We have discussed reactions in which the final temperature is the same as the initial temperature. The enthalpy change will have a different value if the temperature of the system changes during the reaction. One case of interest is that the chemical reaction takes place adiabatically at constant pressure. In this case the enthalpy change is equal to the heat transferred, which is equal to zero. In order to compute the final temperature
Process 3: Temperature change of products (and possibly remaining reactants) to final temperature of adiabatic reaction
Process 2: Isothermal reaction at temperature T1 T2
T1
Process 1: Adiabati c reaction
Reactants Products
T
Figure 2.11 The Process to Calculate the Final Temperature of an Adiabatic Reaction.
of a system in which a chemical reaction takes place adiabatically, we consider the processes shown in Figure 2.11. Process 1 is the adiabatic reaction, for which∆His equal to zero. Process 2 is the reaction carried out isothermally. Process 3 is the change in temperature of the products plus any remaining reactants to the same final state as process 1. Because enthalpy is a state function, the enthalpy change of process 1 is equal to the sum of the enthalpy changes of processes 2 and 3:
∆H1∆H2+∆H30 (2.7-22)
If a stoichiometric mixture is taken and if the reaction proceeds to completion, there are no remaining reactants, and we can write
∆H3 T2
T1
s
i1
νiCP,m(i)dT
(products only in sum)
−∆H2 (2.7-23)
This equation can be solved forT2if the values of the heat capacities are known as a function of temperature. If there are remaining amounts of some reactants, their heat capacities must be added to the sum in Eq. (2.7-23). If the reaction comes to equilibrium short of completion, the equilibrium can shift as the temperature
changes (see Chapter 7). The calculation is more complicated. We will not discuss that case.
E X A M P L E 2.32
Find the final temperature if the reaction of Eq. (2.7-1) is carried out adiabatically at constant pressure beginning at 298.15 K. Assume that a stoichiometric mixture is present before the reaction and that the reaction proceeds to completion. Assume that the heat capacity of water vapor is constant and equal to its value at 2000 K.
Solution
The reaction as written in Eq. (2.7-1) includes liquid water as the product. In our case the water produced will be a vapor, so we calculate∆H◦ at 298.15 K for the gaseous product:
∆H◦(298.15 K)2∆fH◦(H2O(g))−0−0
2(−241.818 kJ mol−1) −483.636 kJ mol−1 −483636 J mol−1 CP(products)2(51.18 J K−1mol−1)102.36 J K−1mol−1
T2 298.15 K
102.36 J K−1mol−1
dT−483636 J mol−10
102.36 J K−1mol−1
(T2−298.15 K)483636 J mol−1 T25023 K≈5000 K
This result might be inaccurate because of the actual dependence of the heat capacity of water vapor on temperature.
Exercise 2.28
Find the final temperature if a stoichiometric mixture of methane and oxygen is ignited at 298.15 K and allowed to react adiabatically at a constant pressure. Assume that the reaction proceeds to completion and that the heat capacities of the products are constant and equal to their values at 2000 K.
If heat capacities are represented by polynomials as in Table A.6, a more nearly accurate final temperature can be calculated. This leads to a nonlinear equation, which can be solved by trial and error or by other numerical techniques.
Exercise 2.29
Using the parameters from Table A.6, find the final temperature after the adiabatic combustion of the stoichiometric mixture of hydrogen and oxygen in Example 2.32.
P R O B L E M S
Section 2.7: Calculation of Enthalpy Changes of a Class of Chemical Reactions
2.48 Calculate∆H◦and∆U◦for the reactions at 298.15 K:
a. C3H8(g) + 5 O2(g)−→3 CO2(g) + 4 H2O(l) b. 2 SO2(g) + O2(g)−→2 SO3(g)
c. SO3(g) + H2O(g)−→H2SO4(l)
2.49 Calculate∆H◦and∆U◦for the reactions at 298.15 K:
a. 4 CuO(s)−→2 Cu2O(s) + O2(g) b. 2 CO(g) + O2(g)−→2 CO2(g)
c. C2H5OH(l) + 3 O2(g)−→2 CO2(g) + 3 H2O(l) 2.50 Calculate∆H◦for the reactions of Problem 2.48 at
75◦C.
2.51 Calculate∆H◦for the reactions of Problem 2.49 at 75◦C.
2.52 a. Find the values of the enthalpy changes of formation of methane, carbon dioxide, and liquid water at 373.15 K, using heat capacity values from Table A.8.
b. Using the values from part a, find the standard-state enthalpy change for the reaction at 373.15 K:
CH4(g)+2O2(g)−→CO2(g)+2H2O(l) c. Find the standard-state enthalpy change of the reaction
of part b, using Eq. (2.7-19). Comment on the comparison of your answer with that of part b.
