Q20.92 CP - Chemistry: Molecular Nature Of Matter And Change

Question

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 ${\mathbf{\text{ΔG}}}_{\mathbf{\text{rxn}}}\mathbf{\text{ at 298 K}}$if ${\mathbf{\text{P}}}_{{\mathbf{\text{N}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{5}}}}{\mathbf{\text{= P}}}_{{\mathbf{\text{F}}}_{\mathbf{\text{2}}}}{\mathbf{\text{= 0.20 atm, P}}}_{{\mathbf{\text{NF}}}_{\mathbf{\text{3}}}}{\mathbf{\text{= 0.25 atm, and P}}}_{{\mathbf{\text{O}}}_{\mathbf{\text{2}}}}\mathbf{\text{= 0.50 atm.}}$

Step-by-Step Solution

Answer

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

b) The ${\text{ΔG}}_{\text{rxn}}^{\text{°}}$  of the reaction is -569.20kJ.

c)  The  ${\text{ΔG}}_{\text{rxn}}$ at 298K is -559.625 kJ/mol

Step 1: Concept Introduction

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.

Step 2: Writing a balanced equation for the gaseous reaction between N 2 O 5 and F 2 .

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

The balanced reaction equation is given as:

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

Step 3: Determining ΔG rxn °

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,

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

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

The  of the reaction is -569.20 kJ.

Step 4: Finding ΔG rxn  at 298 K

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

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

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

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

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