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