EBK CHEMICAL PRINCIPLES
EBK CHEMICAL PRINCIPLES
8th Edition
ISBN: 8220101425812
Author: DECOSTE
Publisher: Cengage Learning US
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Chapter 9, Problem 69E
Interpretation Introduction

Interpretation: The ΔH value for the following reaction needs to be determined:

  C6H4(OH)2(aq)+H2O2(aq)C6H4O2(aq)+2H2O(l)

Concept Introduction: For the overall processes, the enthalpy change can be determined by adding the enthalpy change of all the steps involved in the process is known as Hess’s Law. The equation to show Hess’s law is:

  ΔHtotal= ΔH1+ΔH2+ΔH3+...+ΔHn

Expert Solution & Answer
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Answer to Problem 69E

  ΔH = -202.6 kJ

Explanation of Solution

Given:

  C6H4(OH)2(aq)C6H4O2(aq)+H2(g)             ΔH=177.4 kJ                 -(1)    H2(g)+O2(g)H2O2(aq)                            ΔH=-191.2 kJ                -(2) H2(g)+12O2(g)H2O(g)                               ΔH=241.8 kJ              -(3)              H2O(g)H2O(l)                                ΔH=43.8 kJ                -(4)    

Rules related to enthalpies of the reaction are:

  • When a reaction is inverted, then the sign of enthalpy is also inverted.
  • When a reaction is multiplied by ‘n’ coefficient, then the value of enthalpy is multiplied by ‘n’ value.

The required equation is:

  C6H4(OH)2(aq)+H2O2(aq)C6H4O2(aq)+2H2O(l)

In order to obtain the required equation,reaction (2) is reversedand added to the reaction (1):

                             H2O2(aq)H2(g)+O2(g)                                          ΔH=-(-191.2 kJ)                                C6H4 (OH)2(aq)C6H4O2 (aq)+H2(g)                                 ΔH=177.4 kJ                 _C6H4(OH)2(aq)+H2O2(aq)C6H4O2(aq)+2H2(g)+O2(g)                   ΔH=368.6 kJ                 -(5)

Double the reaction (3) and now adding reaction (5):

                    2 H2(g)+O2(g)2H2O(g)                                                ΔH=2(-241.8 kJ)                C6H4 (OH)2 (aq)+H2O2(aq)C6H4O2 (aq)+2H2 (g)+O2(g)                 ΔH=368.6 kJ                 _C6H4(OH)2(aq)+H2O2(aq)C6H4O2(aq)+2H2O(g)                          ΔH=-115 kJ                 -(6)

Double the reaction (4) and add to the reaction (6):

                               2H2O(g)2H2O(l)                                                 ΔH=2(43.8 kJ)                C6H4 (OH)2 (aq)+H2O2(aq)C6H4O2(aq)+2H2O(g)                          ΔH=-115 kJ                 _C6H4(OH)2(aq)+H2O2(aq)C6H4O2(aq)+2H2O(l)                          ΔH=202.6 kJ  

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Chapter 9 Solutions

EBK CHEMICAL PRINCIPLES

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