Chapter 12, Problem 73AE

Consider the decomposition of the compound C₅H₆O₃ as follows:

${\text{C}}_5{\text{H}}_6{\text{O}}_3\left(g\right)\to {\text{C}}_2{\text{H}}_6\left(g\right)+3\text{CO}\left(g\right)$

When a 5.63-g sample of pure C₅H₆O₃(g) was sealed into an otherwise empty 2.50-L flask and heated to 200°C, the pressure in the flask gradually rose to 1.63 atm and remained at that value. Calculate K for this reaction.

Interpretation Introduction

Interpretation: The decomposition reaction, mass of ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}$, volume of flask and rise in pressure is given. The value of equilibrium constant is to be calculated.

Concept introduction: The state when the reactants involved in a chemical reaction and the products formed in the reaction exist in concentrations having no further tendency to change is known as an equilibrium state of the reaction. When the equilibrium constant is expressed in terms of pressure, it is represented as $K_{\text{p}}$.

To determine: The value of equilibrium constant.

Answer to Problem 73AE

Answer

The calculated value of equilibrium constant is $\underset{\_}{6.67\times 10^{-6}}$.

Explanation of Solution

Explanation

Given

The rise in pressure of given reaction is $1.63\text{atm}$.

Mass of ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}$ is $\text{5}\text{.63}\text{g}$.

Volume of flask is $\text{2}\text{.50}\text{L}$.

The given temperature is $\text{200}°\text{C}$.

The molar mass of ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}$,

$\left[\text{5}\left(\text{12}\text{.01}\right)+\left(\text{6}\times \text{1}\text{.008}\right)+\left(\text{3}\times \text{15}\text{.999}\right)\right]\text{g}=\text{114}\text{.10}\text{g/mol}$

The conversion of degree Celsius $\left(°\text{C}\right)$ into Kelvin $\left(\text{K}\right)$ is done as,

$T\left(\text{K}\right)=T\left(°\text{C}\right)+\text{273}$

Hence,

The conversion of $\text{200}°\text{C}$ into Kelvin is,

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+\text{273}\\ T\left(\text{K}\right)=\left(\text{200}+\text{273}\right)\text{K}\\ =\text{473}\text{K}\end{array}$

The stated reaction is,

${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}{}_{\left(g\right)}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{}_{\left(g\right)}+{\text{3CO}}_{\left(g\right)}$

The number of moles is calculated using the formula,

$\text{Number}\text{of}\text{moles}=\text{Given}\text{mass}\times \frac{\text{1}\text{mole}\text{of}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{compound}}$

Substitute the values of given mass and molar mass of ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}$ in above equation.

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Given}\text{mass}\times \frac{\text{1}\text{mole}\text{of}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{compound}}\\ =\text{5}\text{.63}\text{g}\times \frac{\text{1}\text{mole}\text{of}{\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}}{\text{114}\text{.10}\text{g/mol}}\\ =\text{0}\text{.0493}\text{mol}\end{array}$

It is the initial moles of ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}$.

The initial moles of product is,

$\begin{array}{l}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)=0\\ \text{3CO}\left(g\right)=0\end{array}$

The formula of ideal gas law is,

$PV=n_{\text{total}}RT$

Where,

• $P$ is total pressure.
• $V$ is volume.
• $n_{\text{total}}$ is total moles.
• $R$ is the universal gas constant $(\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol})$.
• $T$ is absolute temperature.

Substitute the values of $P,V,R$ and $T$ in above equation.

$\begin{array}{c}PV=n_{\text{total}}RT\\ \text{1}\text{.63}\text{atm}\times \text{2}\text{.50}\text{L}=n_{\text{total}}\times \text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol}\times \text{473}\text{K}\\ n_{\text{total}}=\frac{\text{1}\text{.63}\text{atm}\times \text{2}\text{.50}\text{L}}{\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol}\times \text{473}\text{K}}\\ n_{\text{total}}=\text{0}\text{.105}\text{mol}\end{array}$

Simplify the above equation,

$\begin{array}{c}{n}_{\text{total}}=\frac{\text{1}\text{.63}\text{atm}\times \text{2}\text{.50}\text{L}}{\text{0}\text{.08206}\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol}\times \text{473}\text{K}}\\ n_{\text{total}}=\text{0}\text{.105}\text{mol}\end{array}$

It is assumed that the change in pressure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)$ is $x$.

