can only calculate the value of W,
I1U
andI1H per mole
of the gas.7 . 1 8 A P P LI CATI O N O F T H E F I R ST lAW O F T H E R M O DY N A M I C S TO C H E M I CA L R EACTI O N S . T H E H EAT OF R EACTI O N
If a chemical reaction takes place in a system, the temperature of the system immediately after the reaction generally is different from the temperature immediately before the reaction. To restore the system to its initial temperature, heat must flow either to or from the surroundings. If the system is hotter after the reaction than before, heat must flow
to
the surroundings to restore the system to the initial temperature. In this event the reaction isexothermic ;
by the convention for heat flow, the heat of the reaction is negative. If the system is colder after the reaction than before, heat must flowfrom
the surroundings to1 30 Energy and the F i rst law of Thermodynamics
restore the system to the initial temperature. In this event the reaction is
endothermic,
and the heat of the reaction is positive. Theheat of a reaction
is the heat withdrawn from the surroundings in the transformation of reactants atT
andp
to products at thesame
T
and p.In the laboratory the majority of chemical reactions are performed under a constant pressure ; therefore the heat withdrawn from the surroundings is equal to the change in enthalpy of the system. To avoid mixing the enthalpy change associated with the chemical reaction and that associated with a temperature or pressure change in the the initial and final states of the system must have the same temperature and pressure.
For example, in the reaction
Fe203(s) + 3 HzCg) ---+ 2 Fe(s) + 3 H20(l),
the initial and final states are : Initial state
T, p 1 mole solid Fez03
3 moles gaseous Hz
Final state
T, p 2 moles solid Fe
3 moles liquid HzO
Since the state of aggregation of each substance must be specified, the letters s, 1, and g appear in parentheses after the formulas of the substances. Suppose that we think of the change in state as occurring in two distinct steps. In the first step, reactants at
T
andp
are transformed adiabatically to products atT'
andp.
Step 1 .
Fe203(s) + 3 HzCg) ---+ 2 Fe(s) + 3 H20(l).
T, p
T', p
At constant pressure, I1H = Qp ; but, since this first step is adiabatic, (Qp)l = 0 and I1H 1 =
O.
In the second step, the system is placed in a heat reservoir at the initial temperature T. Heat flows into or out of the reservoir as the products of the reaction come to the initial temperature.
Step 2.
T', p
T, p
for which I1H 2 = Qp . The sum of the two steps is the overall change in state Fe203(S) + 3 HzCg) ---+ 2 Fe(s) + 3 H20(l)
and the I1H for the overall reaction is the sum of the enthalpy changes in the two steps : I1H = I1Hl + I1H2 = 0 + Qp ,
(7.60) where Qp is the heat of the reaction, the increase in enthalpy of the system resulting from the chemical reaction.
The Format i o n React i o n 1 31
The increase in enthalpy in a chemical reaction can be viewed in a different way. At a specified temperature and pressure, the molar enthalpy H of each substance has a definite value. For any reaction, we can write
I1H = Hfina1 - Hinitial ' (7.61)
But the enthalpy of the initial or the final state is the sum of the enthalpies of the sub stances present initially or finally. Therefore, for the example,
Hfina1
=
2H(Fe, s) + 3H(H20, 1), Hinitial = H(Fe203 ' s) + 3H(H2 ' g),and Eq. (7.61) becomes
M = [2.H(Fe, s) + 3.H(H20, 1)] - [.H(Fe203 , s) + 3.H(H2 ' g)]. (7.62) It seems reasonable that measuring M could lead ultimately to the evaluation of the four molar enthalpies in Eq. (7.62). However, there are four " unknowns " and only one equa tion. We could measure the heats of s�veral different reactions, but this would introduce more " unknowns." We deal with this difficulty in the next two sections.
