Physical Chemistry Third Edition

84 2 Work, Heat, and Energy: The First Law of Thermodynamics

Exercise 2.27

a.Find∆Ufor the process of the previous example.

b.Find∆H,q, andwfor the process in which the system of the previous example is first

heated at constant volume from 298.15 K to 373.15 K and then expanded isothermally from

a volume of 5.000 L to a volume of 10.000 L.

Enthalpy Changes for Irreversible Processes

Consider an irreversible process that begins with an equilibrium or metastable state

and ends with an equilibrium state. To calculate∆Hfor such an irreversible process,

we find a reversible process with the same initial and final states, calculate∆Hfor that

process, and assign that value to the irreversible process, using the fact thatHis a state

function.

EXAMPLE2.28

Find∆Handqif 2.000 mol of supercooled liquid water at− 15. 00 ◦C freezes irreversibly at

a constant pressure of 1.000 atm and a temperature of− 15. 00 ◦C. Assume thatCP, mof liquid

water is constant and equal to 75.48 J K−^1 mol−^1 , and thatCP, mof ice is constant and equal

to 37.15 J K−^1 mol−^1.

Solution

We assume that the supercooled water is metastable and treat it as though it were at equilib-

rium. We calculate∆Halong the reversible path shown in Figure 2.9. Step 1 is the reversible

heating of the supercooled liquid to 0.00◦C, the equilibrium freezing temperature. Step 2 is

the reversible freezing of the system at 0.00◦C, and step 3 is the reversible cooling of the

solid to− 15. 00 ◦C. The enthalpy change for the irreversible process is equal to the enthalpy

change of this reversible process.

∆H 1 

∫ 273 .15 K

258 .15 K

CP(l)dTCP(l)∆T

( 2 .000 mol)

(

75 .48JK−^1 mol−^1

)

( 15 .00 K)2264 J

The enthalpy change of step 2 was calculated in an earlier example:

∆H 2 −12020 J

The enthalpy change of step 3 is similar to that of step 1.

∆H 3 

∫ 258 .15 K

273 .15 K

CP(s)dTCP(s)∆T

( 2 .000 mol)

(

37 .15JK−^1 mol−^1

)

( 15 .00 K)1114 J