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Definition of Heat of Reaction: The enthalpy change when stoichiometric quantities of reactants react completely in a single reaction to form products at the same temperature and pressureFor example, suppose we have 3 moles/min of A flowing to a reactor and 2 moles/min of B. In the reactor, the following reaction takes place, going to completion:Definition of Standard Heat of Reaction: The heat of reaction when both reactants and products are at 1 atm and a specified reference temperature (almost always 25^{o}C)
3A + 2B = 2DBecause A and B are being fed in stoichiometric amounts and the reaction goes to completion, 2 moles/min of compound D are produced. From what we learned about reacting systems when we did mass balances, the extent of reaction is therefore 2x. If heat to the reactor is controlled such that the outlet temperature of the product is the same as that for the feed streams, then the heat that must be added or taken out of the reactor in order to maintain isothermal conditions is a direct measure of the heat of reaction. Thus,
Q = H_{out}  SH_{in }= DH_{r}Normally, the heat of reaction is given in units of energy per mole; e.g., kJ/mol. Per mole of which compound? None of them. Remember that the extent of reaction is in units of moles of reaction for the reaction as written. Thus, the molar enthalpy of reaction means
Dh_{r} = DH_{r}/xThe enthalpy difference between the products and the reactants can be either positive or negative. It takes energy to break bonds and energy is released when bonds are formed. Thus, if the net result of the reaction is the formation of more tightly bonded molecules, then the products are at a lower energy than the reactants and H_{out}  SH_{in }is negative; i.e., the heat of reaction is negative and heat must be taken out of the system. This is an exothermic reaction because heat is given off and must be removed from the reactor if the temperature is to be kept constant. Likewise, if the products are at a higher energy than the reactants, then H_{out}  SH_{in }= DH_{r} is positive. Reactions with a positive heat of reaction are endothermic; they absorb heat as the reaction proceeds, so heat must be added to the reactor to maintain isothermal conditions. In summary:
DH_{r} > 0 endothermic, heat absorbed
DH_{r} < 0 exothermic, heat evolved
Obtaining
Heats of Reaction
You have learned in your freshman chemistry classes that you can add
two or more chemical equations to give another equation, and that heats
of reaction add in exactly the same manner. This is called Hess's
Law. This procedure can be used with heats of formation, heats
of combustion, other heats of reaction, etc. We will use this method of
combining reactions to obtain heats of reaction from heats of formation,
combustion, and other reactions throughout the remainder of the course,
so we should review the methods here.
1. From Heats of Formation
Definition: The heat of formation is defined as the heat of reaction when a compound is formed from its elements (in their natural state of aggregation) at 25^{o}C and 1 atm. The element's natural state of aggregation is the phase and molecular structure that is most stable at 25^{o}C and 1 atm.As an example, the heat of formation for CO(g) would be the heat of reaction for C(graph) + ½O_{2}(g) = CO(g). Note that O_{2} is the naturally occurring element, not O, at 25^{o}C and 1 atm. Similarly, graphite is the naturally occurring form of carbon at these conditions. Some additional examples of reactions for which the heat of reaction is the heat of formation are:
C(graph) + O_{2}(g) = CO_{2}(g) DH_{r} = DH_{f,CO2}Values for heats of formation are available in Table B.1 in the textbook or from the DIPPR database. Remember that enthalpies are relative quantities: we must always measure changes in enthalpy because there is no absolute value of enthalpy. By building up tables of heats of formation, we are really setting up a system of measuring enthalpies of compounds relative to the elements in their naturally occurring states of aggregation. This is equivalent to assigning an enthalpy of zero to all of the elements in their normal states of aggregation and then measuring all enthalpies relative to that reference.
6C(graph) + 3H_{2}(g) = C_{6}H_{6}(l) DH_{r} = DH_{f,benzene}
H_{2}(g) + ½O_{2}(g) = H_{2}O(l) DH_{r} = DH_{f,H2O(l)}
To obtain the heat of reaction from heats of formation, we simply add and subtract formation equations (and their corresponding heats of formation in the same manner) in concert with Hess's Law to obtain the desired reaction. For example, if I want the heat of reaction for the following reaction:
2CO(g) + O_{2}(g) = 2CO_{2}(g)I select formation reactions for each of the compounds and write them down, multiply each by an appropriate factor so that they can be added to produce the desired equation, and then do the same arithmetic with the heats of formation. In this case, I would write:
C(graph) + O_{2}(g) = CO_{2}(g) DH_{f,CO2} = 393.5 kJ/molNow in order to combine these two equations together to get what I want, I must multiply the second equation by 1 and add the two together. Thus,
C(graph) + ½O_{2}(g) = CO(g) DH_{f,CO} = 110.52 kJ/mol
C(graph) + O_{2}(g) = CO_{2}(g) DH_{f,CO2} = 393.5 kJ/molNow in order to combine these two equations together to get what I want, I must multiply the first equation by 2 and the second equation by 2, and add the two together. Thus,
C(graph) + ½O_{2}(g) = CO(g) DH_{f,CO} = 110.52 kJ/mol
Notice that the two 2C(graph) terms cancel out when we add the two equations together. Likewise, one of the O_{2}(g) terms in the first equation cancels with the O_{2}(g) in the second equation. When adding together the heats of formation, we use the same arithmetic for adding those quantities that we used to add together the equations. Thus, we take 2 times the heat of formation for CO_{2}(g) and subtract from it (or add a negative) 2 times the heat of formation of CO(g).
