(1)已知:N2(g)+O2(g)=2NO(g);△H=180.5KJ?mol-l4NH3(g)+5O2(g)=4NO(g)+6H2O(g);△H=-

2025-05-14 18:57:51
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回答(1):

(1)①N2(g)+O2(g)=2NO(g);△H=180.5KJ?mol-l
②4NH3(g)+5O2(g)=4NO(g)+6H2O(g);△H=-905KJ?mol-l
③2H2(g)+O2(g)=2H2O(g);△H=-483.6KJ?mol-l
依据盖斯定律计算得到,N2(g)+3H2(g)?2NH3(g)的△H=-184.8KJ/mol;
则N2(g)+3H2(g)?2NH3(g)的△H=-92.4KJ/mol;
故答案为:-92.4KJ/mol;
(2)①一定温度下在恒容密闭容器中通入惰气总压增大分压不变,平衡不动,则N2O5的转化率不变,故答案为:不变;
②500s内N2O5的分解速率=

5.00mol/L?3.52mol/L
500S
=2.96×10-3mol/L?S;
故答案为:2.96×10-3mol/L?S;
③2N2O5(g)?4NO2(g)+O2(g);△H>O,反应是吸热反应,反应的N2O5浓度变化=5.00-2.48=2.52mol/L,二氧化氮浓度为5.04mol/L,在T2温度下,反应1000s时测得NO2的浓度为4.98mol?L-1,二氧化氮浓度减小,平衡逆向进行,说明温度降低,T2<T1
故答案为:<;
(3)从电解原理来看,N2O4制备N2O5为氧化反应,则N2O5应在阳极区生成,反应式为N2O4+2HNO3-2e-=2N2O5+2H+
故答案为:阳极,N2O4+2HNO3-2e-=2N2O5+2H+