Question

Given the thermochemical equations

X
2
+
3
Y
2
⟶
2
XY
3
Δ
H
1
=
−
340
kJ

X
2
+
2
Z
2
⟶
2
XZ
2
Δ
H
2
=
−
140
kJ

2
Y
2
+
Z
2
⟶
2
Y
2
Z
Δ
H
3
=
−
240
kJ

Calculate the change in enthalpy for the reaction.

4
XY
3
+
7
Z
2
⟶
6
Y
2
Z
+
4
XZ
2

Answer 1

((2*2nd reaction) +(3*3rd reaction)) - (2*1st reaction) = The reaction in the problem

2X2+4Z2 --> 4XZ2

6Y2+3Z2 --> 6Y2Z

4XY3 --> 2X2+6Y2

-------------------------------

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

Therefore the change in enthalpy=

(-280 - 720 +680) kj = - 320 kj