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Given the thermochemical equations X2 + 3Y2 → 2 XY3 ΔH1 = -340 kJ X2 + 2Z2 → 2 XZ2 ΔH2 = -180 kJ 2Y2 + Z2 → 2Y2Z ΔH3 = -240 kJ Calculate the change in enthalpy for the reaction. 4XY3+7Z2⟶6Y2Z+4XZ2