REVIEW PROBLEMS
5.4 DETERMINATION OF STANDARD ENTHALPIES OF FORMATION AND REACTION
143 Reactants aA + bB
in their standard states at T
Products cC + dD in their standard states
at T
Elements in their standard states
at T (1)
(2) (3)
Figure 5.1
Steps used to relate H° of a reaction to fH° of reactants and products.
where niis the stoichiometric number of substance i in the reaction and fH°T,iis the standard enthalpy of formation of substance i at temperature T.
To prove (5.6), consider the reaction aA bB → cC dD, where a, b, c, and d are the unsigned stoichiometric coefficients and A, B, C, and D are substances. Figure 5.1 shows two different isothermal paths from reactants to products in their standard states. Step 1 is a direct conversion of reactants to products. Step 2 is a conversion of reactants to standard-state elements in their reference forms. Step 3 is a conversion of elements to products. (Of course, the same elements produced by the decomposition of the reactants will form the products.) Since H is a state function, H is independent of path and H1 H2 H3. We have H1 H°Tfor the reaction. The reverse of process 2 would form aA bB from their elements; hence,
where fH°T(A) is the standard enthalpy of formation of substance A at temperature T.
Step 3 is the formation of cC dD from their elements, so
The relation H1 H2 H3becomes
which is Eq. (5.6) for the reaction aA bB → cC dD, since the stoichiometric numbers niare negative for reactants.
There are many more chemical reactions than there are chemical substances.
Rather than having to measure and tabulate H° for every possible chemical reaction, we can use (5.6) to calculate H° from tabulated fH° values of the substances in-volved, provided we have determined fH° of each substance. The next section tells how fH° is measured.
5.4 DETERMINATION OF STANDARD ENTHALPIES OF FORMATION AND REACTION
Measurement of fH°
The quantity fH°T,i is H° for isothermally converting pure standard-state elements in their reference forms to one mole of standard-state substance i. To find fH°T,i, we carry out the following steps:
1. If any of the elements involved are gases at T and 1 bar, we calculate H for the hypothetical transformation of each gaseous element from an ideal gas at T and 1 bar to a real gas at T and 1 bar. This step is necessary because the standard state
¢H°T a¢fH°T1A2 b¢fH°T1B2 c¢fH°T1C2 d¢fH°T1D2
¢H3 c¢fH°T1C2 d¢fH°T1D2
¢H2 a ¢fH°T1A2 b¢fH°T1B2
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∆fH°298
Figure 5.2
fH°298values. The scales are logarithmic.
Chapter 5
Standard Thermodynamic Functions of Reaction
144
of a gas is the hypothetical ideal gas at 1 bar, whereas only real gases exist at 1 bar.
The procedure for this calculation is given at the end of this section.
2. We measure H for mixing the pure elements at T and 1 bar.
3. We use H 兰21CP dT 兰21(V TVa) dP [Eq. (4.63)] to find H for bringing the mixture from T and 1 bar to the conditions under which we plan to carry out the reaction to form substance i. (For example, in the combustion of an element with oxygen, we might want the initial pressure to be 30 atm.)
4. We use a calorimeter (see after Example 5.1) to measure H for the reaction in which the compound is formed from the mixed elements.
5. We use (4.63) to find H for bringing the compound from the state in which it is formed in step 4 to T and 1 bar.
6. If compound i is a gas, we calculate H for the hypothetical transformation of i from a real gas to an ideal gas at T and 1 bar.
The net result of these six steps is the conversion of standard-state elements at T to standard-state i at T. The standard enthalpy of formation fH°T,iis the sum of these six H’s. The main contribution by far comes from step 4, but in precise work one in-cludes all the steps.
Once fH°i has been found at one temperature, its value at any other temperature can be calculated using CPdata for i and its elements (see Sec. 5.5). Nearly all ther-modynamics tables list fH° at 298.15 K (25°C). Some tables list fH° at other tem-peratures. Some values of fH°298are plotted in Fig. 5.2. A table of fH°298is given in the Appendix. Once we have built up such a table, we can use Eq. (5.6) to find H°298
for any reaction whose species are listed.
