Question

ΔH for the reaction IF5 (g) → IF3 (g) + F2 (g) is ________ kJ, give the data below. IF (g) + F2 (g) → IF3 (g) ΔH = -390 kJ IF (g) + 2F2 (g) → IF5 (g) ΔH = -745 kJ

Answer 1

We need ΔH for the reaction IF5 (g) → IF3 (g) + F2 (g)

Given

IF (g) + F2 (g) → IF3 (g) ΔH = -390 kJ - - - - - (1)

IF (g) + 2F2 (g) → IF5 (g) ΔH = -745 kJ - - - - - - (2)

To get the desired solution, reverse the second reaction and add that to the first reaction.

Hence the second reaction will be

IF5 (g) -> IF (g) + 2F2 (g)  ΔH = +745 kJ - - - - - - (3)

Since we reversed the reaction ΔH will become positive ​​​​​​

Now

Add (3) and (1)

IF5 (g) -> IF (g) + 2F2 (g)  ΔH = +745 kJ

IF (g) + F2 (g) → IF3 (g) ΔH = -390 kJ

IF (g) and one F2 (g) will be cancelled on reactants and products side

Hence

IF5 (g) → IF3 (g) + F2 (g) ΔH = +745 KJ - 390 KJ = +355 KJ

Therefore

ΔH for the reaction IF5 (g) → IF3 (g) + F2 (g) is +355 KJ