7 . 1 9 T H E F O R M ATI O N R EACTI O N
We can simplify the result in Eq. (7.62) by considering the formation reaction of a com pound. The formation reaction of a compound has
one mole
of the compound andnothing
else
on the product side ; only elements in their stable states of aggregation appear on the reactant side. The increase in enthalpy in such a reaction is theheat of formation,
orenthalpy of formation�
of the compound, I1H f ' The following reactions are examples of formation reactions.H2(g) + t02(g) --* H2O(l)
2 Fe(s) + !02(g) --* Fe203(S)
tH2(g) + tBril)
--*
HBr(g) tN2(g) + 2 Hig) + tClig)--*
NH4CI(s)If the I1H for these reactions is written in terms of the molar enthalpies of the sub stances, we obtain, using the first two as examples,
I1Hf(H20, I) = .H(H20, 1) - .H(H2 ' g) - t.H(02 , g) I1H(Fe203 , s) = .H(Fe203 , s) - 2.H(Fe, s) - !.H(02 , g) Solving for the molar enthalpy of the compound in each example, we have
H(H20, I) = H(H2 , g) + tH(02 ' g) + I1HtCH20, 1)
- - 3 - .
H(Fe203 , s) = 2H(Fe, s) + 2H(02 , g) + MtCFe203 , s) (7.63) These equations show that the molar enthalpy of a compound is equal to the total enthalpy of the elements that compose the compound plus the enthalpy of formation of the com pound. Thus we can write for any compound,
1 32 E nergy a n d the F i rst law of Thermodyn a m i cs
in which LH(elements) is the total enthalpy of the elements (in their stable states of aggregation) in the compound.
Next we insert the values of H(H20, 1) and H(Fe203 , s) given by Eq. (7.63) into Eq. (7.62) ; this yields
i1.H = 2H(Fe, s) + 3[H(H2 , g) + tH(02 , g) + i1.HiH20, I)J - [2B(Fe, s) + tH(02 , g) + LlHiFe203 , s)J - 3H(H2 , g) Collecting like terms, this becomes
(7.65) Equation (7.65) states that the change in enthalpy of the reaction depends only on the heats offormation of the
compounds
in the reaction. The change in enthalpy is independent of the enthalpies of the elements in their stable states of aggregation.A moment's reflection on Eq. (7.64) tells us that this independence of the values of the enthalpies of the elements must be correct for all chemical reactions. If, in the expression for the i1.H of a reaction, we replace the molar enthalpy of every compound by the expres sion in Eq. (7.64), then it is clear that the sum of the enthalpies of the elements composing the reactants must be equal to the sum of the enthalpies of the elements composing the products. The balanced chemical equation requires this. Therefore the enthalpies of the elements must drop out of the expression. We are left only with the proper combination of the enthalpies of formation of the compounds. This conclusion is correct at every temperature and pressure.
The enthalpy of formation of a compound at 1 atm pressure is the
standard
enthalpy of formation, LlHi . Values of LlHi at 25 °C are tabulated in Appendix V, Table A-V, for a number of compounds.III EXAMPLE 7.4 Using the values of LlHi given in Table A-V, calculate the heat of the
reaction
Fe203(s) + 3 H2(g) ---+ 2 Fe(s) + 3 H20(l),
From Table A-V we find
LlHi(H20, 1) = - 285.830 kJ/mol ; LlHi(Fe203 , s) = - 824.2 kJ/mol. Then
i1.H = 3( - 285.830 kJ/mol) - 1( - 824.2 kJ/mol) = ( - 857.5 + 824.2) kJ/mol
= - 33.3 kllmol.
The negative sign indicates that the reaction is exothermic. Note that the stoichiometric coefficients in these expressions are
pure numbers.
The unit for i1.H is kJ/mol. This means permole of reaction.
Once we balance the chemical equation in a particular way, as above, this defines the mole of reaction. If we had balanced the equation differently, as, for example,tFe203(S) + tHz(g) ---+ Fe(s) + tH20(l)
then this amount of reaction would be one mole of reaction and LlH would be
Conventional Va l u es of M o l a r Enthalpies 1 33
7 . 20 C O N VE N T I O N A L VA L U E S O F M O LA R E NT H A l P I E S
The molar enthalpy II of any substance is a function of T and
p;
II = H(T,p).
Choosingp
= 1 atm as thestandard
pressure, we define thestandard
molar enthalpy Jr of a sub stance byHO = H(T, 1 atm). (7.66)
From this it is clear that HO is a function only of temperature. The degree superscript on any thermodynamic quantity indicates the value of that quantity at the standard pressure. (Because the dependence of enthalpy on pressure is very slight-compare with Section 7. 15-we will often use standard enthalpies at pressures other than one atm ; the error will not be serious unless the pressure is very large, for example, 1000 atm.)
