Enthalpy Changes for Irreversible Processes

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.

273.15

Step 2: Reversible process

Irreversible process Liquid

T/K

Solid

Process 1: Heating of supercooled liquid Process 3: Cooling of solid

258.15

Figure 2.9 Irreversible and Reversible Paths for Example 2.28.

The total enthalpy change is

∆H∆H1+∆H2+∆H3 −10870 J −10.87 kJ Because the process is at constant pressure,q∆H.

P R O B L E M S

Section 2.6: Calculation of Enthalpy Changes of Processes without Chemical Reactions

2.42 a. Calculate∆Hand∆Ufor heating 1.00 mol of argon from 100 K to 300 K at a constant pressure of 1.00 atm.

State any assumptions.

b. Calculate∆Hand∆Ufor heating 1.00 mol of argon from 100 K to 300 K at a constant volume of 30.6 L.

c. Explain the differences between the results of parts a and b.

2.43 a. Findq,w,∆U, and∆H for heating 1.000 mol of neon gas from 273.15 K to 373.15 K at a constant pressure of 1.000 atm. State any approximations and assumptions.

b. Findq,w,∆U, and∆Hfor heating 1.000 mol of neon gas from 273.15 K to 373.15 K at a constant

volume of 22.4 L. State any approximations and assumptions.

2.44 Supercooled steam is condensed irreversibly but at a constant pressure of 1.000 atm and a constant temperature of 96.5^◦C. Find the molar enthalpy change. State any assumptions and approximations.

2.45 The enthalpy change of fusion of mercury is

2331 J mol⁻¹. Find∆H for converting 100.0 g of solid mercury at−75.0^◦C to liquid mercury at 25.0^◦C at a constant pressure of 1.000 atm. Assume that the heat capacities are constant and equal to their values in Table A.6 of the appendix.

2.46 Find∆Hif 100.0 g of supercooled liquid mercury at

−50.0^◦C freezes irreversibly at constant temperature and a constant pressure of 1.000 atm. The enthalpy change of fusion at the normal melting temperature is 2331 J mol⁻¹.

Assume that the heat capacities are constant and equal to their values in Table A.6 of the appendix.

2.47 Find the value ofqand the value of∆Hif 2.000 mol of solid water (ice) at−10.00^◦C is turned into liquid water at

80.00^◦C, with the process at a constant pressure of 1.000 atm. Assume that the heat capacities are constant and equal to their values in Table A.6 of the appendix.

2.7 Calculation of Enthalpy Changes of a Class of Chemical Reactions

A chemical reaction involves the breaking of some chemical bonds and/or the formation of other chemical bonds. The breaking of bonds requires an input of energy and the formation of bonds gives off energy, so that nearly every constant-temperature chemical reaction is accompanied by energy and enthalpy changes. If the system gives off heat when a reaction takes place at constant temperature, the reaction is calledexothermic.

If the system absorbs heat at constant temperature, the reaction is calledendothermic.

In this chapter we will consider only chemical reactions in which every reactant or product is either a gas or a pure liquid or solid, and we will assume that all gases are ideal gases. Three reactions in this class are:

2H2(g)+O2(g)−→2H2O(l) (2.7-1) CaCO3(s)−→CaO(s)+CO2(g) (2.7-2)

N2O4(g)−→2NO2(g) (2.7-3)

The label “s” refers to solid, the label “l” refers to liquid, and the label “g” refers to gas. We will consider solution reactions in a later chapter.

Physical chemists like to write a single equation that can apply to every possible case. In order to do this, we write chemical reaction equations with the symbols for all substances on the right-hand side and replace the−→symbol by an equals sign. The three reactions of Eqs. (2.7-1) through (2.7-3) become

02H2O(l)−2H2(g)−O2(g) (2.7-4) 0CaO(s)+CO2(g)−CaCO3(s) (2.7-5)

02NO2(g)−N2O4(g) (2.7-6)

We denote the stoichiometric coefficient of substance number ibyν_i or by ν(Fi), whereFirepresents the chemical formula of the substance. If substance number i is a product,ν_i is positive, and if substance number j is a reactant, ν_j is negative.

If NO2 is called substance number 1 and N2O4 is called substance number 2, then ν1ν(NO2)2 andν2ν(N2O4) −1 in Eq. (2.7-6). If the number of substances involved in the reaction is represented bys, we number the substances from 1 tos, and represent any chemical reaction by the equation

0 s

i1

ν_iFi (2.7-7)

In a mixture of ideal gases, each substance behaves as though the others were absent. If all substances are either pure liquids or solids or ideal gases the enthalpy of the system is a sum of contributions of the separate substances

H s i1

niHm(i) (2.7-8)

whereHm(i) is the molar enthalpy (enthalpy per mole) of substance numberiandn_i is the amount (in moles) of that substance.

Consider a chemical reaction that begins with reactants in equilibrium or metastable states at some particular temperature and pressure and ends with products in equilibrium states at the same temperature and pressure. The enthalpy change of the reaction is given by

∆HH(final)−H(initial) s i1

∆n_iH_m(i) (2.7-9) where∆n_i is the change in the amount of substance numberi. We say thatone mole of reactionoccurs if

∆n_i ν_i (2.7-10)

for each substance. That is, a number of moles of a product appears that is equal to that product’s stoichiometric coefficient, and a number of moles of a reactant disappear that are equal to the magnitude of its stoichiometric coefficient. For 1 mol of reaction,

∆H s

i1

ν_iHm(i) (2.7-11)

We use dimensionless stoichiometric coefficients. One can think of the units of ν_i as moles of substancei per mole of reaction. The enthalpy change in Eq. (2.7-11) has the units of J mol⁻¹ (meaning joules per mole of the reaction as written). If all stoichiometric coefficients are doubled,∆Hfor the reaction doubles. When we give a value of∆Hfor a reaction, it is always for 1 mol of the reaction as the reaction equation is written.

