Answer:
ΔH = 125.94kJ
Explanation:
It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law). In the problem:
1. 2W(s) + 3O2(g) → 2WO3(s) ΔH = -1685.4 kJ
2. 2H2(g) + O2(g) → 2H2O(g) ΔH = -477.84 kJ
-1/2 (1):
WO3(s) → W(s) + 3/2O2(g) ΔH = 842.7kJ
3/2 (2):
3H2(g) + 3/2O2(g) → 3H2O(g) ΔH = -716.76kJ
The sum of last both reactions:
WO3(s) + 3H2(g) → W(s) + 3H2O(g)
ΔH = 842.7kJ -716.76kJ
The enthalpy change (ΔH) for the reaction is 125.94kJ.
Hess's Law:
The Hess's law states that the total enthalpy change during a complete chemical reaction is the same regardless of the path taken by the chemical reaction.
It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law).
It is given that:
1. 2W(s) + 3O₂(g) → 2WO₃(s) ΔH = -1685.4 kJ
2. 2H₂(g) + O₂(g) → 2H₂O(g) ΔH = -477.84 kJ
On dividing equation 1. by - 1/2 for simplification we get:
W(s) + 3/2 O₂(g) → WO₃(s) ΔH = 842.7kJ....................(a)
On dividing equation 2. by 3/2 for simplification we get:
3H₂(g) + 3/2O₂(g) → 3H₂O(g) ΔH = -716.76kJ...................(b)
On adding equations (a) and (b)
WO₃(s) + 3H₂(g) → W(s) + 3H₂O(g)
ΔH = 842.7kJ -716.76kJ
ΔH = 125.94kJ
The enthalpy change for the reaction is 125.94 kJ.
Learn more:
brainly.com/question/1445358