As we showed in Section 7.19, the enthalpy change in any chemical reaction does not depend on the numerical values of the enthalpies of the elements that compose the com pound. Because this is so we may assign any arbitrary, convenient values to the molar enthalpies of the elements in their stable states of aggregation at a selected temperature and pressure. Clearly, if we chose the required values randomly from the numbers in a telephone directory this could introduce a good deal of unnecessary numerical clutter into our work. Since the numbers do not matter, they can all be the same ; if they can all be the same, they all might as well be zero and eliminate the clutter entirely.
The enthalpy of every
element
in its stable state of aggregation at 1 atm pressure and at 298.1 5 K is assigned the value zero. For example, at 1 atm and 298.1 5 K the stable state of aggregation of bromine is the liquid state. Hence, liquid bromine, gaseous hydrogen, solid zinc, solid (rhombic) sulfur, and solid (graphite) carbon all have H�98. 1 5 =o.
(We will write H 298 as an abbreviation for H298.1 5 ')
For elementary solids that exist in more than one crystalline form, the modification that is stable at 25 °C and 1 atm is assigned HO = 0 ; for example, the zero assignment goes to rhombic sulfur rather than to monoclinic sulfur, and to graphite rather than to diamond. In cases in which more than one molecular species exists (for example, oxygen atoms, 0 ; diatomic oxygen, O2 ; and ozone, 03) the zero enthalpy value is assigned to the most stable form at 25 °C and 1 atm pressure ; for oxygen, H�98(02 ' g) =
O.
Once the value of the standard enthalpy of the elements at 298.1 5 K has been assigned, the value at any other temperature can be calculated. Since at constant pressure,d
Ho
=C� d
T, thenIT
298d
HO =IT
298c�
dT, H�-
H�98 =IT
298c� dT,
H� = H�98 +
IT c� d
T, (7.67)298
which is correct for both elements and compounds ; for elements, the first term on the right-hand side is zero.
Given the definition ofthe formation reaction, if we introduce the conventional assign ment, HO(elements) = 0, into the expression for the heat of formation, Eq. (7.63) or Eq. (7.64), we find that for any compound
HO = I1H'} . (7.68)
The standard heat of formation I1H'} is the conventional molar enthalpy of the compound relative to the elements that compose it. Accordingly, if the heats of formation 11H'} of all
1 34 Energy and the F i rst law of Thermodynamics
the compounds in a chemical reaction are known, the heat of the reaction can be calcula ted from equations formulated in the manner of Eq. (7.62).
7 . 21 T H E D ET E R M I N AT I O N O F H EATS O F F O R M AT I O N
In some cases it is possible to determine the heat of formation of a compound directly by carrying out the formation reaction in a calorimeter and measuring the heat effect produced. Two important examples are
C(graphite)
+
Gig) ---+ COz(g),HzCg)
+
!oig) ---+ Hz0(l),/';;.H'} = - 393.51 kJ/mol /';;.H'} = - 285.830 kJ/mol.
These reactions can be conducted easily in a calorimeter ; the reactions go to completion, and conditions can easily be arranged so that only one product is formed. Because of the importance of these two reactions, the values have been determined quite accurately.
The majority of formation reactions are unsuitable for calorimetric measurement ; these heats offormation must be determined by indirect methods. For example,
C(graphite)
+
2 Hz(g) ---+ CH4(g)·This reaction has three strikes against it as far as its use in calorimetry is concerned. The combination of graphite with hydrogen does not occur readily ; if we did manage to get these materials to react in a calorimeter, the product would not be pure methane, but an exceedingly complex mixture of hydrocarbons. Even if we succeeded in analyzing the product mixture, the result of such an experiment would be impossible to interpret.
There is one method that is generally applicable if the compound burns easily to form definite products. The heat of formation of a compound can be calculated from the measured value of the heat of combustion of the compound. The combustion reaction has one mole of the substance to be burned on the reactant side, with as much oxygen as is necessary to burn the substance completely ; organic compounds containing only carbon, hydrogen, and oxygen are burned to gaseous carbon dioxide and liquid water.
