Chapter 13.5, Problem 2.4ACP

Interpretation Introduction

Interpretation: It should be determined that what mass of dissolved $C{O}_2$ which is released to the atmosphere in order to reach equilibrium.

Concept introduction:

• Henry’s law: The solubility of a gas in liquid is directly proportional to the gas pressure.

  $\begin{array}{l}{\text{S}}_{\text{g}}{\text{= k}}_{\text{H}}{\text{P}}_{\text{g}}\text{,}\\ \text{where,}{\text{S}}_{\text{g}}\text{is}\text{the}\text{gas}\text{solubility}\text{(in}\text{mol/kg)}\\ {\text{k}}_{\text{H}}\text{is Henry's constant }\\ {\text{P}}_{\text{g}}\text{ is the partial pressure of the gaseous solute}\text{.}\end{array}$

• $Density=\frac{Mass}{Volume}$

Answer to Problem 2.4ACP

$\text{6}\text{.0}\text{g}{\text{CO}}_2$ will be released to the atmosphere in order to reach equilibrium.

Explanation of Solution

$\begin{array}{l}\text{Given,}\\ \\ \text{partial}\text{pressure}\text{of}{\text{CO}}_{\text{2}}\text{=}\text{4}\text{bar}\\ \text{Henry's}\text{constant}\text{=}\text{0}\text{.034}\text{mol/kg}\text{.bar}\\ \\ \text{According}\text{to}\text{Henry's}\text{law,}\\ {\text{S}}_{\text{g}}\text{=}{\text{k}}_{\text{H}}{\text{P}}_{\text{g}}\\ \text{solubility}\text{of}{\text{CO}}_{\text{2}}\text{before}\text{opening}\text{=}\text{0}\text{.034}\text{mol/kg}\text{.bar}\text{×}\text{4}\text{bar}\\ \text{=}\text{0}\text{.136}\text{mol}\text{/kg}\\ \end{array}$

Assume that, the density of the solution is $1.0g/c{m}^3$. Therefore $1.0L$ corresponds to $1kg$. The amount of $C{O}_2$ dissolved is $0.136mol$.

Solubility of $C{O}_2$after opening can be calculated as follows:

$\begin{array}{l}\text{partial}\text{pressure}\text{of}{\text{CO}}_{\text{2}}\text{=}\text{3}{\text{.7×10}}^{\text{-4}}\text{bar}\\ \text{Henry's}\text{constant}\text{=}\text{0}\text{.034}\text{mol/kg}\text{.bar}\end{array}$

According to Henry’s law:

${\text{S}}_{\text{g}}\text{=}{\text{k}}_{\text{H}}{\text{P}}_{\text{g}}$

Substituting the given values in the above equation, the solubility can be calculated.

$\begin{array}{l}{\text{S}}_{\text{g}}\text{=}\text{0}\text{.034mol/kg}\text{.bar}\text{×3}{\text{.7×10}}^{\text{-4}}\text{bar}\\ \text{=}\text{1}\text{.26}{\text{×10}}^{\text{-5}}\text{mol/kg}\end{array}$ $\begin{array}{l}\text{Solubility}\text{of}{\text{CO}}_{\text{2}}\text{after}\text{opening}\text{=}\text{1}\text{.26}\text{×}{\text{10}}^{\text{-5}}\text{mol/Kg}\text{.}\\ \text{The}\text{amount}\text{of}{\text{CO}}_{\text{2}}\text{in}\text{1}\text{.0}\text{kg}\text{is}\text{1}\text{.26}\text{×}{\text{10}}^{\text{-5}}\text{mol}\text{.}\end{array}$

$\begin{array}{l}\text{Mass}\text{of}{\text{CO}}_{\text{2}}\text{released}\text{=}\text{(Amount}\text{of}{\text{CO}}_{\text{2}}\text{present}\text{before}\text{opening}\text{soda}\text{-}\\ \text{Amount}\text{of}{\text{CO}}_{\text{2}}\text{present}\text{after}\text{opening}\text{soda)}\text{×}\text{Molar}\text{mass}\text{of}{\text{CO}}_{\text{2}}\\ \\ \text{=}\text{(0}\text{.136}\text{mol}{\text{CO}}_{\text{2}}\text{-}\text{1}\text{.26}\text{×}{\text{10}}^{\text{-5}}\text{mol}{\text{CO}}_{\text{2}}\text{)}\text{×}\text{44}\text{.01}\text{g/mol}{\text{CO}}_{\text{2}}\\ \text{=}\text{6}\text{.0}\text{g}{\text{CO}}_{\text{2}}\\ \end{array}$

Conclusion

$\text{6}\text{.0}\text{g}{\text{CO}}_2$ would be released to the atmosphere in order to reach equilibrium.