Eqn 1: X2 + 3 Y2 ⟶ 2 XY3 Δ𝐻1 = −370 kJ
Eqn 2: X2 + 2 Z2 ⟶ 2 XZ2 Δ𝐻2 = −120 kJ
Eq. 3: 2 Y2 + Z2 ⟶ 2 Y2Z Δ𝐻3 = −270 kJ
Calculate the change in enthalpy for the reaction.
4 XY3 + 7 Z2 ⟶ 6 Y2Z + 4 XZ2
In order to solve this Hess's Law problem, we need to look for the components of the equation in our given equation. We can find the compound XY3 in eqn 1, 2 XY3 as a product, however, to get it to match our desired equation, 4 XY3 as a reactant, we need to flip equation 1 and multiply by two. When we do that, we will also flip the sign of Δ𝐻1 and multiply it by two:
4 XY3 ⟶ 2 X2 + 6 Y2 Δ𝐻1 = -(-370 kJ) x 2 = + 740 kJ
Since the next substance, Z2, is in more than one equation, we will skip it for now. We then find the compound Y2Z in eqn 3, 2 Y2Z as a product, which is as 6 Y2Z and a product, too, so we only need to multiply eqn 3 by three, and therefore multiply Δ𝐻3 by three:
6 Y2 + 3 Z2 ⟶ 6 Y2Z Δ𝐻3 = (−270 kJ) x 3 = - 810 kJ
The last compound, XZ2, is found in eqn 2, as the product 2 XZ2, which means we will multiply eqn 2 and Δ𝐻2 by two to match our desired product, 4 XZ2:
2 X2 + 4 Z2 ⟶ 4 XZ2 Δ𝐻2 = (−120 kJ) x 2 = - 240 kJ
Adding all three equations (making sure to sum and cancel substances where appropriate) and all three enthalpies, we get:
4 XY3 ⟶ 2 X2 + 6 Y2 Δ𝐻 = + 740 kJ
6 Y2 + 3 Z2 ⟶ 6 Y2Z Δ𝐻 = - 810 kJ
2 X2 + 4 Z2 ⟶ 4 XZ2 Δ𝐻 = - 240 kJ
4 XY3 + 7 Z2 ⟶ 6 Y2Z + 4 XZ2 Δ𝐻 = - 310 kJ