EXAMPLE 5.1
Calculation of H° from fH° dataFind H°298for the combustion of one mole of the simplest amino acid, glycine, NH2CH2COOH, according to
(5.7) Substitution of AppendixfH°298values intoH°298
inifH°298,i[Eq. (5.6)]givesH°298as
Exercise
Use Appendix data to find H°298for the combustion of one mole of sucrose, C12H22O11(s), to CO2(g) and H2O(l). (Answer:5644.5 kJ/mol.)
Calorimetry
To carry out step 4 of the preceding procedure to find fH° of a compound, we must measure H for the chemical reaction that forms the compound from its elements. For certain compounds, this can be done in a calorimeter. We shall consider measurement of H for chemical reactions in general, not just for formation reactions.
The most common type of reaction studied calorimetrically is combustion. One also measures heats of hydrogenation, halogenation, neutralization, solution, dilution, mixing, phase transitions, etc. Heat capacities are also determined in a calorimeter.
Reactions where some of the species are gases (for example, combustion reactions)
973.49 kJ>mol 312102 521285.8302 21393.5092 1528.102 94102 4 kJ>mol
NH2CH2COOH1s2 94 O21g2 S 2CO21g2 52 H2O1l2 12 N21g2
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are studied in a constant-volume calorimeter. Reactions not involving gases are stud-ied in a constant-pressure calorimeter.
The standard enthalpy of combustioncH°Tof a substance is H°Tfor the reac-tion in which one mole of the substance is burned in O2. For example, cH° for solid glycine is H° for reaction (5.7). Some cH°298values are plotted in Fig. 5.3.
An adiabatic bomb calorimeter (Fig. 5.4) is used to measure heats of combus-tion. Let R stand for the mixture of reactants, P for the product mixture, and K for the bomb walls plus the surrounding water bath. Suppose we start with the reactants at 25°C. Let the measured temperature rise due to the reaction be T. Let the system be the bomb, its contents, and the surrounding water bath. This system is thermally insu-lated and does no work on its surroundings (except for a completely negligible amount of work done by the expanding water bath when its temperature rises). Therefore q 0 and w 0. Hence U 0 for the reaction, as noted in step (a) of Fig. 5.4.
After the temperature rise T due to the reaction is accurately measured, one cools the system back to 25°C. Then one measures the amount of electrical energy Uel that must be supplied to raise the system’s temperature from 25°C to 25°C T; this is step (b) in Fig. 5.4. We have Ub Uel VIt, where V, I, and t are the voltage, cur-rent, and time.
The desired quantity rU298(where r stands for reaction) is shown as step (c).
The change in the state function U must be the same for path (a) as for path (c) (b), since these paths connect the same two states. Thus Ua Uc Ub and 0
rU298 Uel. Hence rU298 Uel, and the measured Uel enables rU298 to be found.
Instead of using Uel, we could use an alternative procedure. We have seen that
rU298 Ub(Fig. 5.4b). If we imagine carrying out step (b) by supplying heat qb to the system K P (instead of using electrical energy), then we would have Ub qb CKPT, where CKPis the average heat capacity of the system K P over the temperature range. Thus
(5.8) To find CKP, we repeat the combustion experiment in the same calorimeter using ben-zoic acid, whose U of combustion is accurately known. For burning the benzoic acid, let rU298, P, and T denote U298of reaction, the reaction products, and the tem-perature rise. Similar to Eq. (5.8), we have rU298 CKPT. Measurement of
¢rU298 CKP ¢T
Section 5.4 Determination of Standard Enthalpies of Formation and Reaction 145
∆cH298°
Figure 5.3
Standard enthalpies of combustion at 25°C. The scale is logarithmic.
The products are CO2(g) and H2O(l).
(a)
Figure 5.4
(a) An adiabatic bomb calorimeter.
The shaded walls are adiabatic.
(b) Energy relations for this calorimeter.
