Question

Question: Consider the following reaction at

${\mathbf{N}}_2\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{F}}_2\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}{\mathbf{NF}}_3$

An equilibrium mixture contains the following partial pressures

${\mathbf{P}}_{{\mathbf{N}}_2}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{021}\mathbf{atm}\mathbf{,}{\mathbf{P}}_{{\mathbf{F}}_2}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{063}\mathbf{atm}\mathbf{,}\mathbf{and}{\mathbf{P}}_{{\mathbf{NF}}_3}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{48}\mathbf{atm}\mathbf{.}$

Calculate ${\mathbf{\Delta G}}^o$for the reaction at 800 K.

Short answer

The standard free energy change for the reaction is -71 kJ/mol

Step-by-step solution

Definition of standard free energy change

The amount of energy released in the conversion of reactants to products under standard conditions is defined as the standard free energy change.

Calculation of  ΔGo

The reaction is ${\text{N}}_{\text{2}}\text{(g)}+{\text{3F}}_{\text{2}}\text{(g)}\rightleftharpoons 2{\text{NF}}_{\text{3}}\text{(g)}$

$\begin{array}{l}{\text{ΔG}}_{\text{R}}^{\text{0}}=-\text{RTln}\frac{{\left[{\text{P}}_{{\text{NF}}_{\text{3}}}\right]}^2}{\left[{\text{P}}_{{\text{N}}_{\text{2}}}\right]{\left[{\text{P}}_{{\text{F}}_2}\right]}^3}\\ =-\text{0.008314\hspace{0.17em}kJ/mol}-\text{K}\times \text{800\hspace{0.17em}K}\times \text{ln}\frac{{\left[0.48\right]}^2}{\left[0.021\right]{\left[0.063\right]}^3}\\ =-71\text{\hspace{0.17em}kJ/mol}\end{array}$

The standard free energy change is-71 kJ/mol.