Chemistry
Chemistry
13th Edition
ISBN: 9781259911156
Author: Raymond Chang Dr., Jason Overby Professor
Publisher: McGraw-Hill Education
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Chapter 24, Problem 24.44QP

Given these data

C 2 H 4 ( g ) + 3 O 2 ( g ) 2 CO 2 ( g ) + 2 H 2 O( l ) Δ H ° = 1411 kJ/mol 2C 2 H 2 ( g ) + 5 O 2 ( g ) 4 CO 2 ( g ) + 2 H 2 O( l ) Δ H ° = 2599 kJ/mol H 2 ( g ) + 1 2 O 2 ( g ) H 2 O( l ) Δ H ° = 285.8 kJ/mol

calculate the heat of hydrogenation for acetylene:

C 2 H 2 ( g ) + H 2 ( g ) C 2 H 4 ( g )

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

The standard enthalpy change in kilojoules per mole for the given reaction has to be calculated.

Concept introduction:

Enthalpy change of reaction: The enthalpy change of a reaction is roughly equivalent to the amount of energy lost or gained during the reaction.

ΔHro=ΣnΔHo(products)-ΣnΔHo(reactants)where,nisStoichiometriccoefficient

Or

The enthalpy change of a overall reaction is equal to the sum of multiple steps.

Explanation of Solution

Net overall reaction is as follows,

In order to obtain ethene as product reversing first equation changes the sign but not the magnitude of enthalpy value,

2CO2(g)+2H2O(l)C2H4(g)+3O2(g)ΔHo=+1411kJ/mol

1 mol of acetylene as reactant undergoes the reaction, hence dividing the second given equation, enthalpy value becomes half of the given value as,

C2H2(g)+52O2(g)2CO2(g)+H2O(l)ΔHo=1299.5kJ/mol

Equating the all the given equations as,

2CO2(g)+2H2O(l)C2H4(g)+3O2(g)ΔHo=+1411kJ/molC2H2(g)+52O2(g)2CO2(g)+H2O(l)ΔHo=1299.5kJ/molH2(g)+12O2(g)H2O(l)ΔHo=285.8kJ/mol_overall:C2H2(g)+H2(g)C2H4(g)ΔHo=-174.3kJ/mol_

The standard enthalpy change in kilojoules per mole for the given reaction is -174.3kJ/mol

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Chapter 24 Solutions

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