根据能量变化示意图，下列热化学方程式正确的是( ) A．N2(g)＋3H2(g)===2NH3(g)ΔH＝－(b－a)kJ·mol－1 B．N2(g)＋3H2(g)===2NH3(g)ΔH＝－(a－b)kJ·mol－1 C．2NH3(l)===N2(g)＋3H2(g)ΔH＝2(a＋b－c)kJ·mol－1 D．2NH3(l)===N2(g)＋3H2(g)ΔH＝2(b＋c－a)kJ·mol－1 答案：解析：选D 由图可得N2(g)＋H2(g)===NH3(g) ΔH＝－(b－a)kJ·mol－1，则N2(g)＋3H2(g)===2NH3(g) ΔH＝－2(b－a)kJ·mol－1，故A、B两项错误；再由图可得NH3(l)===N2(g)＋H2(g) ΔH＝(b＋c－a)kJ·mol－1，则2NH3(l)===N2(g)＋3H2(g) ΔH＝2(b＋c－a)kJ·mol－1，故C项错误，D项正确。