Question

Given the following thermochemical equations:

X2 + 3Y2 ----> 2XY3 ^H1= -370 kJ

X2 + 2Z2 ----> 2XZ2 ^H2= -170 kJ

2Y2 + 2Y2Z -----> 2Y2Z ^H3= -280 kJ

Calculate the change in enthalpy for the following reaction:

4XY3 + 7Z2 -----> 6Y2Z + 4XZ2

change H= ________ kJ

Answer 1

The change in enthalpy for the reaction equals: delta H = sum (delta H) of products - sum (delta H) of reagents.
For reagents and products of the following reaction:
delta H (XY3) = -370 kJ / 2 = -185 kJ.
delta H (XZ2) = -170 kJ / 2 = -85 kJ.
delta H (Y2Z) = -280 kJ / 2 = -140 kJ.
delta H (Z2) = 0

delta H (for the reaction) = 6*(-140 kJ) + 4*(-85 kJ) - 4*(-185)-0 = -840 kJ - 340 kJ + 740 kJ = -440 kJ.

Answer: -440 kJ.