燃燒，生成N2、氣態(tài)H2O。已知：N2(g)+2O2(g)=2NO2(g)△H1=+67.2kJ/molN2H4(g)+O2(g)=N2(g)+2H2O(g)△H2=-534kJ/mol假如都在相同狀態(tài)下，請算出發(fā)射嫦娥二號衛(wèi)星所用燃料反應(yīng)的反應(yīng)熱。2N2H4