Chapter 18, Problem 18.125QP

Interpretation Introduction

Interpretation:

At equilibrium the percent of ammonia dissociation has to be calculated at $\text{402}{}^{\text{o}}\text{C}\text{.}$

Concept introduction:

Standard free energy change:

Standard free energy change is measured by subtracting the product of temperature and standard entropy change from the standard enthalpy change of a system.

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy change}\\ {\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy change}\text{and T}\text{-}\text{temperature}\text{.}\end{array}$

To calculate: the percent of ammonia dissociation at equilibrium

Answer to Problem 18.125QP

• The percent of ammonia dissociation at equilibrium and $\text{402}{}^{\text{o}}\text{C}$ is $\text{93}\text{\%}\text{.}$

Explanation of Solution

Given reaction and information

$\begin{array}{l}{\text{2NH}}_{\text{3}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{3H}}_{\text{2}}{}_{\left(\text{g}\right)}\text{+}{\text{N}}_{\text{2}}{}_{\left(\text{g}\right)}\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\text{:}\text{2}\text{×}\text{-45}\text{.90}\text{0}\text{0}\text{kJ}\\ {\text{S}}^{\text{o}}\text{:}\text{2}\text{×}\text{192}\text{.7}\text{3}\text{×}\text{130}\text{.6}\text{191}\text{.6}\text{J/K}\\ \text{Temperature is 402}{}^{\text{o}}\text{C}\end{array}$

Calculate the value of ${\text{ΔH}}^{\text{o}}$

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{reactants}\right)\\ \text{=}\left[\text{0}\text{-}\left(\text{2}\text{×}\text{-45}\text{.90}\right)\right]\text{kJ}\\ \text{=}\text{91}\text{.80}\text{kJ}\text{.}\end{array}$

Calculate the value of ${\text{ΔS}}^{\text{o}}$

$\begin{array}{l}{\text{ΔS}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nS}}^{\text{o}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mS}}^{\text{o}}}\left(\text{reactants}\right)\\ \text{=}\left[\left(\text{3}\text{×}\text{130}\text{.6}\right)\text{+}\text{191}\text{.6}\text{-}\left(\text{2}\text{×}\text{192}\text{.7}\right)\right]\text{J/K}\\ \text{=}\text{198}\text{.0}\text{J/K}\text{.}\end{array}$

Calculate the value of ${\text{ΔG}}^{\text{o}}{}_{\text{T}}$

$\begin{array}{l}{\text{ΔG}}^{\text{o}}{}_{\text{T}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{=}\text{91}\text{.8}\text{kJ}\text{-}\left(\text{675}\text{K}\right)\left(\text{0}\text{.1980}\text{kJ/K}\right)\\ \text{=}\text{-41}\text{.85}\text{kJ}\\ \text{=}\text{-41}\text{.85}\text{×}{\text{10}}^{\text{3}}\text{J}\text{.}\end{array}$

Calculate the value of $\text{K}$ using the relationship between $\text{K}\text{and}{\text{ΔG}}_{\text{T}}^{\text{o}}$

$\begin{array}{l}\text{ln}\text{K}\text{=}\frac{{\text{-ΔG}}^{\text{o}}}{\text{RT}}\\ \text{=}\frac{\text{-41}\text{.85}\text{×}{\text{10}}^{\text{3}}}{\text{-8}\text{.31}\text{×}\text{675}}\\ \text{=}\text{7}\text{.460}\\ \text{K}\text{=}{\text{K}}_{\text{P}}\text{=}{\text{e}}^{\text{7}\text{.460}}\text{=}\text{1738}\text{=}\text{1}\text{.7}\text{×}{\text{10}}^{\text{3}}\text{.}\end{array}$

