Question

【题目】化学反应N2＋3H2===2NH3的能量变化如图所示，该反应的热化学方程式是(　　)

A. N2(g)＋3H2(g)===2NH3(l) ΔH＝2(a－b－c) kJ·mol－1

B. N2(g)＋3H2(g)===2NH3(g) ΔH＝2(b－a) kJ·mol－1

C. N2(g)＋H2(g)===NH3(l) ΔH＝(b＋c－a) kJ·mol－1

D. N2(g)＋H2(g)===NH3(g) ΔH＝(a＋b) kJ·mol－1

Answer 1

【答案】A

【解析】由图示可知:(1)断裂 mol N2(g)和mol H2(g)中的化学键共吸收akJ的能量,由1 mol N和3 mol H生成1 mol NH3(g)放出bkJ的能量,由1 mol NH3(g)变为1 mol NH3(l)放出ckJ能量;(2)由mol N2(g)和mol H2(g)生成1 mol NH3(g)或1 mol NH3(l)均放出能量。故有①N2(g)+H2(g)NH3(g) ΔH=-(b-a)kJ·mol-1=(a-b)kJ·mol-1或②N2(g)+H2(g)NH3(l) ΔH=-(b+c-a)kJ·mol-1=(a-b-c)kJ·mol-1,因此C、D项均错误。将①式乘以2得:N2(g)+3H2(g)2NH3(g) ΔH=2(a-b)kJ·mol-1,B项错误;将②式乘以2得:N2(g)+3H2(g)2NH3(l) ΔH=2(a-b-c)kJ·mol-1,A项正确。