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Determine, at 298 K, Δr1H0, the standard enthalpy of the reaction (1) between hydroquinone with hydrogen peroxide to produce quinone.   C6H4(OH)2(aq) + H2O2(aq) → C6H4O2(aq) + 2 H2O (ℓ) (1)   Knowing the standard reaction enthalpies, at 298 K, of the following reactions:     (2)       C6H4(OH)2 (aq) → C6H4O2 (aq) + H2 (g)        ;           Δr2H0 = + 177.4 kJ mol−1 (3)        H2 (g) + O2 (g) → H2O2 (aq)                           ;           Δr3H0 = −191.2 kJ mol−1 (4)        H2 (g) + ½O2 (g) → H2O (g)                           ;           Δr4H0 = −241.8 kJ mol−1 (5)        H2O (g) → H2O (ℓ)                                         ;           Δr5H0 = −43·8 kJ mol−1

Determine, at 298 K, Δr1H0, the standard enthalpy of the reaction (1) between hydroquinone with hydrogen peroxide to produce quinone.   C6H4(OH)2(aq) + H2O2(aq) → C6H4O2(aq) + 2 H2O (ℓ) (1)   Knowing the standard reaction enthalpies, at 298 K, of the following reactions:     (2)       C6H4(OH)2 (aq) → C6H4O2 (aq) + H2 (g)        ;           Δr2H0 = + 177.4 kJ mol−1 (3)        H2 (g) + O2 (g) → H2O2 (aq)                           ;           Δr3H0 = −191.2 kJ mol−1 (4)        H2 (g) + ½O2 (g) → H2O (g)                           ;           Δr4H0 = −241.8 kJ mol−1 (5)        H2O (g) → H2O (ℓ)                                         ;           Δr5H0 = −43·8 kJ mol−1

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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  • Determine, at 298 K, Δr1H0, the standard enthalpy of the reaction (1) between hydroquinone with hydrogen peroxide to produce quinone.

 

C6H4(OH)2(aq) + H2O2(aq) → C6H4O2(aq) + 2 H2O (ℓ) (1)

 

Knowing the standard reaction enthalpies, at 298 K, of the following reactions:

 

 

(2)       C6H4(OH)2 (aq) → C6H4O2 (aq) + H2 (g)        ;           Δr2H0 = + 177.4 kJ mol−1

(3)        H2 (g) + O2 (g) → H2O2 (aq)                           ;           Δr3H0 = −191.2 kJ mol−1

(4)        H2 (g) + ½O2 (g) → H2O (g)                           ;           Δr4H0 = −241.8 kJ mol−1

(5)        H2O (g) → H2O (ℓ)                                         ;           Δr5H0 = −43·8 kJ mol−1

 

 

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