Question

【題目】根據能量變化示意圖，下列熱化學方程式正確的是( )

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

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

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

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

Answer 1

【答案】C

【解析】

焓變等于反應物斷裂化學鍵吸收的能量減去形成化學鍵釋放的能量，由圖可知，N2(g)+H2(g)=NH3(g) △H=(a-b)kJmol-1，N2(g)+H2(g)=NH3(l) △H=(a-b-c)kJmol-1，據此分析判斷。

A．N2(g)+H2(g)=NH3(g) △H=(a-b)kJmol-1，則N2(g)+3H2(g)=2NH3(g) △H=-2(b-a)kJmol-1，故A錯誤；

B．N2(g)+H2(g)=NH3(g) △H=(a-b)kJmol-1，則N2(g)+3H2(g)=2NH3(g)△H═-2(b-a)kJmol-1，故B錯誤；

C．N2(g)+H2(g)=NH3(l) △H=(a-b-c)kJmol-1，則N2(g)+3H2(g)=2NH3(l) △H=2(a-b-c)kJmol-1，則2NH3(1)=N2(g)+3H2(g)△H=2(b+c-a)kJmol-1，故C正確；

D．N2(g)+H2(g)=NH3(l) △H=(a-b-c)kJmol-1，則N2(g)+3H2(g)=2NH3(l) △H=2(a-b-c)kJmol-1，則2NH3(1)=N2(g)+3H2(g)△H=2(b+c-a)kJmol-1，故D錯誤；

故選C。