Now, calculate the value of ${\text{K}}_{\text{c}}$

$\begin{array}{l}{\text{K}}_{\text{C}}\text{=}{\text{K}}_{\text{P}}{\left(\text{RT}\right)}^{\text{-2}}\\ \text{=}\left(\text{1738}\right){\left(\text{0}\text{.0821}\text{×}\text{675}\right)}^{\text{-2}}\\ \text{=}\text{0}\text{.5661}\text{.}\end{array}$

The starting concentration of ammonia is $\text{1}\text{.00}\text{mol/20}\text{.0}\text{L}\text{=}\text{0}\text{.0500}\text{M}\text{.}$

$\begin{array}{l}{\text{2NH}}_{\text{3}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{3H}}_{\text{2}}{}_{\left(\text{g}\right)}\text{+}{\text{N}}_{\text{2}}{}_{\left(\text{g}\right)}\\ \text{Starting}\text{0}\text{.0500}\text{0}\text{0}\\ \text{Change}\text{-2x}\text{+}\text{3x}\text{+x}\\ \text{Equilibrium}\text{0}\text{.0500}\text{-}\text{2x}\text{3x}\text{x}\end{array}$

The equilibrium expression is given by

$\begin{array}{l}{\text{k}}_{\text{C}}\text{=}\frac{{\left[{\text{H}}_{\text{2}}\right]}^{\text{3}}\left[{\text{N}}_{\text{2}}\right]}{{\left[{\text{NH}}_{\text{3}}\right]}^{\text{2}}}\\ \text{=}\frac{{\left(\text{3x}\right)}^{\text{3}}\text{x}}{{\left(\text{0}\text{.0500}\text{-}\text{2x}\right)}^{\text{2}}}\text{=}\text{0}\text{.5661}\\ \text{=}\frac{{\text{x}}^{\text{4}}}{{\left(\text{0}\text{.0500}\text{-}\text{2x}\right)}^{\text{2}}}\text{=}\frac{\text{0}\text{.5661}}{\text{27}}\text{=}\text{2}\text{.096}\text{×}{\text{10}}^{\text{-2}}\text{.}\end{array}$

Take square root on both sides

$\frac{{\text{x}}^{\text{2}}}{\left(\text{0}\text{.0500}\text{-}\text{2x}\right)}\text{=}\text{0}\text{.1448}$

Rearrange the above equation in quadratic form

${\text{x}}^{\text{2}}\text{+}\left(\text{0}\text{.2896}\right)\text{x}\text{-}\left(\text{7}\text{.240}\text{×}{\text{10}}^{\text{-3}}\right)\text{=}\text{0}$

Calculate the value of $\text{x}$

$\begin{array}{l}\text{x}\text{=}\frac{\text{-0}\text{.2896}\text{±}\sqrt{{\left(\text{0}\text{.2896}\right)}^{\text{2}}\text{+}\text{4}\left(\text{7}\text{.240}\text{×}{\text{10}}^{\text{-3}}\right)}}{\text{2}}\\ \text{=}\text{0}\text{.02315}\text{M}\left(\text{positive}\text{root}\right)\\ \text{Hence,}\left[{\text{NH}}_{\text{3}}\right]\text{=}\text{0}\text{.0500}\text{-}\text{2}\left(\text{0}\text{.02315}\right)\text{=}\text{0}\text{.00369}\text{M}\text{.}\end{array}$

Calculate the percent ammonia dissociated

$\begin{array}{l}{\text{percent NH}}_{\text{3}}\text{ dissociated}\text{=}\left(\text{1}\text{-}\frac{\text{0}\text{.00369}\text{M}}{\text{0}\text{.0500}\text{M}}\right)\text{×}\text{100}\text{\%}\\ \text{=}\text{92}\text{.6}\text{=}\text{93}\text{\%}\text{.}\end{array}$

The percent of ammonia dissociation at equilibrium and $\text{402}{}^{\text{o}}\text{C}$ is $\text{93}\text{\%}\text{.}$

Conclusion

The percent of ammonia dissociation at equilibrium and $\text{402}{}^{\text{o}}\text{C}$ was calculated.