Question

given that
$h_{2}(g)+f_{2}(g)\to 2hf(g)$ $delta h_{r m}^{*}=-546.6 mathrm{~kj}$
$2 mathrm{h}_{2}(mathrm{~g})+mathrm{o}_{2}(mathrm{~g}) \to 2 mathrm{h}_{2} mathrm{o}(mathrm{l})$ $delta h_{r m}^{*}=-571.6 mathrm{~kj}$
calculate the value of $delta h_{r m}^{*}$ for
$2 mathrm{~f}_{2}(mathrm{~g})+2 mathrm{h}_{2} mathrm{o}(mathrm{l}) \to 4 mathrm{hf}(mathrm{g})+mathrm{o}_{2}(mathrm{~g})$
$delta h_{r m}^{*}=\square mathrm{kj}$

Explanation:

Step1: Manipulate the given equations

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

Step2: Apply Hess's law

Add the two manipulated equations:
\((2H_{2}(g)+2F_{2}(g)\to4HF(g))+(2H_{2}O(l)\to2H_{2}(g)+O_{2}(g))\)
The \(2H_{2}(g)\) terms cancel out.
The overall equation is \(2F_{2}(g)+2H_{2}O(l)\to4HF(g)+O_{2}(g)\)
By Hess's law, \(\Delta H_{rxn}=\Delta H_{1}+\Delta H_{2}\)
Substitute \(\Delta H_{1}=-1093.2\space kJ\) and \(\Delta H_{2}=+571.6\space kJ\)
\(\Delta H_{rxn}=-1093.2 + 571.6\space kJ\)

Answer:

\(-521.6\space kJ\)