Chapter 18, Problem 18.126QP

Interpretation Introduction

Interpretation:

At equilibrium the number of moles of $\text{CO}\text{and}{\text{H}}_{\text{2}}\text{O}$ formed has to be calculated at $\text{785}{}^{\text{o}}\text{C}\text{.}$

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 number of moles of $\text{CO}\text{and}{\text{H}}_{\text{2}}\text{O}$ formed at $\text{785}{}^{\text{o}}\text{C}$

Answer to Problem 18.126QP

• The number of moles of $\text{CO}\text{and}{\text{H}}_{\text{2}}\text{O}$ formed at $\text{785}{}^{\text{o}}\text{C}$ is $\text{5}\text{.0}\text{×}{\text{10}}^{\text{-4}}\text{mol}\text{.}$

Explanation of Solution

Given reaction and information

$\begin{array}{l}{\text{CO}}_{\left(\text{g}\right)}\text{+}{\text{3H}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{CH}}_{\text{4}}{}_{\left(\text{g}\right)}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\\ {\text{ΔH}}_{\text{f}}^{\text{o}}\text{:}\text{-110}\text{.5}\text{0}\text{-74}\text{.87}\text{-241}\text{.8}\text{kJ}\\ {\text{S}}^{\text{o}}\text{:}\text{197}\text{.5}\text{3}\text{×}\text{130}\text{.6}\text{186}\text{.1}\text{188}\text{.7}\text{J/K}\\ \text{Temperature}\text{785}{}^{\text{o}}\text{C}\text{.}\end{array}$

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

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{reactants}\right)\\ \text{=}\left[\left(\text{-74}\text{.87}\right)\text{+}\left(\text{-241}\text{.8}\right)\text{-}\left(\text{-110}\text{.5}\right)\right]\text{kJ}\\ \text{=}\text{-206}\text{.17}\text{kJ}\text{.}\end{array}$

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

$\begin{array}{l}{\text{ΔS}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nS}}^{\text{o}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mS}}^{\text{o}}}\left(\text{reactants}\right)\\ \text{=}\left[\text{186}\text{.1}\text{+}\text{188}\text{.7}\text{-}\left(\text{197}\text{.5}\right)\text{-}\left(\text{3}\text{×}\text{130}\text{.6}\right)\right]\text{J/K}\\ \text{=}\text{-214}\text{.5}\text{J/K}\text{.}\end{array}$

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

$\begin{array}{l}{\text{ΔG}}^{\text{o}}{}_{\text{T}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{=}\text{-206}\text{.17}\text{kJ}\text{-}\left(\text{1058}\text{K}\right)\left(\text{-0}\text{.21450}\text{kJ/K}\right)\\ \text{=}\text{20}\text{.771}\text{kJ}\\ \text{=}\text{20}\text{.771}\text{×}{\text{10}}^{\text{3}}\text{J}\text{.}\end{array}$

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

$\begin{array}{l}\text{ln}\text{K}\text{=}\frac{{\text{-ΔG}}^{\text{o}}}{\text{RT}}\\ \text{=}\frac{\text{20}\text{.771}\text{×}{\text{10}}^{\text{3}}}{\text{-8}\text{.31}\text{×}\text{1058}}\\ \text{=}\text{-2}\text{.362}\\ \text{K}\text{=}{\text{K}}_{\text{P}}\text{=}{\text{e}}^{\text{-2}\text{.362}}\text{=}\text{0}\text{.09418}\text{.}\end{array}$

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

$\begin{array}{l}{\text{K}}_{\text{C}}\text{=}{\text{K}}_{\text{P}}{\left(\text{RT}\right)}^{\text{2}}\\ \text{=}\left(\text{0}\text{.09418}\right){\left(\text{0}\text{.0821}\text{×}\text{1058}\right)}^{\text{2}}\\ \text{=}\text{710}\text{.6}\text{.}\end{array}$

The starting concentration of $\text{CO}\text{and}{\text{H}}_{\text{2}}$ is

$\begin{array}{l}\left[\text{CO}\right]\text{=}\frac{\text{0}\text{.0100}\text{mol}}{\text{25}\text{.0}\text{L}}\text{=}\text{4}\text{.00}\text{×}{\text{10}}^{\text{-4}}\text{M}\text{.}\\ \left[{\text{H}}_{\text{2}}\right]\text{=}\frac{\text{0}\text{.0300}\text{mol}}{\text{25}\text{.0}\text{L}}\text{=}\text{1}\text{.20}\text{×}{\text{10}}^{\text{-3}}\text{M}\text{.}\end{array}$