(b) R + K
at 25°C
P + K at 25°C
∆rU
298
(b) Ue1
∆U = 0 (a)
(c)
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Chapter 5
Standard Thermodynamic Functions of Reaction
146
T and calculation of rU298from the known U of combustion of benzoic acid then gives us the heat capacity CKP. The temperature ranges over which the two combus-tions are carried out are very similar. Also, the main contribution to CKPand CKP comes from the bomb walls and the water bath. For these reasons, it is an excellent ap-proximation to take CKP CKP. (In precise work, the difference between the two is calculated using the known heat capacities of the combustion products.) Knowing CKP, we find rU298from Eq. (5.8).
To find the standard internal energy change U°298for the reaction, we must allow for the changes in URand UPthat occur when the reactants and products are brought from the states that occur in the calorimeter to their standard states. This correction is typically about 0.1 percent for combustion reactions.
(An analysis similar to Fig. 5.4b enables one to estimate the temperature of a flame. See Prob. 5.60.)
For reactions that do not involve gases, one can use an adiabatic constant-pressure calorimeter. The discussion is similar to that for the adiabatic bomb calorimeter, ex-cept that P is held fixed instead of V, and H of reaction is measured instead of U.
EXAMPLE 5.2
Calculation of cU° from calorimetric data Combustion of 2.016 g of solid glucose (C6H12O6) at 25°C in an adiabatic bomb calorimeter with heat capacity 9550 J/K gives a temperature rise of 3.282°C.Find cU°298of solid glucose.
With the heat capacity of the products neglected, Eq. (5.8) gives U
(9550 J/K)(3.282 K) 31.34 kJ for combustion of 2.016 g of glucose. The experimenter burned (2.016 g)/(180.16 g/mol) 0.01119 mol. Hence U per mole of glucose burned is (31.34 kJ)/(0.01119 mol) 2801 kJ/mol, and this is cU°298if the difference between conditions in the calorimeter and standard-state conditions is neglected.
Exercise
If 1.247 g of glucose is burned in an adiabatic bomb calorimeter whose heat capacity is 11.45 kJ/K, what will be the temperature rise? (Answer: 1.693 K.)
Relation between H° and U°
Calorimetric study of a reaction gives either U° or H°. Use of H ⬅ U PV allows interconversion between H° and U°. For a process at constant pressure,
H U P V. Since the standard pressure P° [Eq. (5.2)] is the same for all sub-stances, conversion of pure standard-state reactants to products is a constant-pressure process, and for a reaction we have
(5.9) Similar to H°
iniH°m, i[Eq. (5.3)], the changes in standard-state volume and internal energy for a reaction are given by V° iniV°m, iand U° iniU°m,i. A sum like iniU°m, ilooks abstract, but when we see ini , we can translate this into“products minus reactants,” since the stoichiometric number niis positive for products and negative for reactants.
The molar volumes of gases at 1 bar are much greater than those of liquids or solids, so it is an excellent approximation to consider only the gaseous reactants and products in applying (5.9). For example, consider the reaction
aA1s2 bB1g2 S cC1g2 dD1g2 eE1l2
¢H° ¢U° P° ¢V°
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Section 5.4 Determination of Standard Enthalpies of Formation and Reaction 147 Neglecting the volumes of the solid and liquid substances A and E, we have V°
cV°m,C dV°m,D bV°m,B. The standard state of a gas is an ideal gas, so V°m RT/P°
for each of the gases C, D, and B. Therefore V° (c d b)RT/P°. The quantity c d b is the total number of moles of product gases minus the total number of moles of reactant gases. Thus, c d b is the change in the number of moles of gas for the reaction. We write c d b ng/mol, where ngstands for moles of gas.
Since c d b is a dimensionless number, we divided ng by the unit “mole” to make it dimensionless. We thus have V° (ng/mol)RT/P°, and (5.9) becomes
(5.10) For example, the reaction C3H8(g) 5O2(g) → 3CO2(g) 4H2O(l) has ng/mol
3 1 5 3 and (5.10) gives H°T U°T 3RT. At 300 K, H° U°
7.48 kJ/mol for this reaction, which is small but not negligible.
EXAMPLE 5.3
Calculation of fU° from fH°For CO(NH2)2(s), fH°298 333.51 kJ/mol. Find fU°298of CO(NH2)2(s).
The formation reaction is
and has ng/mol 0 2 1 . Equation (5.10) gives
Exercise
For CF2ClCF2Cl(g), fH°298 890.4 kJ/mol. Find fU°298of CF2ClCF2Cl(g).
