(1) $\Delta _{r}H_{298}^{o}$= $\Delta _{f}H_{298}^{o}$(NaF) + $\frac{1}{2}\Delta _{f}H_{298}^{o}$(Cl2) - $\Delta _{f}H_{298}^{o}$(NaCl) - $\frac{1}{2}\Delta _{f}H_{298}^{o}$(F2)
=−574 + $\frac{1}{2}$ x 0 - (−411,2) − $\frac{1}{2}$ x 0 = -162,8(kJ).
(2) $\Delta _{r}H_{298}^{o}$= $\Delta _{f}H_{298}^{o}$(NaCl) + $\frac{1}{2}\Delta _{f}H_{298}^{o}$(Br2) - $\Delta _{f}H_{298}^{o}$(NaBr) - $\frac{1}{2}\Delta _{f}H_{298}^{o}$(Cl2)
=−411,2 + $\frac{1}{2}$ x 0 - (−361,1) − $\frac{1}{2}$ x 0 = -50,1(kJ).
(3) $\Delta _{r}H_{298}^{o}$= $\Delta _{f}H_{298}^{o}$(NaBr) + $\frac{1}{2}\Delta _{f}H_{298}^{o}$(I2) - $\Delta _{f}H_{298}^{o}$(NaI) - $\frac{1}{2}\Delta _{f}H_{298}^{o}$(Br2)
= −361,1 + $\frac{1}{2}$ x 0 - (−287,8) − $\frac{1}{2}$ x 0 = - 73,3(kJ).
(4) $\Delta _{r}H_{298}^{o}$= $\Delta _{f}H_{298}^{o}$(Cl- ) + $\frac{1}{2}\Delta _{f}H_{298}^{o}$(Br2) - $\Delta _{f}H_{298}^{o}$(Br- ) - $\frac{1}{2}\Delta _{f}H_{298}^{o}$(Cl2)
= −167,2 + $\frac{1}{2}$ x 0 - (−121,6) − $\frac{1}{2}$ x 0 = - 45,6 (kJ).
(5) $\Delta _{r}H_{298}^{o}$= $\Delta _{f}H_{298}^{o}$(Br-) + $\frac{1}{2}\Delta _{f}H_{298}^{o}$(I2) - $\Delta _{f}H_{298}^{o}$(I- ) - $\frac{1}{2}\Delta _{f}H_{298}^{o}$(Br2)
= -121,6 + $\frac{1}{2}$ x 0 - (-55,2) − $\frac{1}{2}$ x 0 = - 66,4 (kJ).
b) Kết quả này có phù hợp với quy luật biến đổi tính phi kim của dãy halogen trong bảng tuần hoàn các nguyên tố hóa học.