The kinetics and equilibrium of the decomposition of hydrogen iodide have been studied extensively:

$\text{2HI(g)}\rightleftharpoons {\text{H}}_{\text{2}}{\text{(g)+I}}_{\text{2}}\text{(g)}$

(a) At${\text{298K,K}}_{\text{c}}{\text{=1.26×10}}^{\text{-3}}$for this reaction. Calculate ${\text{K}}_{\text{p}}$.

(b) Calculate${\text{K}}_{\text{c}}$for the formation of $\text{HI}$at $\text{298K}$.

(c)Calculate localid="1657006587824" ${\text{ΔH}}_{\text{rxn}}^{\text{o}}$forlocalid="1657006574683" $\text{HI}$decomposition from ${\text{ΔH}}_{\text{f}}^{\text{o}}$values.

(d) At${\text{729K,K}}_{\text{c}}{\text{=2.0×10}}^{\text{-2}}$for $\text{HI}$decomposition. Calculate${\text{ΔH}}_{\text{rxn}}$for this reaction from the Vant-Hoff equation.

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 –

$\text{2HI(g)}\rightleftharpoons {\text{H}}_2{\text{(g)+I}}_{\text{2}}\text{(g)}$

To solve for ${\text{K}}_{\text{p}}$, 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.

localid="1662032772545" $\begin{array}{c}{\text{K}}_c{\text{=1.26×10}}^{\text{-3}}\\ \text{R=0.0821}\frac{\text{L}\times \text{atm}}{\text{mol}\times \text{K}}\\ \text{T=298K}\end{array}$

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

$\begin{array}{c}{\text{Δn}}_{\text{gas}}{\text{=n}}_{\text{products}}{\text{-n}}_{\text{reactants}}\\ \text{=(1+1)-2}\\ \text{=0}\end{array}$

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

localid="1662032813004" role="math" $\begin{array}{c}{\text{K}}_p{\text{=1.26×10}}^{\text{-3}}\left(0.0821\frac{\text{L}\times \text{atm}}{\text{mol}\times \text{K}}\times 298K\right)\text{°}\\ {\text{K}}_p{\text{=1.26×10}}^{\text{-3}}\text{(1)}\\ {\text{K}}_p{\text{=1.26×10}}^{\text{-3}}\end{array}$

Therefore, the value for equilibrium pressure is obtained as.${\text{K}}_p{\text{=1.26×10}}^{\text{-3}}$

(b)

The formation of is the reverse reaction and the equilibrium constant for this reaction will be the inverse of the equilibrium constant for the forward reaction –

$\begin{array}{l}{\text{K'}}_{\text{c}}\text{=}\frac{\text{1}}{{\text{K}}_{\text{c}}}\\ {\text{K'}}_{\text{c}}\text{=}\frac{\text{1}}{{\text{1.26×10}}^{\text{-3}}}\\ {\text{K'}}_{\text{c}}\text{=794}\end{array}$

Therefore, the value for equilibrium concentration constant is obtained as .${\text{K'}}_{\text{c}}\text{=794}$

(c)

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

$\begin{array}{c}{\text{ΔH}}_{\text{f}}^{\text{°}}\text{(HI(g))=25.9kJ/mol}\\ {\text{ΔH}}_{\text{f}}^{\text{°}}\left({\text{H}}_{\text{2}}\text{(g)}\right)\text{=0kJ/mol}\\ {\text{ΔH}}_{\text{f}}^{\text{°}}\left({\text{I}}_{\text{2}}\text{(g)}\right)\text{=0kJ/mol}\end{array}$

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

$\begin{array}{l}\left.{\text{ΔH}}_{\text{rxn}}^{\text{°}}{\text{=ΔH}}_{\text{f}}^{\text{°}}{\text{(products)-ΔH}}_{\text{f}}^{\text{°}}\text{(reactants}\right)\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=}\left(\text{1mol}\cdot {\text{ΔH}}_f^{\text{o}}{H}_2\text{+1mol}\cdot {\text{ΔH}}_f^{\text{o}}{I}_2\right)\text{-}\left(\text{2mol}\cdot {\text{ΔH}}_f^{\text{o}}HI\right)\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=1mol}\cdot \text{0kJ/mol+1mol}\cdot \text{0kJ/mol-1mol}\cdot \text{25.9kJ/mol}\\ {\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=-51.8kJ}\end{array}$

Therefore, the value for standard heat of reaction is obtained as ${\text{ΔH}}_{\text{rxn}}^{\text{°}}\text{=-51.8kJ}$.

(d)

The value for${\text{K}}_{\text{c}}{\text{=1.26×10}}^{\text{-3}}$at temperature${\text{T}}_1\text{=298K}$’

The value for${\text{K}}_{\text{c}}{\text{=2.0×10}}^{\text{-2}}$ at temperature.${\text{T}}_2\text{=729K}$

Use the Vant-Hoff equation to express and calculate${\text{ΔH}}_{\text{rxn}}^{\text{o}}$–

$\begin{array}{c}\text{ln}\frac{{\text{K}}_{\text{2}}}{{\text{K}}_{\text{1}}}\text{=-}\frac{{\text{ΔH}}_{\text{rxn}}^{\text{o}}}{\text{R}}\left[\frac{\text{1}}{{\text{T}}_{\text{2}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{1}}}\right]\\ \text{ln}\frac{\text{2.0}\times {\text{10}}^{\text{-2}}}{\text{1.26}\times {\text{10}}^{-3}}\text{=-}\frac{{\text{ΔH}}_{\text{rxn}}^{\text{o}}}{\text{8.314J/K}\cdot \text{mol}}\left[\frac{\text{1}}{\text{729K}}\text{-}\frac{\text{1}}{\text{298K}}\right]\\ 2.7646\text{=}\frac{{\text{ΔH}}_{\text{rxn}}^{\text{o}}}{\text{8.314J/K}\cdot \text{mol}}\left[\frac{\text{298K-729K}}{\text{298K}\cdot \text{729K}}\right]\\ {\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=}\frac{2.7646\times \text{8.314J/K}\cdot \text{mo}l\times \text{298K}\times \text{729K}}{\text{431K}}\\ {\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=1.2}\times {\text{10}}^{\text{4}}\text{J/mol}\end{array}$

Therefore, the value for standard heat of reaction is obtained as.

${\text{ΔH}}_{\text{rxn}}^{\text{o}}\text{=1.2}\times {\text{10}}^{\text{4}}\text{J/mol}$