Page 167 - Physical Chemistry

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Chapter 5 the heats of combustion of ethane, hydrogen, and graphite, these heats being readily
Standard Thermodynamic measured. The following values are found at 25°C:
Functions of Reaction
C H 1g2 7 2 O 1g2 S 2CO 1g2 3H O1l2 ¢H° 1560 kJ>mol (1)
6
298
2
2
2
2
C1graphite2 O 1g2 S CO 1g2 ¢H° 393 kJ>mol (2)
1
2
2
2
298
H 1g2 1 2 O 1g2 S H O1l2 ¢H° 286 kJ>mol (3)
2
298
2
2
Multiplying the definition H° n H° [Eq. (5.3)] by 1, 2, and 3 for reactions
i
i
m,i
(1), (2), and (3), respectively, we get
1 1560 kJ>mol2 2H° 1CO 2 3H° 1H O2 H° 1C H 2 3.5H° 1O 2
2
6
m
2
2
m
m
m
2
1
21 393 kJ>mol2 2H° 1CO 2 2H° 1O 2 2H° 1C2
m
2
2
2
m
m
31 286 kJ>mol2 3H° 1H O2 3H° 1H 2 1.5H° 1O 2
m
m
2
2
m
2
where the subscript 298 on the H° ’s is understood. Addition of these equations gives
m
85 kJ>mol H° 1C H 2 2H° 1C2 3H° 1H 2 (5.11)
6
2
m
m
2
m
But the quantity on the right side of (5.11) is H° for the desired formation reaction
2C1graphite2 3H 1g2 S C H 1g2 (5.12)
2
2
6
Therefore H° 85 kJ/mol for ethane.
f
298
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
C H on the right and multiply reaction (3) by 3 to give 3 moles of H on the left.
6
2
2
Multiplication of reactions (1), (2), and (3) by 1, 2, and 3, followed by addition,
1
gives reaction (5.12). Hence H° for (5.12) is [ ( 1560) 2( 393 ) 3( 286)]
2
298
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
elements oxygen S combustion products S ethane oxygen
H Since the reactants and products are not ordinarily in their standard states when we
carry out a reaction, the actual enthalpy change H for a reaction differs somewhat
T
from H°. However, this difference is small, and H and H° are unlikely to have dif-
Adia- T T T
Re(T) Pr(T ∆T)
batic ferent signs. For the discussion of this paragraph, we shall assume that H and H° T
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
∆H T Isothermal in a system with P-V work only, H equals q , the heat flowing into the system.
P
The quantities H and H° correspond to enthalpy differences between products
T
T
Pr(T)
and reactants at the same temperature T; H H products,T H reactants,T . Therefore,
T
when a reaction is run under constant-T-and-P conditions (in a constant-temperature
bath), the heat q absorbed by the system equals H . For an exothermic reaction
T
Figure 5.5 ( H 0) run at constant T and P, q is negative and the system gives off heat to its
T
surroundings. When an endothermic reaction is run at constant T and P, heat flows
Enthalpy changes for adiabatic into the system. If an exothermic reaction is run under adiabatic and constant-P con-
versus isothermal transformations
of reactants (Re) to products (Pr) ditions, then q 0 (since the process is adiabatic) and H H products H reactants 0
at constant pressure. The reaction (since H q ); here, the products will be at a higher temperature than the reactants
P
is exothermic. (Fig. 5.5). For an exothermic reaction run under conditions that are neither adiabatic