For example, the combustion reaction for methane is
CH4(g)
+
2 0ig) ---+ COz(g)+
2 HzO(l).The measured heat of combustion is /';;.H�omb = - 890.36 kJ Imol. In terms of the enthalpies of the individual substances,
/';;.H�omb = HO(COz , g)
+
2HO(HzO, 1) - HO(CH4 ' g). Solving this equation for HO(CH4 ' g),HO(CH4 ' g) = HO(COz ' g)
+
2HO(HzO, 1) - /';;.H�omb' (7.69) The molar enthalpies of COz and HzO are known to a high accuracy ; from this knowledge and the measured value of the heat of combustion, the molar enthalpy of methane (the heat of formation) can be calculated by using Eq. (7.69).HO(CH4 ' g) = - 393.51
+
2( - 285.83) - ( - 890.36)= - 965. 17
+
890.36 = - 74.8 1 kllmol.The measurement of the heat of combustion is used to determine the heats of forma tion of all organic compounds that contain only carbon, hydrogen, and oxygen. These
Seq uences of R eactions 1 35
compounds burn completely to carbon dioxide and water in the calorimeter. The combus tion method is used also for organic compounds containing sulfur and nitrogen ; however, in these cases the reaction products are not so definite. The sulfur may end up as sulfurous acid or sulfuric acid ; the nitrogen may end up in the elementary form or as a mixture of oxy-acids. In these cases considerable skill and ingenuity are required in the determination ofthe conditions for the reaction and in the analysis ofthe reaction products. The accuracy of the values obtained for this latter class of compounds is very much less than that ob tained for the compounds containing only carbon, hydrogen, and oxygen.
The problem of determining the heat offormation of any compound resolves into that of finding some chemical reaction involving the compound which is suitable for calori metric measurement, then measuring the heat of this reaction. If the heats of formation of all of the other substances involved in this reaction are known, then the problem is solved. If the heat offormation of one of the other substances in the reaction is not known, then we must find a calorimetric reaction for that substance, and so on.
Devising a series of reactions from which an accurate value of the heat of formation of a particular compound can be obtained can be a challenging problem. A calorimetric reaction must take place quickly (that is, be completed within a few minutes at most), with as few side reactions as possible and preferably none at all. Very few chemical reactions take place without concomitant side reactions, but their effects can be minimized by controlling the reaction conditions so as to favor the main reaction as much as possible. The final product mixture must be carefully analyzed, and the thermal effect of the side - reactions must be subtracted from the measured value. Precision calorimetry is demanding work.
7 . 22 S EQ U E N C ES O F R EACTI O N S ; H ES S ' S LAW
The change in state of a system produced by a specified chemical reaction is definite. The corresponding enthalpy change is definite, since the enthalpy is a function ofthe state. Thus, if we transform a specified set of reactants to a specified set of products by more than one sequence of reactions, the total enthalpy change must be the same for every sequence. This rule, which is a consequence of the first law of thermodynamics, was originally known as Hess's law of constant heat summation. Suppose that we compare two different methods of synthesizing sodium chloride from sodium and chlorine.
Method 1 .
Na(s) + H2O(l) ---+ NaOH(s) + !H2(g), A.H = - 139.78 kJ/mol
! H2(g) + !CI2(g) ---+ HCI(g), A.H = - 92.3 1 kJ/mol
HCI(g) + NaOH(s) ---+ NaCI(s) + H2O(l), A.H = - 179.06 kJ/mol
Net change : Na(s) + !CI2(g) ---+ NaCI(s), A.Hnet = - 41 1. 15 kJ/mol Method 2.
! HzCg) + !CI2(g) ---+ HCI(g), A.H = - 92.31 kJ/mol
Na(s) + HCI(g) ---+ NaCI(s) + !H2(g), A.H = - 3 1 8.84 kJ/mol
1 36 Energy a n d t h e F i rst Law of Thermodynam i cs
The net chemical change is obtained by adding together all the reactions in the sequence ; the net enthalpy change is obtained by adding together all the enthalpy changes in the sequence. The net enthalpy change must be the same for every sequence which has the same net chemical change. Any number of reactions can be added or subtracted to yield the desired chemical reaction ; the enthalpy changes of the reactions are added or sub tracted algebraically in the corresponding way.