Now, construct the concentration table

$\begin{array}{l}{\text{CO}}_{\left(\text{g}\right)}\text{+}{\text{3H}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{CH}}_{\text{4}}{}_{\left(\text{g}\right)}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}\\ \text{Starting}\text{4}\text{.00}\text{×}{\text{10}}^{\text{-4}}\text{1}\text{.20}\text{×}{\text{10}}^{\text{-3}}\text{0}\text{0}\\ \text{Change}\text{-x}\text{-3x}\text{+x}\text{+x}\\ \text{Equilibrium}\left(\text{4}\text{.00}\text{×}{\text{10}}^{\text{-4}}\right)\text{-x}\left(\text{1}\text{.20}\text{×}{\text{10}}^{\text{-3}}\right)\text{-3x}\text{x}\text{x}\end{array}$

The equilibrium expression is given by

$\begin{array}{l}{\text{k}}_{\text{C}}\text{=}\frac{\left[{\text{CH}}_{\text{4}}\right]\left[{\text{H}}_{\text{2}}\text{O}\right]}{\left[\text{CO}\right]{\left[{\text{H}}_{\text{2}}\right]}^{\text{3}}}\\ \text{=}\frac{{\text{x}}^2}{\left[\left(4.00\times {10}^{-4}\right)-x\right]{\left[\left(1.20\times {10}^{-3}\right)-3x\right]}^3}\text{=}710.6\end{array}$

The denominator can be written as ${\text{3}}^{\text{3}}{\left[\left(\text{4}\text{.00}\text{×}{\text{10}}^{\text{-4}}\right)\text{-x}\right]}^{\text{4}}$ so this equation can be reduced to quadratic equation.

Multiply both sides of the equation by $\text{27}$, and the take square rrot.

$\frac{\text{x}}{{\left(\text{4}\text{.00}\text{×}{\text{10}}^{\text{-4}}\text{-x}\right)}^{\text{2}}}\text{=}\text{138}\text{.51}$

Rearrange the above equation in quadratic form

${\text{x}}^{\text{2}}\text{+}\left(\text{-8}\text{.019}\text{×}{\text{10}}^{\text{-3}}\right)\text{x}\text{+}\left(\text{1}\text{.60}\text{×}{\text{10}}^{\text{-7}}\right)\text{=}\text{0}$

Calculate the value of $\text{x}$

$\begin{array}{l}\text{x}\text{=}\frac{\text{-}\left(\text{-8}\text{.019}\text{×}{\text{10}}^{\text{-3}}\right)\text{±}\sqrt{{\left(\text{-8}\text{.019}\text{×}{\text{10}}^{\text{-3}}\right)}^{\text{2}}\text{-}\text{4}\left(\text{1}\text{.60}\text{×}{\text{10}}^{\text{-7}}\right)}}{\text{2}}\\ \text{=}\frac{\text{8}\text{.019×}{\text{10}}^{\text{-3}}\text{±}\text{7}\text{.979}\text{×}{\text{10}}^{\text{-3}}}{\text{2}}\text{=}\text{2}\text{.00}\text{×}{\text{10}}^{\text{-5}}\left(\text{negative}\text{root}\right)\\ \text{x}\text{=}\left[{\text{CH}}_{\text{4}}\right]\text{=}\left[{\text{H}}_{\text{2}}\text{O}\right]\text{=}\text{2}\text{.00}\text{×}{\text{10}}^{\text{-5}}\text{M}\text{.}\end{array}$

Calculate the mole of $\text{CO}\text{and}{\text{H}}_{\text{2}}\text{O}$

$\begin{array}{l}\text{mol}{\text{CH}}_{\text{4}}\text{=}\text{mol}{\text{H}}_{\text{2}}\text{O}\text{=}\text{25}\text{.0}\text{L}\text{×}\left(\text{2}\text{.00}\text{×}{\text{10}}^{\text{-5}}\text{mol/L}\right)\\ \text{=5}\text{.00}\text{×}{\text{10}}^{\text{-4}}\text{mol}\text{.}\end{array}$

The number of moles of $\text{CO}\text{and}{\text{H}}_{\text{2}}\text{O}$ formed at $\text{785}{}^{\text{o}}\text{C}$ is $\text{5}\text{.0}\text{×}{\text{10}}^{\text{-4}}\text{mol}\text{.}$

Conclusion

The number of moles of $\text{CO}\text{and}{\text{H}}_{\text{2}}\text{O}$ formed at $\text{785}{}^{\text{o}}\text{C}$ was calculated at equilibrium.