①A(g)+2B(g)⇌2C(g)+2D(l)△H1=﹣250.kJ•mol﹣1 K1=0.2
②E(s)+B(g)⇌C(g)△H2=﹣310kJ•mol﹣1K2=2
③F(g)+ B(g)⇌D(l)△H3=﹣200kJ•mol﹣1K3=0.8