(Answer:885.4 kJ/mol.)
Exercise
In Example 5.2, cU°298of glucose was found to be 2801 kJ/mol. Find cH°298 of glucose. (Answer:2801 kJ/mol.)
For reactions not involving gases, ngis zero, and H° is essentially the same as U° to within experimental error. For reactions involving gases, the difference between H° and U°, though certainly not negligible, is usually not great. The quantity RT in (5.10) equals 2.5 kJ/mol at 300 K and 8.3 kJ/mol at 1000 K, and
ng/mol is usually a small integer. These RT values are small compared with typi-cal H° values, which are hundreds of kJ/mol (see the fH° values in the Appendix). In qualitative reasoning, chemists often don’t bother to distinguish be-tween H° and U°.
Hess’s Law
Suppose we want the standard enthalpy of formation fH°298of ethane gas at 25°C.
This is H°298for 2C(graphite) 3H2(g) → C2H6(g). Unfortunately, we cannot react graphite with hydrogen and expect to get ethane, so the heat of formation of ethane cannot be measured directly. This is true for most compounds. Instead, we determine
324.83 kJ>mol
¢fU°298 333.51 kJ>mol 1722 18.314 103 kJ>mol-K2 1298.15 K2
7 2 1 2
C1graphite2 12O21g2 N21g2 2H21g2 S CO1NH2221s2
¢H°T ¢U°T ¢ng RT>mol
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H
Re(T ) Pr(T ∆T)
Pr(T )
Adia-batic Isothermal
∆HT
Figure 5.5
Enthalpy changes for adiabatic versus isothermal transformations of reactants (Re) to products (Pr) at constant pressure. The reaction is exothermic.
the heats of combustion of ethane, hydrogen, and graphite, these heats being readily measured. The following values are found at 25°C:
(1) (2) (3) Multiplying the definition H°
iniH°m, i[Eq. (5.3)] by 1, 2, and 3 for reactions (1), (2), and (3), respectively, we getwhere the subscript 298 on the H°m’s is understood. Addition of these equations gives (5.11) But the quantity on the right side of (5.11) is H° for the desired formation reaction
(5.12) Therefore fH°298 85 kJ/mol for ethane.
We can save time in writing if we just look at chemical reactions (1) to (3), figure out what factors are needed to multiply each reaction so that they add up to the desired reaction (5.12), and apply these factors to the H° values. Thus, the desired reaction (5.12) has 2 moles of C on the left, and multiplication of reaction (2) by 2 will give 2 moles of C on the left. Similarly, we multiply reaction (1) by 1 to give 1 mole of C2H6on the right and multiply reaction (3) by 3 to give 3 moles of H2on the left.
Multiplication of reactions (1), (2), and (3) by 1, 2, and 3, followed by addition, gives reaction (5.12). Hence H°298for (5.12) is [(1560) 2(393 ) 3(286)]
kJ/mol. The procedure of combining heats of several reactions to obtain the heat of a desired reaction is Hess’s law. Its validity rests on the fact that H is a state function, so H° is independent of the path used to go from reactants to products. H° for the path elements → ethane is the same as H° for the path
Since the reactants and products are not ordinarily in their standard states when we carry out a reaction, the actual enthalpy change HTfor a reaction differs somewhat from H°T. However, this difference is small, and HTand H°Tare unlikely to have dif-ferent signs. For the discussion of this paragraph, we shall assume that HTand H°T
have the same sign. If this sign is positive, the reaction is said to be endothermic; if this sign is negative, the reaction is exothermic. For a reaction run at constant pressure in a system with P-V work only, H equals qP, the heat flowing into the system.
