Question

given that

h₂(g) + f₂(g) → 2hf(g) δh°_rxn = -546.6 kj

2h₂(g) + o₂(g) → 2h₂o(l) δh°_rxn = -571.6 kj

calculate the value of δh°_rxn for

2f₂(g) + 2h₂o(l) → 4hf(g) + o₂(g)

δh°_rxn = kj

Explanation:

Step1: Manipulate the given equations

Given equation (1): \(H_{2}(g)+F_{2}(g)\to2HF(g)\), \(\Delta H_{rxn}^{\circ}=- 546.6\space kJ\). Multiply it by \(2\): \(2H_{2}(g)+2F_{2}(g)\to4HF(g)\), \(\Delta H_{1}=2\times(-546.6)\space kJ=-1093.2\space kJ\)
Given equation (2): \(2H_{2}(g)+O_{2}(g)\to2H_{2}O(l)\), \(\Delta H_{rxn}^{\circ}=-571.6\space kJ\). Reverse it: \(2H_{2}O(l)\to2H_{2}(g)+O_{2}(g)\), \(\Delta H_{2}=+ 571.6\space kJ\)

Step2: Apply Hess's law

Add the two manipulated equations:
\((2H_{2}(g)+2F_{2}(g)\to4HF(g))+(2H_{2}O(l)\to2H_{2}(g)+O_{2}(g))\) gives \(2F_{2}(g)+2H_{2}O(l)\to4HF(g)+O_{2}(g)\)
By Hess's law, \(\Delta H=\Delta H_{1}+\Delta H_{2}\)
\(\Delta H=-1093.2 + 571.6\)

Answer:

\(-521.6\)