ACTIVITY: Answer completely Calculate the standard enthalpies of combustion at 25ᵒC of the following:`(a) C4H8(g) + H2(g) à C4H10(g)ΔHᵒ = -2878.6kJ(b)C4H6(g) + 2H2(g) à C4H10(g)ΔHᵒ = -2543.5 kJThe photo included is for reference HESS LAW (Law of Heat Summation)"The enthalpy change of the overall process is the sum of theenthalpy changes of the individual component steps."- It is the standard enthalpy of Reaction, AH°.- It can be illustrated in the following exampleConsider the following reactions:(a) Fe, + O, à FeO(b) 2 Fe, + 02 , à Fe,OAH° = -272 kJAH° = -82.5 kJ(g)1. Make sure to write the balance equation.2. The final equation requires 2 moles of Fe0 which means we willmultiply equation (a) and its enthalpy change by 2.3. Reverse equation (a), which means, the product goes to the reactantside,(2) (FeO„à Fe, + 0,)AH° = (-272 kJ)(2) =(s)544 kJ2 FeOà 2 Fe, + O)AH° = -825.5 kJ4. Combine the final equation (a) and the initial equation (b). And thencancel the same components with same number of particles. In thiscase, cancel 2 Fe(a) 2 FeOà 2-Fe, + O)(b) 2 Fe + 02 à Fe,O().AH° = 544 kJAH° = -825.5 kJ(g)5. Get the sum of the equation and its enthalpy.(a) 2 FeO„à 2-Fe +Og)(b) 2 Fe + 02 à Fe,OAH° = 544 kJAH° = -825.5 kJ(g)2 FeO, Og) à Fe,O,AH° = 281.5 kJ(s) +

Given: C4H8(g)+ H2(g) →C4H10 (g) (∆H = -2878.6 KJ) C4H6(g) + 2H2(g) →C4H10 (g) (∆H = -2543.5 KJ)

Answered: ACTIVITY: Answer completelyCalculate…

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

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ACTIVITY: Answer completely Calculate the standard enthalpies of combustion at 25ᵒC of the following: `(a) C4H8(g) + H2(g) à C4H10(g)ΔHᵒ = -2878.6kJ (b)C4H6(g) + 2H2(g) à C4H10(g)ΔHᵒ = -2543.5 kJ The photo included is for reference

HESS LAW (Law of Heat Summation)
"The enthalpy change of the overall process is the sum of the
enthalpy changes of the individual component steps."
- It is the standard enthalpy of Reaction, AH°.
- It can be illustrated in the following example
Consider the following reactions:
(a) Fe, + O, à FeO
(b) 2 Fe, + 02 , à Fe,O
AH° = -272 kJ
AH° = -82.5 kJ
(g)
1. Make sure to write the balance equation.
2. The final equation requires 2 moles of Fe0 which means we will
multiply equation (a) and its enthalpy change by 2.
3. Reverse equation (a), which means, the product goes to the reactant
side,
(2) (FeO„à Fe, + 0,)
AH° = (-272 kJ)(2) =
(s)
544 kJ
2 FeOà 2 Fe, + O)
AH° = -825.5 kJ
4. Combine the final equation (a) and the initial equation (b). And then
cancel the same components with same number of particles. In this
case, cancel 2 Fe
(a) 2 FeOà 2-Fe, + O)
(b) 2 Fe + 02 à Fe,O
().
AH° = 544 kJ
AH° = -825.5 kJ
(g)
5. Get the sum of the equation and its enthalpy.
(a) 2 FeO„à 2-Fe +Og)
(b) 2 Fe + 02 à Fe,O
AH° = 544 kJ
AH° = -825.5 kJ
(g)
2 FeO, Og) à Fe,O,
AH° = 281.5 kJ
(s) +

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