Given the thermochemical equations x 2 + 3 y 2 ⟶ 2 xy 3 δ h 1 = − 380 kj x 2 + 2 z 2 ⟶ 2 xz 2 δ h 2 = − 130 kj 2 y 2 + z 2 ⟶ 2 y 2 z δ h 3 = − 260 kj calculate the change in enthalpy for the reaction. 4 xy 3 + 7 z 2 ⟶ 6 y 2 z + 4 xz 2

The enthalphy change of the given reaction is -280 KJ.

Explanation:

X 2 + 3 Y 2 ⟶ 2 XY 3                            Δ H 1 = − 380 kJ     Eq-1

X 2 + 2 Z 2 ⟶ 2 XZ 2                            Δ H 2 = − 130 kJ     Eq-2

2 Y 2 + Z 2 ⟶ 2 Y 2 Z                           Δ H 3 = − 260 kJ    Eq-3

4 XY 3 + 7 Z 2 ⟶ 6 Y 2 Z + 4 XZ 2       ΔH                           Eq-4

By Hess law,

      The heat of any reaction  ΔH  for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction:

Eq-4 can be manipulated as,

    Eq-4 = -2(Eq-1) +2(Eq-2) +3(Eq-3)

By Hess law, same happen with their enthalphy

Therefore,

    ΔH = -2(ΔH1) + 2(ΔH2) + 3(ΔH3)

          = -2 × -380 + 2 × -130 + 3 × -260

          = -280 KJ