a) Write a balanced equation for the gaseous reaction between N₂O₅ and F₂ to form NF₃ and O₂.

b) Determine ${\mathbf{\text{ΔG}}}_{\mathbf{\text{rxn}}}^{\mathbf{\text{°}}}\mathbf{\text{.}}$

c) Find role="math" localid="1663390048995" ${\mathbf{\text{ΔG}}}_{\mathbf{\text{rxn}}}\mathbf{\text{at298K}}$if ${\mathbf{\text{P}}}_{{\mathbf{\text{N}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{5}}}}{\mathbf{\text{=P}}}_{{\mathbf{\text{F}}}_{\mathbf{\text{2}}}}{\mathbf{\text{=0.20atm,P}}}_{{\mathbf{\text{NF}}}_{\mathbf{\text{3}}}}{\mathbf{\text{=0.25atm,andP}}}_{{\mathbf{\text{O}}}_{\mathbf{\text{2}}}}\mathbf{\text{=0.50atm.}}$

The chemical reaction of vapours of metallic compounds is used in the gaseous reaction method. Thermal decomposition or reactions involving more than two chemical species are examples of these kinds of reactions.

a) Whendinitrogen pentoxide reacts with difluorine it gives rise to nitrogen trifluoride and dioxygen

The balanced reaction equation is given as:

${\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}{\text{(g)+6F}}_{\text{2}}\text{(g)}\to {\text{4NF}}_{\text{3}}{\text{(g)+5O}}_{\text{2}}\text{(g)}$

b) Since Gibb's energy is a state function, it only depends on the system's starting and final states, we can calculate ${\text{ΔG}}_{\text{rxn}}^{\text{°}}$

${\text{ΔG}}_{\text{rxn}}^{\text{°}}{\text{=ΔG}}_{\text{products}}^{\text{°}}{\text{-ΔG}}_{\text{reactants}}^{\text{°}}$

Using appendix B,

$\begin{array}{l}{\text{ΔG}}_{\text{rxn}}^{\text{°}}\text{=}\left[{\text{4ΔG}}^{\text{°}}\left({\text{NF}}_{\text{3}}\text{(g)}\right){\text{+5ΔG}}^{\text{°}}\left({\text{O}}_{\text{2}}\text{(g)}\right)\right]\text{-}\left[{\text{2ΔG}}^{\text{°}}\left({\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\text{(g)}\right){\text{+6×ΔG}}^{\text{°}}\left({\text{F}}_{\text{2}}\text{(g)}\right)\right]\\ {\text{ΔG}}_{\text{rxn}}^{\text{°}}\text{=}\left[\text{4mol×}\left(\text{-83.30}\frac{\text{kJ}}{\text{mol}}\right)\text{+5mol×}\left(\text{0.00}\frac{\text{kJ}}{\text{mol}}\right)\right]\text{-}\left[\text{2mol×}\left(\text{118.00}\frac{\text{kJ}}{\text{mol}}\right)\text{+6mol×}\left(\text{0.00}\frac{\text{kJ}}{\text{mol}}\right)\right]\\ \end{array}$

${\text{ΔG}}_{\text{ran}}^{\text{°}}\text{=-569.20kJ}$

The of the reaction is -569.20 kJ.

c) The partial pressures are given, the reaction quotient Q can be calculated:

$\begin{array}{c}\text{Q=}\frac{{\left[{\text{NF}}_{\text{3}}\right]}^{\text{4}}{\left[{\text{O}}_{\text{2}}\right]}^{\text{5}}}{{\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\right]}^{\text{2}}{\left[{\text{F}}_{\text{2}}\right]}^{\text{6}}}\\ \text{Q=}\frac{{\text{[0.25atm]}}^{\text{4}}{\text{[0.50atm]}}^{\text{5}}}{{\text{[0.20atm]}}^{\text{2}}{\text{[0.20atm]}}^{\text{6}}}\\ \text{Q=47.68372}\end{array}$

Then, ${\text{ΔG}}_{\text{rxn}}$can be expressed and calculated as:

$\begin{array}{c}{\text{ΔG}}_{\text{rxn}}{\text{=ΔG}}_{\text{rxn}}^{\text{°}}{\text{+R×T×lnQΔG}}_{\text{rxn}}\\ {\text{=-569.20×10}}^{\text{3}}\frac{\text{J}}{\text{mol}}\text{+8.314}\frac{\text{J}}{\text{mol×K}}\text{×298K×ln(47.68372)}\\ {\text{ΔG}}_{\text{rxn}}\text{=-559625.199}\frac{\text{J}}{\text{mol}}\\ \text{=-559.625kJ/mol}\end{array}$

The${\text{ΔG}}_{\text{rxn}}$ at the given partial pressures is -559.625 kJ/mol.