The quantities HTand H°Tcorrespond to enthalpy differences between products and reactants at the same temperature T ; HT Hproducts, T Hreactants, T. Therefore, when a reaction is run under constant-T -and-P conditions (in a constant-temperature bath), the heat q absorbed by the system equals HT. For an exothermic reaction (HT 0) run at constant T and P, q is negative and the system gives off heat to its surroundings. When an endothermic reaction is run at constant T and P, heat flows into the system. If an exothermic reaction is run under adiabatic and constant-P con-ditions, then q 0 (since the process is adiabatic) and H ⬅ Hproducts Hreactants 0 (since H qP); here, the products will be at a higher temperature than the reactants (Fig. 5.5). For an exothermic reaction run under conditions that are neither adiabatic
elements oxygen S combustion products S ethane oxygen
1 2
2C1graphite2 3H21g2 S C2H61g2
85 kJ>mol H°m1C2H62 2H°m1C2 3H°m1H22 31286 kJ>mol2 3H°m1H2O2 3H°m1H22 1.5H°m1O22 2139312 kJ>mol2 2H°m1CO22 2H°m1O22 2H°m1C2
11560 kJ>mol2 2H°m1CO22 3H°m1H2O2 H°m1C2H62 3.5H°m1O22 H21g2 12O21g2 S H2O1l2 ¢H°298 286 kJ>mol C1graphite2 O21g2 S CO21g2 ¢H°298 39312 kJ>mol
C2H61g2 72O21g2 S 2CO21g2 3H2O1l2 ¢H°298 1560 kJ>mol
Chapter 5
Standard Thermodynamic Functions of Reaction
148
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Section 5.4 Determination of Standard Enthalpies of Formation and Reaction 149 nor isothermal, some heat flows to the surroundings and the temperature of the system
rises by an amount that is less than T under adiabatic conditions.
EXAMPLE 5.4
Calculation of fH° and fU° from cH°The standard enthalpy of combustion cH°298of C2H6(g) to CO2(g) and H2O(l) is 1559.8 kJ/mol. Use this cH° and Appendix data on CO2(g) and H2O(l) to find fH°298and fU°298of C2H6(g).
Combustion means burning in oxygen. The combustion reaction for one mole of ethane is
The relation H° inifH°i[Eq. (5.6)] gives for this combustion
Substitution of the values offH° of CO2(g) and H2O(l) andcH° gives at 298 K
Note that this example essentially repeats the preceding Hess’s law calculation.
Reactions (2) and (3) above are the formation reactions of CO2(g) and H2O(l).
To findfU°298fromfH °298, we must write the formation reaction for C2H6, which is 2C(graphite) 3H2(g) → C2H6(g). This reaction hasng/mol 1 3
2, and Eq. (5.10) gives
fU°298 84.7 kJ兾mol (2)(0.008314 kJ兾mol-K)(298.1 K) 79.7 kJ兾mol A common student error is to find ngfrom the combustion reaction instead of from the formation reaction.
Exercise
cH°298 of crystalline buckminsterfullerene, C60(cr), is 2.589 104 kJ/mol [H. P. Diogo et al., J. Chem. Soc. Faraday Trans., 89, 3541 (1993)]. With the aid of Appendix data, find fH°298of C60(cr). (Answer: 2.28 103kJ/mol.)
Calculation of Hidⴚ Hre
The standard state of a gas is the hypothetical ideal gas at 1 bar. To findfH° of a gaseous compound or a compound formed from gaseous elements, we must calculate the differ-ence between the standard-state ideal-gas enthalpy and the enthalpy of the real gas (steps 1 and 6 in the first part of Sec. 5.4). Let Hre(T, P°) be the enthalpy of a (real) gaseous substance at T and P°, and let Hid(T, P°) be the enthalpy of the corresponding fictitious ideal gas at T and P°, where P°⬅ 1 bar. Hid(T, P°) is the enthalpy of a hypo-thetical gas in which each molecule has the same structure (bond distances and angles and conformation) as in the real gas but in which there are no forces between the mole-cules. To find Hid Hre, we use the following hypothetical isothermal process at T :
(5.13) Real gas at P° S1a2 real gas at 0 bar S1b2 ideal gas at 0 bar S1c2 ideal gas at P°
¢fH°1C2H6, g2 84.7 kJ>mol
¢fH°1C2H6, g2 0
1559.8 kJ>mol 21393.51 kJ>mol2 31285.83 kJ>mol2
¢cH° 2 ¢fH°1CO2, g2 3 ¢fH°1H2O, l2 ¢fH°1C2H6, g2 72 ¢fH°1O2, g2 C2H61g2 72O21g2 S 2CO21g2 3H2O1l2
