Question

【題目】化學反應N2+3H22NH3的能量變化如圖所示,該反應的熱化學方程式是(　　)

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】由圖示可知:(1)斷裂 mol N2(g)和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;(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項均錯誤。將①式乘以2得:N2(g)+3H2(g)2NH3(g)　ΔH=2(a-b)kJ·mol-1,B項錯誤;將②式乘以2得:N2(g)+3H2(g)2NH3(l)　ΔH=2(a-b-c)kJ·mol-1,A項正確。