Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 5, Problem 133AP

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 1 .2  ×  10 4 kJ of energy every day and assuming that this energy is totally supplied by the metabolism of glucose ( C 6 H 12 O 6 ) , calculate the amounts of CO 2 produced and LiOH required to purify the air.

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

The balanced equation for the given process is to be written and the amounts of CO2 produced and 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:

ΔH°rxn=nΔHf°(products)mΔHf°(reactants)

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 ΔHf°.

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

Answer to Problem 133AP

Solution:

(a) 2 LiOH(aq)+CO2(g)Li2CO3(aq)+H2O(l)

(b) 1.1 kg CO21.2 kg LiOH

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:

LiOH(aq)+CO2(g)Li2CO3(aq)+H2O(l)

Here, the salt formed is lithium carbon trioxide.

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

2 LiOH(aq)+CO2(g)Li2CO3(aq)+H2O(l)

Hence, the balanced chemical equation is; 2 LiOH(aq)+CO2(g)Li2CO3(aq)+H2O(l).

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:

C6H12O6(s)+6 O2(g)6 CO2(g)+6 H2O(l)

From appendix 2,

ΔH°f(C6H12O6)=1274.5 kJ/molΔH°f(H2O)=285.8 kJ/molΔH°f(CO2)=393.5 kJ/mol

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:

ΔH°rxn=[6 ΔH°f(CO2)+6 ΔH°f(H2O)][ΔH°f(C6H12O6)+6 ΔH°f(O2)]

Substitute 1274.5 kJ/mol for ΔH°f(C6H12O6), 285.8 kJ/mol for ΔH°f(H2O), 393.5 kJ/mol for ΔH°f(CO2), and 0 for  ΔH°f(O2) in the above equation as follows:

ΔH°rxn=[6(393.5 kJ/mol)+6(285.8 kJ/mol)][(1274.5 kJ/mol)+6(0)]=2801 kJ/mol

This enthalpy is required for the formation of 6 mol of carbon dioxide.

The molar mass of carbon dioxide is 44.01 g/mol.

Calculate the mass of carbon dioxideas follows:

(1.2×104kJ)(mol CO22801 kJ)(44.01 g CO21 mol CO2)=1131 g

In one gram, there are 0.001 kg.

Convert gram to kilogram as follows:

1131 g=(1131 g)(0.001 kg1 g)=1.1 kg CO2

So, the mass of carbon dioxide is 1.1 kg.

The molar mass of LiOH is 23.9 g/mol.

Calculate the mass of lithium hydroxideas follows:

(1131 g CO2)(mol CO244.01 g CO2)(2 mol LiOH1 mol CO2)(23.9 g LiOH1 mol LiOH)=1228 g

Convert gram to kilogram as follows:

1228 g=(1228 g)(0.001 kg1 g)=1.2 kg LiOH

Hence, the amount of CO2 produced is 1.1 kg and the amount of LiOH required to purify the air is 1.2 kg.

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Chapter 5 Solutions

Chemistry

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