Chapter 10, Problem 10.117QP

(a)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction of aqueous Lithium hydroxide and Carbon dioxide and the necessary amount of Carbon dioxide and Lithium hydroxide for the purification of air has to be calculated.

Concept Introduction:

The change in enthalpy that is associated with the formation of one mole of a substance from its related elements being in standard state is called standard enthalpy of formation ( ${\text{ΔH}}_{\text{f}}\text{°}$).  The standard enthalpy of formation is used to determine the standard enthalpies of compound and element.

The standard enthalpy of reaction is the enthalpy of reaction that takes place under standard conditions.

 The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

Answer to Problem 10.117QP

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

Explanation of Solution

Lithium hydroxide reacts with Carbon dioxide to give Lithium carbonate (soluble product) and Water. The equation can be given as,

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

Since, the equation is unbalanced. 2moles of Lithium are multiplied by 2 on the reactant side in order to given Lithium carbonate. The balanced equation can be given as,

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

(b)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction of aqueous Lithium hydroxide and Carbon dioxide and the necessary amount of Carbon dioxide and Lithium hydroxide for the purification of air has to be calculated.

Concept Introduction:

The change in enthalpy that is associated with the formation of one mole of a substance from its related elements being in standard state is called standard enthalpy of formation ( ${\text{ΔH}}_{\text{f}}\text{°}$).  The standard enthalpy of formation is used to determine the standard enthalpies of compound and element.

The standard enthalpy of reaction is the enthalpy of reaction that takes place under standard conditions.

 The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

Answer to Problem 10.117QP

The amount of Carbon dioxide required to purify air is $\text{1}\text{.1}\text{kg}$

Explanation of Solution

The reaction can be given as,

${\text{C}}_{\text{6}}{\text{H}}_{\text{1}}{}_{\text{2}}{\text{O}}_{\text{6(aq)}}{\text{+6O}}_{\text{2(g)}}\to {\text{6CO}}_{\text{2(g)}}{\text{+6H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Standard enthalpy of formation of ${\text{CO}}_{\text{2}}$= $\text{-393}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of Water = $\text{-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of Oxygen = $\text{0}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of Glucose = $\text{-1274}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}$

$\begin{array}{l}{\text{ΔH°}}_{\text{reaction}}\text{=}\left[{\text{6ΔH°}}_{\text{f}}{\text{(CO}}_{\text{2}}{\text{)+6ΔH°}}_{\text{f}}{\text{(H}}_{\text{2}}\text{O)}\right]{\text{-ΔH°}}_{\text{f}}{\text{(C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{)}\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=}\left[\text{6(-393}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}\text{)+6(-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\right]\text{-(-1274}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=-2801}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

To calculate the amount of Carbon dioxide

Amount of carbon dioxide = $\text{1}{\text{.2×10}}^{\text{4}}\text{kJ×}\frac{\text{6}\text{mol}{\text{CO}}_{\text{2}}}{\text{2801}\text{kJ}}\text{×}\frac{\text{44}\text{.01}\text{g}{\text{CO}}_{\text{2}}}{\text{1}\text{mol}{\text{CO}}_{\text{2}}}$

= $\text{1131}\text{g=1}\text{.1}\text{kg}$

(c)

Interpretation Introduction

Interpretation:

The balanced equation for the reaction of aqueous Lithium hydroxide and Carbon dioxide and the necessary amount of Carbon dioxide and Lithium hydroxide for the purification of air has to be calculated.

Concept Introduction:

The change in enthalpy that is associated with the formation of one mole of a substance from its related elements being in standard state is called standard enthalpy of formation ( ${\text{ΔH}}_{\text{f}}\text{°}$).  The standard enthalpy of formation is used to determine the standard enthalpies of compound and element.

The standard enthalpy of reaction is the enthalpy of reaction that takes place under standard conditions.

 The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

Answer to Problem 10.117QP

The amount of Lithium hydroxide required to purify air is $\text{1}\text{.2}\text{kg}$

Explanation of Solution

Amount of Lithium hydroxide = $\text{1131}\text{g}{\text{CO}}_{\text{2}}\text{×}\frac{\text{1 mol}{\text{CO}}_{\text{2}}}{\text{44}\text{.01}\text{g}\text{}{\text{CO}}_{\text{2}}}\text{×}\frac{\text{2}\text{mol}\text{LiOH}}{\text{1}\text{mol}{\text{CO}}_{\text{2}}}\text{×}\frac{\text{23}\text{.9}\text{g}\text{LiOH}}{\text{1}\text{mol}\text{LiOH}}$

= $\text{1229}\text{g=1}\text{.2}\text{kg}$

The amount of Lithium hydroxide is calculated by plugging in the values of molar masses of Carbon dioxide and Lithium Hydroxide and the amount of Carbon dioxide.  The amount of Lithium hydroxide is found to be $\text{1229}\text{g=1}\text{.2}\text{kg}$.