Chapter 5, Problem 127AP

The carbon dioxide exhaled by sailors in a submarine is often removed by reaction with an aqueous lithium hydroxide solution. (a) Write a balanced equation for this

process. (Hint: The products are water and a soluble salt.) (b) If every sailor consumes $\text{1}\text{.2 }\times {\text{ 10}}^{\text{4}}\text{kJ}$ of energy every day and assuming that this energy is totally supplied by the metabolism of glucose $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\right)$, calculate the amounts of ${\text{CO}}_{\text{2}}$ produced and $\text{LiOH}$ required to purify the air.

Interpretation Introduction

Interpretation:

The balanced equation for the given process is to be written and the amounts of ${\text{CO}}_{\text{2}}$ produced and $\text{LiOH}$ requiredto purify the air are to be calculated.

Concept introduction:

The standard enthalpy for a reaction is the amount of enthalpy change which occurs at standard conditions.

The standard enthalpy of a reaction is to be determined by using the equation as given below:

$\Delta H{°}_{\text{rxn}}={\displaystyle \sum n}\Delta H_{\text{f}}^{\text{°}}\left(\text{products}\right)-{\displaystyle \sum m}\Delta H_{\text{f}}^{°}\left(\text{reactants}\right)$

Here, the stoichiometric coefficients are represented by m for the reactants and n for the products. The enthalpy of formation at standard conditions is represented by $\Delta H_{\text{f}}^{°}$.

The value of enthalpy of formation of an element is zero at its most stable state.

Answer to Problem 127AP

Solution:

(a) $2\text{ LiOH}\left(aq\right)+{\text{CO}}_{\text{2}}\left(g\right)\to {\text{Li}}_2{\text{CO}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

(b) $\begin{array}{l}1.1{\text{ kg CO}}_2\\ 1.2\text{ kg LiOH}\end{array}$

Explanation of Solution

a) A balance equation for the given process.

The reaction of carbon dioxide with aqueous lithium hydroxide solution and the products formed which are water and a soluble salt are as follows:

$\begin{array}{l}\text{LiOH}\left(aq\right)+{\text{CO}}_2\left(g\right)\to {\text{Li}}_2{\text{CO}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\\ \end{array}$

Here, the salt formed is lithium carbon trioxide.

Now, balance this equation to obtain the balanced chemical equationas shown below:

$2\text{ LiOH}\left(aq\right)+{\text{CO}}_{\text{2}}\left(g\right)\to {\text{Li}}_2{\text{CO}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Hence, the balanced chemical equation is; $2\text{ LiOH}\left(aq\right)+{\text{CO}}_{\text{2}}\left(g\right)\to {\text{Li}}_2{\text{CO}}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

b) The amount of carbon dioxide and lithium hydroxide.

The metabolism of glucose, which is also known as combustion of glucose, takes place as shown below:

${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(s\right)+6{\text{ O}}_{\text{2}}\left(g\right)\to 6{\text{ CO}}_2\left(g\right){\text{+6 H}}_{\text{2}}\text{O}\left(l\right)$

From appendix 2,

$\begin{array}{c}\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\right)=-1274.5\text{ kJ}/mol\\ \Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)=-285.8\text{ kJ}/mol\\ \Delta H{°}_{\text{f}}\left({\text{CO}}_{\text{2}}\right)=-393.5\text{ kJ}/mol\end{array}$

The enthalpy of formation of oxygen gas is zero because it is in the most stable form.

Calculate the standard enthalpy of reaction as follows:

$\Delta H{°}_{\text{rxn}}=\left[6\Delta H{°}_{\text{f}}\left({\text{CO}}_2\right)+6\Delta H{°}_{\text{f}}\left({\text{H}}_2\text{O}\right)\right]-\left[\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\right)+6\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)\right]$

Substitute $-1274.5\text{ kJ}/mol$ for $\Delta H{°}_{\text{f}}\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\right)$, $-285.8\text{ kJ}/mol$ for $\Delta H{°}_{\text{f}}\left({\text{H}}_{\text{2}}\text{O}\right)$, $-393.5\text{ kJ}/mol$ for $\Delta H{°}_{\text{f}}\left({\text{CO}}_{\text{2}}\right)$, and 0 for $\Delta H{°}_{\text{f}}\left({\text{O}}_2\right)$ in the above equation as follows:

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[6\left(-393.5\text{ kJ}/mol\right)+6\left(-285.8\text{ kJ}/mol\right)\right]-\left[\left(-1274.5\text{ kJ}/mol\right)+6\left(0\right)\right]\\ =-2801\text{ kJ}/\text{mol}\end{array}$

This enthalpy is required for the formation of $6\text{ mol}$ of carbon dioxide.

The molar mass of carbon dioxide is $44.01\text{ g}/\text{mol}$.

Calculate the mass of carbon dioxideas follows:

$\left(1.2\times {10}^4\cancel{\text{kJ}}\right)\left(\frac{\text{6 }\cancel{{\text{mol CO}}_2}}{2801\cancel{\text{kJ}}}\right)\left(\frac{44.01{\text{ g CO}}_2}{1\cancel{{\text{ mol CO}}_2}}\right)=1131\text{ g}$

In one gram, there are $0.001\text{ kg}$.

Convert gram to kilogram as follows:

$\begin{array}{c}1131\text{ g}=\left(1131\cancel{\text{g}}\right)\left(\frac{0.001\text{ kg}}{1\cancel{\text{ g}}}\right)\\ =1.1{\text{ kg CO}}_2\end{array}$

So, the mass of carbon dioxide is $1.1\text{ kg}$.

The molar mass of $\text{LiOH}$ is $23.9\text{ g/mol}$.

Calculate the mass of lithium hydroxideas follows:

$\left(1131\cancel{{\text{g CO}}_2}\right)\left(\frac{\text{1 }\cancel{{\text{mol CO}}_2}}{44.01\cancel{{\text{g CO}}_2}}\right)\left(\frac{2\cancel{\text{mol LiOH}}}{1\cancel{{\text{mol CO}}_2}}\right)\left(\frac{\text{23}\text{.9 g LiOH}}{1\cancel{\text{ mol LiOH}}}\right)=1228\text{ g}$

Convert gram to kilogram as follows:

$\begin{array}{c}1228\text{ g}=\left(1228\cancel{\text{g}}\right)\left(\frac{0.001\text{ kg}}{1\cancel{\text{ g}}}\right)\\ =1.2\text{ kg LiOH}\end{array}$

Hence, the amount of ${\text{CO}}_{\text{2}}$ produced is $1.1\text{ kg}$ and the amount of $\text{LiOH}$ required to purify the air is $1.2\text{ kg}$.