2.53 a. Calculate∆H◦for the reaction at 298.15 K:
2C2H6(g)+7O2(g)−→4CO2(g)+6H2O(l) b. Using the value of∆vapHm(H2O) and the result of part
a, find∆H◦of this reaction for the case that the water is vapor.
c. Recalculate this quantity using∆fHm(H2O(g)) and compare with your result of part a. Comment on any difference.
2.54 The molar enthalpy change of combustion of glycerol at 298.15 K is equal to−393.73 kcal mol−1. Convert this to kJ mol−1and find the value of∆fH◦of glycerol at this temperature.
2.55 a. The molar enthalpy change of combustion of sucrose, C12H22O11, at 298.15 K is equal to−5640.9 kJ mol−1. Calculate its enthalpy change of formation at 298.15 K.
Calculate∆U◦for the combustion reaction.
b. The molar enthalpy change of combustion of stearic acid, C18H36O2, at 298.15 K is equal to –11280.6 kJ mol−1. Calculate its enthalpy change of formation at 298.15 K. Compare its enthalpy change of combustion per gram with that of sucrose.
2.56 Find the value of∆H◦ for the combustion of methane at 125◦C. State any assumptions. Assume that the heat capacities are constant between 25◦C and 125◦C.
2.8 Calculation of Energy Changes of Chemical Reactions
It would be possible to make tables of energy changes of formation and to calculate
∆Uvalues in the same way as ∆H values are calculated from enthalpy changes of formation. However, adequate accuracy can be achieved without constructing a table of∆fUvalues. From the definition of the enthalpy we can write an expression for∆U for a chemical reaction:
∆U∆H−∆(PV) (2.8-1)
Ordinarily ∆(PV) is much smaller than∆H, so that a less accurate calculation of
∆(PV) might suffice. For example, if∆His 1000 times larger than∆(PV) and if five significant digits are desired in∆U, then five significant digits are required in∆Hbut only two significant digits are needed in∆(PV).
The productPVis given by
PV P(V(s)+V(l)+V(g)) (2.8-2)
whereV(s) is the volume of all of the solid phases,V(l) is the volume of all of the liquid phases, andV(g) is the volume of the gas phase. Under ordinary conditions the molar volume of a gas is several hundred to a thousand times larger than the molar volume of a solid or liquid. If there is at least one gaseous product or reactant we can ignore the volume of the solid and liquid phases to an adequate approximation,
PV≈PV(g) so that
∆(PV)≈∆PV(g) (2.8-3) If the products and reactants are at the same temperature and if we use the ideal gas equation as an approximation,
∆U∆H−∆PV(g)≈∆H−∆n(g)RT (2.8-4) where∆n(g) is the change in the number of moles of gaseous substances in the reaction equation. If 1 mol of reaction occurs, then
∆n(g)∆ν(g) s
i1
νi
(gases only)
(2.8-5)
which defines the quantity∆ν(g), equal to the number of moles of gas in the product side of the balanced chemical equation minus the numbers of moles of gas in the reactant side of the balanced equation.
E X A M P L E 2.33
a. Find∆(PV) and∆U◦for the reaction of Eq. (2.7-1) at 298.15 K.
b. Using the fact that the molar volume of liquid water is 18 cm3mol−1at 298.15 K, make a more accurate calculation of∆(PV) for the reaction of the previous example.
Solution a. The reaction is
2H2(g)+O2(g)−→2H2O(l)
Since the single product, H2O, is liquid,∆ν(g) −3 and∆n(g) −3.000 mol
∆(PV)∆ν(g)RT(−3)(8.3145 J K−1mol−1)(298.15 K) −7437 J mol−1 −7.437 kJ mol−1
∆H◦2∆fH◦(H2O)−0−0 −571.660 kJ mol−1
∆U◦ −571.660 kJ mol−1−(−7.437 kJ mol−1) −564.223 kJ mol−1
b. V2(2.000 mol)(18 cm3mol−1)36 cm33.6×10−5m3
V1 (3.000 mol)(8.3145 J K−1mol−1)(298.15 K)
100000 Pa 0.0743690 m3
∆V 3.6×10−5m3−0.0743690 m3 −0.07433 m3 Since the pressure is constant atP◦100000 Pa,
∆(PV)P∆V(100000 Pa)(−0.07340 m3) −7433 J mol−1 This value compares with the value of−7437 J mol−1in part a.
Exercise 2.30
Find∆U◦for the reaction of Eq. (2.7-6) at 298.15 K.