The molar enthalpy of any substance depends on its state. Thestandard stateof a liquid or solid substance is specified to be the pure substance at a fixed pressure of exactly 1 bar (100,000 Pa), which we denote byP^◦. The standard state for a gas is defined to be the corresponding ideal gas at pressureP^◦. The difference between the molar enthalpy of a real gas at 1 bar pressure and the corresponding ideal gas at 1 bar is numerically very small, but we will discuss this difference in a later chapter.

If substance numberiis in its standard state, its molar enthalpy is denoted byH_m^◦(i).

Astandard-state reactionis one in which all substances are in their standard states before and after the reaction. The enthalpy change for a standard-state reaction is denoted by∆H^◦. The standard-state pressure was at one time defined to equal 1 atm (101,325 Pa). The difference in numerical values is small, and the formulas involving P^◦ are the same with either choice. For highly accurate work, one must determine which standard pressure was used for an older set of data.

Actual values for standard-state molar enthalpies are not available because an arbi- trary constant can be added to each internal energy without any physical effect. We use thestandard-state enthalpy change of formationto calculate∆H^◦for chemical reactions. The standard-state enthalpy change of formation of substanceiis denoted

by∆fH^◦(i) and is defined to be the standard-state enthalpy change of the chemical reaction that forms 1 mol of substanceiin the specified phase from the appropriate elements in the standard state in their most stable forms. For example, the most stable form of carbon at 1 bar is graphite, not diamond, so that standard enthalpy changes of formation are relative to graphite, not diamond.⁵Standard-state enthalpy changes of formation for a number of substances are listed in Table A.8 of Appendix A.

The enthalpy change for 1 mol of any standard-state reaction in our restricted class is given by

∆H^◦ s i1

ν_i∆fH^◦(i) (2.7-12)

We can show this equation to be correct from the fact that the enthalpy is a state function. Let process 1 convert the reactants into elements in their most stable form.

The standard-state enthalpy change for process 1 is

∆H₁^◦Helements−Hreactants s i1

ν_i∆fH^◦(i)

(reactants only)

(2.7-13)

This process is equivalent to the reverse of all of the formation reactions multi- plied by the magnitude of the stoichiometric coefficients. The sign in front of the sum in Eq. (2.7-13) is positive because the stoichiometric coefficients are negative.

The products of the reaction must contain the same elements in the same amounts as in the reactants. Let process 2 be the production of the products of the reaction of interest from the elements produced in process 1. The standard-state enthalpy change of process 2 is

∆H₂^◦ s i1

ν_i∆fH^◦(i)

(products only)

(2.7-14)

We now invokeHess’s law, which states:The enthalpy change of any process that is equivalent to successively carrying out two other processes is equal to the sum of the enthalpy changes of those two processes.This law is a consequence of the fact that enthalpy is a state function, so that its change is path-independent.

Germain Henri Hess, 1802–1850, was a Swiss-Russian chemist whose law first showed that thermodynamics applies to chemistry.

Our reaction is equivalent to the sum of processes 1 and 2. By Hess’s law,

∆H^◦ s i1

ν_i∆fH^◦(i)

(reactants only)

+ s

i1

ν_i∆fH^◦(i)

(products only)

s

i1

ν_i∆fH^◦(i) (2.7-15)

where the final sum includes all substances involved in the reaction. This equation is the same as Eq. (2.7-12). This equation applies only if the final temperature is equal to the initial temperature.

5An exception is made for phosphorus, which has several solid forms (called allotropes). The less reactive red form is specified instead of the more reactive white form, which is more stable in the absence of other substances.

E X A M P L E 2.29

Find the standard-state enthalpy change of the reaction of Eq. (2.7-6) at 298.15 K, using values of enthalpy changes of formation from Table A.8.

Solution

∆H^◦2∆fH^◦(NO₂)+(−1)∆fH^◦(N₂O₄)

2(33.095 kJ mol⁻¹)−(9.079 kJ mol⁻¹)57.11 kJ mol⁻¹

If the formation reaction for a substance cannot actually be carried out, the enthalpy change of formation can be calculated from the enthalpy change of other reactions that can be combined to give the reaction of interest, using Hess’s law.

E X A M P L E 2.30

The standard-state enthalpy change of combustion of methane at 298.15 K equals

−890.36 kJ mol⁻¹, with liquid water as one of the products. Find the enthalpy change of formation of methane at 298.15 K using the enthalpy changes of formation of H₂O and CO₂. Solution

The balanced reaction equation is

0CO₂(g)+2H₂O(l)−CH₄(g)−2O₂(g) so that

−890.36 kJ mol⁻¹∆fH^◦(CO₂)+2∆fH^◦(H₂O) +(−1)∆fH^◦(CH₄)+(−2)∆fH^◦(O₂)

Because gaseous O₂is the most stable form of oxygen at 298.15 K,∆fH^◦(O₂)0. Using values of the enthalpy changes of formation of the other substances,

∆fH^◦(CH₄)890.36 kJ mol⁻¹+(−393.522 kJ mol⁻¹) +2(−285.830 kJ mol⁻¹)+(−2)(0) −74.82 kJ mol⁻¹

This value agrees fairly well with the value in Table A.8,−74.873 kJ mol⁻¹.