Question

Calculate the standard enthalpy of formation of liquid water (H₂O) using the following thermochemical information:

2 B2O3(s) 4 B(s) + 3 O2(g)		H = +2509.1 kJ
B2O3(s) + 3 H2O(l) B2H6(g) + 3 O2(g)		H = +2147.5 kJ
2 B(s) + 3 H2(g) B2H6(g)		H = +35.4 kJ

H = ______ kJ

Answer 1

Reducing the coefficients of first equation to half then ∆H also become half and we get :

B2O3 (s) -> 2B(s) + 3/2 O2(g) ∆H = 2509.1/2 = 1254.55 KJ

Reversing equation second we get :

B2H6(g) + 3 O2(g) -> B2O3(s) + 3H2O(l) ∆H = -2147.5 KJ

Adding above equations and third equation we get :

3 H2(g) + 3/2 O2(g) -> 3 H2O(l)

∆H = 1254.55-2147.5+35.4

= -857.55 KJ

Now, ∆Hf(H2O) = -857.55/3 = -285.85 KJ