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At 25°C, the enthalpy of formation of gaseous nitric acid is -134 KiloJoules per mole. Calculate the change in internal energy (in KiloJoules per mole) for the reaction given below: (1/2)H2g)+ (1/2)N2g) + (3/2)029) -> HNO3(g)

At 25°C, the enthalpy of formation of gaseous nitric acid is -134 KiloJoules per mole. Calculate the change in internal energy (in KiloJoules per mole) for the reaction given below: (1/2)H2g)+ (1/2)N2g) + (3/2)029) -> HNO3(g)

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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At 25°C, the enthalpy of formation of gaseous nitric acid is-134 KiloJoules per mole. Calculate the change in internal energy (in KiloJoules per mole) for the reaction given below: --> HNO3(g) (1/2)H29) + (1/2)N2(9) + (3/2)0219) NOTE: Express answer in the NEAREST WHOLE NUMBER. 12 DOINTS
Transcribed Image Text:At 25°C, the enthalpy of formation of gaseous nitric acid is-134 KiloJoules per mole. Calculate the change in internal energy (in KiloJoules per mole) for the reaction given below: --> HNO3(g) (1/2)H29) + (1/2)N2(9) + (3/2)0219) NOTE: Express answer in the NEAREST WHOLE NUMBER. 12 DOINTS
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