The oxidation of nitrogen monoxide is favoured at :$\text{457K}$

${\text{2NO(g)+O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2NO}}_{\text{2}}{\text{(g)K}}_{\text{p}}{\text{=1.3×10}}^{\text{4}}$

(a) Calculate ${\text{K}}_{\text{c}}$at .$\text{457K}$

(b) Find${\text{ΔH}}_{\text{rxn}}^{\text{o}}$ from standard heats of formation.

(c) At what temperature does ${\text{K}}_{\text{c}}{\text{=6.4×10}}^{\text{9}}$?

The vapour pressure, also known as equilibrium vapour pressure, is the pressure exerted by vapour at a particular temperature in a closed system when it is in thermodynamic equilibrium with the condensed phase (solid or liquid). The equilibrium vapour pressure is a measurement of a liquid's evaporation rate.

The heat released or absorbed (enthalpy change) during the production of pure material from its ingredients at constant pressure is referred to as heat of formation in chemistry (in their standard states).

(a)

The reaction given is –

${\text{2NO(g)+O}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{2NO}}_2{\text{(g)K}}_p{\text{=1.3×10}}^4$

To solve for ${\text{K}}_{\text{c}}$, use the formula for the relation of${\text{K}}_{\text{c}}$and${\text{K}}_p$which is ${\text{K}}_{\text{p}}{\text{=K}}_{\text{c}}{\text{(RT)}}^{\text{Δn}}$. Identify the given values first.

$\begin{array}{c}{\text{K}}_p\text{=1.3}\times {\text{10}}^{\text{4}}\\ \text{R=0.0821}\frac{\text{L}\times \text{atm}}{\text{mol}\times \text{K}}\\ \text{T=457K}\end{array}$

Calculate the value for –${\text{Δn}}_{gas}$

$\begin{array}{c}{\text{Δn}}_{\text{gas}}{\text{=n}}_{\text{products}}{\text{-n}}_{\text{reactants}}\\ \text{=2-(2+1)}\\ \text{=-1}\end{array}$

Now calculate the value for –${\text{K}}_{\text{c}}$

localid="1662032602520" $\begin{array}{c}\text{1.3}\cdot {\text{10}}^{\text{4}}{\text{=K}}_{\text{c}}{\left(0.0821\frac{L\times atm}{mol\times k}\times 457K\right)}^{-1}\\ {\text{K}}_{\text{c}}\text{=}\frac{\text{1.3}\times {\text{10}}^{\text{4}}}{{\text{(0.0821}\frac{\text{L}\times \text{atm}}{\text{mol}\times \text{K}}\text{×457K)}}^{\text{-1}}}\\ {\text{K}}_{\text{c}}{\text{=4.9×10}}^{\text{5}}\end{array}$

Therefore, the value for the equilibrium constant is obtained as .${\text{K}}_{\text{c}}\text{=4.9}\times {\text{10}}^{\text{5}}$

(b)

Use the standard heats of reaction to solve for.${\text{ΔH}}_{\text{rxn}}^{\text{o}}$ Refer to Appendix B and list all the reaction components.

$\begin{array}{c}{\text{ΔH}}_{\text{f}}^{\text{°}}\text{(NO(g))=90.29kJ/mol}\\ {\text{ΔH}}_{\text{f}}^{\text{°}}\left({\text{O}}_{\text{2}}\text{(g)}\right)\text{=0kJ/mol}\\ {\text{ΔH}}_{\text{f}}^{\text{°}}\left({\text{NO}}_{\text{2}}\text{(g)}\right)\text{=33.2kJ/mol}\end{array}$

Substitute the values and calculate–${\text{ΔH}}_{\text{rxn}}^{\text{o}}$

$\begin{array}{l}\left.{\text{ΔH}}_{\text{rxn}}^{\text{°}}{\text{=ΔH}}_{\text{f}}^{\text{°}}{\text{(products)-ΔH}}_{\text{f}}^{\text{°}}\text{(reactants}\right)\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\left(\text{2mol}\cdot {\text{ΔH}}_f^{\text{o}}{\text{NO}}_{\text{2}}\right)\text{-}\left(\text{2mol}\cdot {\text{ΔH}}_{\text{f}}^{\text{o}}\text{NO+1mol}\cdot {\text{ΔH}}_f^{\text{o}}{\text{O}}_{\text{2}}\right)\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=2mol}\cdot \text{33.2kJ/mol-2mol}\cdot \text{90.29kJ/mol-1mol}\cdot \text{0kJ/mol}\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=-114.18kJ}\end{array}$

Therefore, the value for standard heat of reaction is obtained as.${\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=-114.18kJ}$

(c)

The value for${\text{K}}_{\text{c}}{\text{=4.9×10}}^{\text{5}}$at temperature${\text{T}}_1\text{=457K}$.

It is needed to obtain${\text{T}}_2$, the temperature at which .${\text{K}}_{\text{c}}{\text{=6.4×10}}^9$

Use the Vant-Hoff equation to express and calculate ${\text{T}}_2$–

$\begin{array}{l}\text{ln}\frac{{\text{K}}_{\text{2}}}{{\text{K}}_{\text{1}}}\text{=-}\frac{{\text{ΔH}}_{\text{rxn}}^{\text{o}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]\\ \text{ln}\frac{\text{6.4}\times {\text{10}}^{\text{9}}}{\text{4.9}\times {\text{10}}^{\text{5}}}\text{=-}\frac{\text{-114.18kJ×}\left(\frac{{\text{10}}^{\text{3}}\text{J}}{\text{1kJ}}\right)}{\text{8.314J/K}\cdot \text{mol}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{\text{457K}}\right]\\ \text{ln}\frac{\text{6.4}\times {\text{10}}^{\text{9}}}{\text{4.9}\times {\text{10}}^{\text{5}}}\text{=}\frac{\text{114.18}\times {\text{10}}^{\text{3}}\text{J}}{\text{8.314J/K}\cdot \text{mol}}\left[\frac{{\text{457K-T}}_{\text{2}}}{\text{457K}\cdot {\text{T}}_{\text{2}}}\right]\\ {\text{T}}_{\text{2}}\text{=348K}\end{array}$

Therefore, the value for temperature is obtained as .${\text{T}}_2\text{=348K}$