91Ó°ÊÓ

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:

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

b) Since Gibb's energy is a state function, it only depends on the system's starting and final states, we can calculate ${\text{Δ³Ò}}_{\text{rxn}}^{\text{°}}$

${\text{Δ³Ò}}_{\text{rxn}}^{\text{°}}{\text{=Δ³Ò}}_{\text{products}}^{\text{°}}{\text{-Δ³Ò}}_{\text{reactants}}^{\text{°}}$

Using appendix B,

$\begin{array}{l}{\text{Δ³Ò}}_{\text{rxn}}^{\text{°}}\text{=}\left[{\text{4Δ³Ò}}^{\text{°}}\left({\text{NF}}_{\text{3}}\text{(g)}\right){\text{+5Δ³Ò}}^{\text{°}}\left({\text{O}}_{\text{2}}\text{(g)}\right)\right]\text{-}\left[{\text{2Δ³Ò}}^{\text{°}}\left({\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\text{(g)}\right){\text{+6×Δ³Ò}}^{\text{°}}\left({\text{F}}_{\text{2}}\text{(g)}\right)\right]\\ {\text{Δ³Ò}}_{\text{rxn}}^{\text{°}}\text{=}\left[\text{4³¾´Ç±ô×}\left(\text{-83.30}\frac{\text{kJ}}{\text{mol}}\right)\text{+5³¾´Ç±ô×}\left(\text{0.00}\frac{\text{kJ}}{\text{mol}}\right)\right]\text{-}\left[\text{2³¾´Ç±ô×}\left(\text{118.00}\frac{\text{kJ}}{\text{mol}}\right)\text{+6³¾´Ç±ô×}\left(\text{0.00}\frac{\text{kJ}}{\text{mol}}\right)\right]\\ \end{array}$

${\text{Δ³Ò}}_{\text{ran}}^{\text{°}}\text{=-569.20kJ}$

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, ${\text{Δ³Ò}}_{\text{rxn}}$can be expressed and calculated as:

$\begin{array}{c}{\text{Δ³Ò}}_{\text{rxn}}{\text{=Δ³Ò}}_{\text{rxn}}^{\text{°}}{\text{+R×T×lnQΔ³Ò}}_{\text{rxn}}\\ {\text{=-569.20×10}}^{\text{3}}\frac{\text{J}}{\text{mol}}\text{+8.314}\frac{\text{J}}{\text{³¾´Ç±ô×°­}}\text{×298°­Ã—±ô²Ô(47.68372)}\\ {\text{Δ³Ò}}_{\text{rxn}}\text{=-559625.199}\frac{\text{J}}{\text{mol}}\\ \text{=-559.625kJ/mol}\end{array}$

The${\text{Δ³Ò}}_{\text{rxn}}$ at the given partial pressures is -559.625 kJ/mol.