Given the thermochemical equations

Question

X2+3Y2⟶2XY3Δ𝐻1=−390 kJ

X2+2Z2⟶2XZ2Δ𝐻2=−160 kJ

2Y2+Z2⟶2Y2ZΔ𝐻3=−270

Calculate the change in enthalpy for the reaction.

4XY3+7Z2⟶6Y2Z+4XZ2

Answer 1

Hess' Law

4XY₃ + 7Z₂⟶6Y₂Z + 4XZ₂ TARGET EQUATION

Given:

(1) X₂ + 3Y₂⟶ 2XY₃ ... Δ𝐻1=−390 kJ

(2) X₂ + 2Z₂⟶ 2XZ₂ ... Δ𝐻2=−160 kJ

(3) 2Y₂ + Z₂⟶ 2Y₂Z ... Δ𝐻3=−270

Procedure:

reverse(1) and x2: 4XY₃ ==> 2X₂ + 6Y₂ ... ∆H = +780 kJ

copy(2) and x2: 2X₂ + 4Z₂⟶ 4XZ₂ ... ∆H = -320 kJ

copy(3) and x3: 6Y₂ + 3Z₂⟶ 6Y₂Z ... ∆H = -810 kJ

Add these equations and combine and cancel like items:

4XY₃ + 2X₂ + 4Z₂ + 6Y₂ + 3Z₂ ==> 2X₂ + 6Y₂ + 4XZ₂ + 6Y₂Z

4XY₃ + 7Z₂ ==> 4XZ₂ + 6Y₂Z TARGET EQUATION

∆H = +780 + (-320) + (-810) = -350 kJ