If a certain chemical reaction is combined in a sequence with the reverse of the same reaction, there is no net chemical effect, and
AH
= 0 for the combination. It follows immediately that theAH
of the reverse reaction is equal in magnitude but opposite in sign to that of the forward reaction.The utility of this property of sequences, which is really nothing more than the fact that the enthalpy change in a system is independent of the path, is illustrated by the sequence
1)
2)
C(graphite) + tOzCg) ---+ CO(g),
CO(g) + tOzCg) ---+ COzCg),
The net change in the sequence is
3) C(graphite) + OzCg) ---+ COzCg),
Therefore
AH 3
=AH
1 +AH
2 ' In this particular instance,AH
2 andAH 3
are readilymeasurable in the calorimeter, while
AH
1 is not. Since the value ofAH
1 can be computedfrom the other two values, there is no need to measure it.
Similarly, by subtracting reaction (2) from reaction (1) we obtain
4) C(graphite) + CO2(g) ---+ 2 CO(g),
and the heat of this reaction can also be obtained from the measured values.
* 7 . 23 H EATS O F S O L U TI O N A N D D I L U TI O N
The
heat of solution
is the enthalpy change associated with the addition of a specified amount of solute to a specified amount of solvent at constant temperature and pressure. For convenience we shall use water as the solvent in the illustrations, but the argUment can be applied to any solvent with slight modification. The change in state is represented byX +
n
Aq ---+ X'n
Aq,AHs .
One mole of solute X is added to
n
moles of water. The water is given the symbol Aq in this equation; it is convenient to assign a conventional enthalpy of zero to the water in these solution reactions.Consider the examples
HCI(g) + lOAq ---+ HCI · lOAq,
AHl
= - 69.01 kJ/molHCl(g) + 25 Aq ---+ HCI · 25 Aq,
AH
2 = - 72.03 kJ ImolHCl(g) + 40Aq ---+ HCI · 40Aq,
AH3
= - 72.79 kJ/molHCI(g) + 200Aq ---+ HCI · 200Aq,
AH4
= - 73.96 kJ/molHCI(g) + oo Aq ---+ HCI · oo Aq,
AHs
= - 74.85 kJ/molThe values of
AH
show the general dependence of the heat of solution on the amount of solvent. As more and more solvent is used, the heat of solution approaches a limitingH eats of R eaction at Constant Vol u me 1 37
value, the value in the " infinitely dilute " solution ; for HCI this limiting value is given by tllIs ·
If we subtract the first equation from the second in the above set, we obtain
HCI · lOAq + 1 5 Aq � HCI · 25 Aq, L1H = tllI2 -L1Hl = - 3.02 kJ/mol.
This value of L1H is a heat of dilution: the heat withdrawn from the surroundings when
additional solvent is added to a solution. The heat of dilution of a solution is dependent on the original concentration of the solution and on the amount of solvent added.
The heat of formation of a solution is the enthalpy associated with the reaction (using hydrochloric acid as an example) :
tHig) + tCl2(g) + nAq � HCI ·
n
Aq, L1H'j ,where the solvent Aq i s counted a s having zero enthalpy.
The heat of solution defined above is the
integral
heat of solution. This distinguishes it from thedifferential
heat of solution, which is defined in Section 1 1.24.7 . 24 H EATS O F R EACTI O N AT C O N STA N T VO L U M E
If any of the reactants or products of the calorimetric reaction are gaseous, it is necessary to conduct the reaction in a sealed bomb. Under this condition the system is initially and finally in a constant volume rather than being under a constant pressure. The measured heat of reaction at constant volume is equal to an energy increment, rather than to an enthalpy increment :
Qv = L1 U (7.70)
The corresponding change in state is
R(T, V, p) � P(T, V, p'),
where R(T, V, p) represents the reactants in the initial condition T, V, p ; and P(T, V, p') represents the products in the final condition T, V, p'. The temperature and volume remain constant, but the pressure may change from p to p' in the transformation.
To relate the L1U in Eq. (7.70) to the corresponding tllI, we apply the defining equation
for H to the initial and final states :