Consider the following processes: ∆H (kJ/mol) 1/2A → B +150 3B → 2C + D –125 E + A → 2D +350 For B + D → E + 2C, ∆H will be (Mains 2011)

Adding all the equations, we get ∆H A → 2B 300 kJ/mol 3B → 2C + D –125 kJ/mol 2D → A + E –350 kJ/mol B + D → E + 2C; ∆H = (300 – 125 – 350) = –175 kJ/mol