Question

【題目】化學反應N2＋3H2===2NH3的能量變化如圖所示，該反應的熱化學方程式是(　　)

A. N2(g)＋H2(g)＝NH3(l)　ΔH＝(a－b－c) kJ/mol

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

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

D. N2(g)＋H2(g)＝NH3(g)　ΔH＝(a＋b) kJ/mol

Answer 1

【答案】A

【解析】由圖示可知斷裂0.5 mol N2(g)和1.5 mol H2(g)中的化學鍵共吸收a kJ的能量,由1 mol N和3 mol H生成1 mol NH3(g)放出b kJ的能量，由1 mol NH3(g)變為1 mol NH3(l)放出c kJ能量；由0.5 mol N2(g)和1.5 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項均錯誤，A選項正確；將①式乘以2得：N2(g)+3H2(g)2NH3(g)ΔH=2(a-b)kJ·mol-1,B錯誤；答案選A。