On the reaction with one molecule of ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}$, the equilibrium reaction is,

$\begin{array}{ccccc}& {\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}{}_{\left(\text{g}\right)}& \text{®}& {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}{}_{\left(\text{g}\right)}& {\text{3CO}}_{\left(\text{g}\right)}\\ \text{Initial}\text{moles}\text{:}& \text{0}\text{.0493}& & \text{0}& \text{0}\\ \text{Change}\left(\text{moles}\right)\text{:}& \text{-x}& & \text{x}& \text{3x}\\ \text{Equilibrium}\left(\text{moles}\right)\text{:}& \text{0}\text{.0493-x}& & \text{x}& \text{3x}\end{array}$

Total moles is calculated using the formula,

$\text{Total}\text{moles}=\text{Moles}\text{of}{\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}+\text{Moles}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}+\text{Moles}\text{of}\text{CO}$

Substitute the values of all calculated moles in above equation.

$\begin{array}{c}\text{Total}\text{moles}=\text{Moles}\text{of}{\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}+\text{Moles}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}+\text{Moles}\text{of}\text{CO}\\ \text{0}\text{.105}\text{mol}=\text{0}\text{.0493}-x+x+\text{3}x\\ x=\text{0}\text{.0186}\text{mol}\end{array}$

Hence, the moles of reactant ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}$ is,

$\begin{array}{c}\text{Moles}\text{of}{\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}=0.0493-x\\ =\left(0.0493-0.0186\right)\text{mol}\\ =\text{0}\text{.0307}\text{mol}\end{array}$

The moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ is,

$\begin{array}{c}\text{Moles}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}=x\\ =0.0186\text{mol}\end{array}$

The moles of $\text{CO}$ is,

$\begin{array}{c}\text{Moles}\text{of}\text{CO}=3x\\ =3\times 0.0186\text{mol}\\ =\text{0}\text{.0558}\text{mol}\end{array}$

The concentration of each compound is calculated by using the formula,

$\text{Concentration}\text{of}\text{compound}=\frac{\text{Moles}\text{of}\text{compound}}{\text{Volume}\text{of}\text{compound}}$ (1)

Substitute the values of moles of ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}$ and volume of flask in equation (1).

$\begin{array}{c}\text{Concentration}\text{of}\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}\right]=\frac{\text{Moles}\text{of}{\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}}{\text{Volume}\text{of}\text{flask}}\\ =\frac{\text{0}\text{.0307}\text{mol}}{\text{2}\text{.50}\text{L}}\\ =\underset{\_}{\text{0}\text{.0123}\text{mol/L}}\end{array}$

Substitute the values of moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ and volume of flask in equation (1).

$\begin{array}{c}\text{Concentration}\text{of}\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\right]=\frac{\text{Moles}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}{\text{Volume}\text{of}\text{flask}}\\ =\frac{\text{0}\text{.0186}\text{mol}}{\text{2}\text{.50}\text{L}}\\ =\underset{\_}{\text{7}{\text{.44×10}}^{\text{-3}}\text{mol/L}}\end{array}$

Substitute the values of moles of $\text{CO}$ and volume of flask in equation (1).

$\begin{array}{c}\text{Concentration}\text{of}\left[\text{CO}\right]=\frac{\text{Moles}\text{of}\text{CO}}{\text{Volume}\text{of}\text{flask}}\\ =\frac{\text{0}\text{.0558}\text{mol}}{\text{2}\text{.50}\text{L}}\\ =\underset{\_}{\text{0}\text{.0223}\text{mol/L}}\end{array}$

At equilibrium, the equilibrium ratio is expressed by the formula,

$K_{\text{p}}=\frac{\text{Concentration}\text{of}\text{products}}{\text{Concentration}\text{of}\text{reactants}}$

Where,

• $K_{\text{p}}$ is the equilibrium constant.

The equilibrium ratio for the given reaction is,

$K=\frac{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\right]{\left[\text{CO}\right]}^3}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}\right]}$

Substitute the values of concentration of $\text{CO}$, ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ and ${\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}$ in above equation.

$\begin{array}{c}K=\frac{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\right]{\left[\text{CO}\right]}^3}{\left[{\text{C}}_{\text{5}}{\text{H}}_{\text{6}}{\text{O}}_{\text{3}}\right]}\\ =\frac{{\left(\text{0}\text{.0223}\text{mol/L}\right)}^3\times \text{7}\text{.44}\times {\text{10}}^{-\text{3}}\text{mol/L}}{\text{0}\text{.0123}\text{mol/L}}\\ =\underset{\_}{6.67\times 10^{-6}}\end{array}$

Conclusion

Conclusion

The calculated value of equilibrium constant is $\underset{\_}{6.67\times 10^{-6}}$