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In step (a), we isothermally reduce the pressure of the real gas from 1 bar to zero. In step (b), we wave a magic wand that eliminates intermolecular interactions, thereby changing the real gas into an ideal gas at zero pressure. In step (c), we isothermally increase the pressure of the ideal gas from 0 to 1 bar. The overall process converts the real gas at 1 bar and T into an ideal gas at 1 bar and T. For this process,
(5.14) The enthalpy change Hafor step (a) is calculated from the integrated form of Eq. (4.48), ( T V TVa [Eq. (4.63) with dT 0]:
For step (b), Hb Hid(T, 0 bar) Hre(T, 0 bar). The quantity Ure Uid(both at the same T ) is Uintermol (Sec. 2.11), the contribution of intermolecular interactions to the internal energy. Since intermolecular interactions go to zero as P goes to zero in the real gas, we have Ure Uidin the zero-pressure limit. Also, as P goes to zero, the equation of state for the real gas approaches that for the ideal gas. Therefore (PV )re equals (PV )idin the zero-pressure limit. Hence Hre⬅ Ure (PV)reequals Hidin the zero-pressure limit:
(5.15) For step (c), Hcis zero, since H of an ideal gas is independent of pressure.
Equation (5.14) becomes
(5.16)
where a⬅ V1( P was used. The integral in (5.16) is evaluated using P-V-T data or an equation of state (Sec. 8.8) for the real gas. The difference Hm,re Hm,idis quite small at 1 bar (since intermolecular interactions are quite small in a 1-bar gas) but is included in precise work. Some values of Hm,re Hm,idat 298 K and 1 bar are
7 J/mol for Ar, 17 J/mol for Kr, and 61 J/mol for C2H6. Figure 5.6 plots Hm,re and Hm,idversus P for N2(g) at 25°C, with Hm,idarbitrarily set equal to zero. Steps (a) and (c) of the process (5.13) are indicated in the figure. Intermolecular attractions make Ureand Hreslightly less than Uidand Hid, respectively, at 1 bar.
Conventional Enthalpies. Instead of tabulating fH° values and using these to find
H° of reactions, one can construct a table of conventional (or relative) standard-state en-thalpies of substances and use these to calculate H° of reactions from H° iniH°m,i, where H°m,iis the conventional standard-state molar enthalpy of substance i. To construct such a table, we begin by arbitrarily assigning the value zero to the conventional molar enthalpy at 25°C and 1 bar of the most stable form of each pure element:
(5.17) Although the actual absolute enthalpies of different elements do differ, the convention (5.17) cannot lead to error in chemical reactions because elements are never interconverted in chemical reactions. Knowing the conventional H°m,298of an element, we can use H 兰21CPdT at constant P to find the conventional H°mof an element at any temperature T. If any phase changes occur between 298.15 K and T, we make separate allowance for them.
H°m,298 0 for each element in its stable form Hid1T, P°2 Hre1T, P°2
冮
0P° c T a00VTbPVd dP const. T Hre1T, 0 bar2 Hid1T, 0 bar2 and ¢Hb 0
¢Ha Hre1T, 0 bar2 Hre1T, P°2
冮
P°01V TVa2 dP¢H Hid1T, P°2 Hre1T, P°2 ¢Ha ¢Hb ¢Hc Chapter 5
Standard Thermodynamic Functions of Reaction
150
N2, 25ºC
Figure 5.6
Change in Hmwith P for the isothermal conversion of real-gas N2to ideal-gas N2at 25°C.
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Section 5.5 Temperature Dependence of Reaction Heats
151 So far, only elements have been considered. Suppose we want the conventional
en-thalpy of liquid water at T. The formation reaction is H2 O2 → H2O. Therefore
fH°T(H2O, l) H°m, T(H2O, l) H°m, T(H2, g) H°m, T(O2, g). Knowing the conventional enthalpies H°m,Tfor the elements H2and O2, we use the experimental fH°Tof H2O(l) (de-termined as discussed earlier) to find the conventional H°m, Tof H2O(l). Similarly, we can find conventional enthalpies of other compounds.