Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 10, Problem 141CP

When I mole of benzene is vaporized at a constant pressure of 1.00 atm and at its boiling point of 353.0 K, 30.79 kJ of energy (heat) is absorbed and the volume change is +28.90 L. What are ∆E and ∆H for this process?

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

          Change in enthalpy ΔH and change in internal energy ΔE for the given process has to be determined.

Concept introduction:

Internal energy of a system is total energy present in the system. In simple words, it is the sum of kinetic and potential energy of the particles in the system. According to First law of Thermodynamics, Energy of a system is conserved. It is only transferred from one state to another that is from system to surroundings and vice versa. So ΔE can be represented as,

                                            ΔEuniverse=ΔEsys+ΔEsurroundings

Further, ΔE is also equivalent to sum of either heat gained or lost and either work done on the system or by the system.

                                            ΔE=q+w

                                  whereΔE=changeininternalenergyq=quantityofheatgainedorheatlostw=workdone

Enthalpy is heat content of the system. The value of enthalpy does not depend on the path of a reaction but depend on state of the system. It has a unique value for each state of the system. Thus, enthalpy is a state function. Enthalpy is represented as,

                                             H=E+PVwhereH=EnthalpyE=InternalenergyP=PressureV=Volume

Enthalpy change, denoted by ΔH , refers to heat evolved or absorbed during a reaction. If heat is evolved in the reaction that is exothermic reaction ΔH has negative value. For an endothermic reaction, ΔH has positive value. ΔH can be represented as,

                                       ΔH=ΔE+PΔV

                             where,ΔH=ChangeinenthalpyΔE=ChangeinInternalenergyΔV=ChangeinvolumeP=Pressure

Further, ΔH also refers to heat absorbed or lost in the reaction at constant pressure. Thus ΔH can have the following expression also –

                                           ΔH=qPwhereqPisheatgainedorlostatconstantpressure.

Answer to Problem 141CP

Answer

Change in enthalpy ΔH is 30.79 kJ.

Change in internal energy ΔE is 27.86 kJ.

Explanation of Solution

Explanation

Determine ΔH of the reaction.

givendata:heatabsorbed,q=30.79kJ

ΔH of the reaction corresponds to heat absorbed in the reaction. Therefore,

                                ΔH=qp=30.79kJ

Enthalpy of a system corresponds to amount of heat present in the system. Change in enthalpy is change in the heat content of the system during the course of reaction. That is, heat content of the system either decreases or increases after the completion of the reaction. Endothermic reaction proceeds by absorption of heat that at the end of the reaction, heat content of the system is increased. So the amount of heat absorbed or evolved in the reaction corresponds to the enthalpy change of a reaction.

Calculate the work done, ‘w’.

givendata:Pressure,P=1atmvolumechange,ΔV=28.90L

           Weknow,ΔH=ΔE+PΔV......(1)ΔH=q......(2)ΔE=q+w......(3)Subtracting(2)from(3),ΔEΔH=q+w-qFrom (1),ΔEΔH=PΔVTherefore,w=PΔV                          w=1.00atm×29.80L=29.80Latm1Latm=101.33JTherefore,w=29.80×101.33J=2928.4J

Using the three equations, ΔH=ΔE+PΔV , ΔH=q and ΔE=q+w work done is calculated by substituting the relevant values.

Calculate ΔE of the reaction.

           knowndata:q=30.79kJw=2928.4J=2.93kJΔE=q+wTherefore,ΔE=30.29kJ2.93kJ=27.86kJ

‘q’and work done ‘w’ values are calculated in the previous steps. By substituting these values in the equation, ΔE=q+w , change in internal energy ΔE is obtained.

Conclusion

Conclusion

Change in enthalpy ΔH and Change in internal energy ΔE are calculated for the given reaction.

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