Chemistry
Chemistry
4th Edition
ISBN: 9780078021527
Author: Julia Burdge
Publisher: McGraw-Hill Education
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Chapter 5, Problem 98AP

At 25 ° C , the standard enthalpy of formation of HF ( a q )  is -320 .1 kJ/mol; of OH - ( a q ) , it is -229 .6 kJ/mol; of F - ( a q ) ,it is -329 .1 kJ/mol; and of H 2 O ( l ) , it is -285 .8 kJ/mol .

(a)

Calculate the standard enthalpy of neutralisation of HF ( a q ) :

HF ( a q ) + OH - ( a q ) F - ( a q ) + H 2 O ( l )

(b)

Using the value of -56.2 kJ as the standard enthalpy change for the reaction

H+ ( a q ) + OH - ( a q ) H 2 O ( l )

calculate the standard enthalpy change for the reaction

HF ( a q ) H + ( a q ) + F - ( a q )

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

The standard enthalpy of neutralization for the given compound and the standard enthalpy change for the given reaction are to be calculated.

Concept introduction:

The standard enthalpy for a reaction is the amount of enthalpy that occurs under standard conditions.

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

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

Here, the stoichiometric coefficients are represented by m for the reactants and n for the products and the enthalpy of formation under standard conditions is represented by ΔHf°.

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

Answer to Problem 98AP

Solution:

(a) 65.2 kJ/mol

(b) 9 kJ/mol

Explanation of Solution

Given information:

ΔH°f[ HF(aq) ]=320.1 kJ/molΔH°f[ OH(aq) ]=229.6 kJ/molΔH°f[ F(aq) ]=329.1 kJ/molΔH°f[ H2O(l) ]=285.8 kJ/mol

a) The standard enthalpy of neutralization of HF(aq)

The given reaction is as follows:

HF(aq)+OH(aq)F(aq)+H2O(l)

Calculate the enthalpy of neutralization for the given reaction as follows:

ΔH°={ ΔH°f[ F(aq) ]+ΔH°f[ H2O(l) ] }{ ΔH°f[ HF(aq) ]+ΔH°f[ OH(aq) ] }

Substitute 320.1 kJ/mol for ΔH°f[ HF(aq) ], 229.6 kJ/mol for ΔH°f[ OH(aq) ], 329.1 kJ/mol for ΔH°f[ F(aq) ], and 285.8 kJ/mol for ΔH°f[ H2O(l) ] in the above equation

ΔH°={ (329.1 kJ/mol)+(285.8 kJ/mol) }[ (320.1 kJ/mol)+(229.6 kJ/mol) ]=(614.9 kJ/mol)(549.7 kJ/mol)=65.2 kJ/mol

Hence, the enthalpy of neutralization for HF is 65.2 kJ/mol.

b) The standard enthalpy change for the reaction

HF(aq)H+(aq)+F(aq)

The given reaction with its enthalpy of reaction is as follows:

H+(aq)+OH(aq)H2O(l)                             ΔH°=56.2 KJ/mol

Calculate the standard enthalpy of reaction as follows:

ΔH°={ ΔH°f[ H2O(l) ] }{ ΔH°f[ H+(aq) ]+ΔH°f[ OH(aq) ] }

Substitute 229.6 kJ/mol for ΔH°f[ OH(aq) ], 56.2 kJ/mol for ΔH°, and 285.8 kJ/mol for ΔH°f[ H2O(l) ] in the above equation

56.2 kJ/mol={ 285.8 kJ/mol }[ ΔH°f[ H+(aq) ]+(229.6 kJ/mol) ]=285.8 kJ/molΔH°f[ H+(aq) ]+229.6 kJ/mol

Rearrange the above equation for ΔH°f[ H+(aq) ]

as follows:

ΔH°f[ H+(aq) ]=285.8 kJ/mol+56.2 kJ/mol+229.6 kJ/mol=285.8 kJ/mol+285.8 kJ/mol=0 kJ/mol

The reaction of hydrogen fluoride is as follows:

HF(aq)H+(aq)+F(aq)

Calculate the standard enthalpy of the reaction as follows:

ΔH°={ ΔH°f[ H+(aq) ]+ΔH°f[ F(aq) ] }{ ΔH°f[ HF(aq) ] }

Substitute 320.1 kJ/mol for ΔH°f[ HF(aq) ], 0 kJ/mol for ΔH°f[ H+(aq) ], and 329.1 kJ/mol for ΔH°f[ F(aq) ] in the above equation

ΔH°=[ (0 kJ/mol)+(329.1 kJ/mol) ]{ 320.1 kJ/mol }=329.1 kJ/mol+320.1 kJ/mol=9 kJ/mol

Hence, the required enthalpy of reaction is 9 kJ/mol.

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Chapter 5 Solutions

Chemistry

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