Question

Consider the reaction: ClF3(g) + 2NH3(g)  N2(g) + Cl2(g) + 6HF(g) ∆H° = -1196 kJ If H°f for NH3(g) and for ClF3(g) are -46 kJ/mol and -405 kJ/ mol respectively. Calculate the H°f for HF(g)

Answer 1

Explanation:

apply the formula dlta H =hproduct -h reactant

-1196=dltah hf -(delta Hnh3 +delta Hclf3)

-1196=h -(-46-405(

h=