2C(graph) + 2O_{2}(g) = 2CO_{2}(g) DH = 2(393.5 kJ/mol) 2CO(g) = 2C(graph) + O_{2}(g) DH = 2(+110.52 kJ/mol) 2CO(g) + O_{2}(g) = 2CO_{2}(g) DH_{r} = 565.96 kJ/mol
We can generalize this procedure in equation form as:
DH_{r} = Sn_{i}DH_{f,i }(Calculating heats of reaction from heats of formation)where, as the student will perhaps recall, n_{i} is the stoichiometric coefficient for that compound in the reaction (don't forget the appropriate sign). Using this equation, we could immediately write down the heat of reaction without having to do the labor of combining the equations. Thus, for the example above,
DH_{r} = +2(393.5)  2(110.52)  1(0) = 565.96 kJ/molwhere the heat of formation of O_{2}(g) is of course zero since it is an element in its normal state of aggregation.
2. From Heats of Combustion
Definition: The heat of combustion is defined as the heat of reaction when a compound is completely burned with O_{2} to form specific combustion products; namely, CO_{2}(g), H_{2}O(l), SO_{2}(g), Cl_{2}(g), N_{2}(g), etc., at 25^{o}C and 1 atm.Note that in order for the heat of reaction to be the heat of combustion, water must be liquid. All of the other elements must end up in the specific compounds specified above. All of the carbon must end up as CO_{2}, all of the sulfur must end up as SO_{2}, etc. That doesn't mean that when you burn something containing C that it will all be converted to CO_{2}, but if it isn't then the enthalpy change is not the standard heat of combustion.
As an example of the use of heats of combustion, consider how we might obtain the heat of reaction for the conversion of methane to propane:
3CH_{4}(g) = C_{3}H_{8}(g) + 2H_{2}(g)using the following combustion reactions:
CH_{4}(g) + 2O_{2}(g) = CO_{2}(g) + 2H_{2}O(l) DH_{C} = 890.36 kJ/molWe seek a way to add the above three equations in such a way as to produce the desired reaction. It is obvious from the stoichiometry of the desired reaction that we must multiply the first equation by 3 and then subtract the second equation. Next, we can figure out how to apply the third equation in order to cancel out the remaining oxygen molecules. Doing this for both the equations and the heats of combustion gives us:
C_{3}H_{8}(g) + 5O_{2}(g) = 3CO_{2}(g) + 4H_{2}O(l) DH_{C} = 2220.0 kJ/mol
H_{2}(g) + ½O_{2}(g) = H_{2}O(l) DH_{C} = 285.84 kJ/mol
Notice that 3 times the first equation minus the second equation leaves 1O_{2}(g) on the reactant side, so we multiply the third equation by 2 and add it to cancel out the oxygen and give the result shown at the bottom of the table. All of the water molecules also cancel out when this is done.
3CH_{4}(g) + 6O_{2}(g) = 3CO_{2}(g) + 6H_{2}O(l) DH = 3(890.36 kJ/mol) 3CO_{2}(g) + 4H_{2}O(l) = C_{3}H_{8}(g) + 5O_{2}(g) DH = (2220.0 kJ/mol) 2H_{2}O(l) = 2H_{2}(g) + O_{2}(g) DH = 2(285.84 kJ/mol) 3CH_{4}(g) = C_{3}H_{8}(g) + 2H_{2}(g) DH = 120.6 kJ/mol)
We can generalize this procedure in equation form as:
DH_{r} = Sn_{i}DH_{c,i }(Calculating heats of reaction from heats of combustion)Notice the negative sign. This means that we write the heats of combustion for the reactants minus the products, whereas we took products minus reactants when using heats of formation. This is because the combustion reaction uses the desired compound as a reactant whereas a formation reaction produces the desired compound as a product. Thus, for the example above, we could simply have performed the arithmetic, without writing the reactions, by using the heats of combustion for each of the terms in the green line at the bottom of the table; i.e.,
DH_{r} = +3(890.36)  (2220.0)  2(285.84) = 120.6 kJ/mol
2. From Other Heats of Reaction
The same procedures illustrated above can be used to combine any reactions
to form another reaction. One simply uses Hess's Law to combine the reactions
and then uses the same arithmetic to combine the heats of reaction.
The exercises below will give you practice in obtaining heats of reaction
from the above three methods.
Practice Exercise on Obtaining Heats of Reaction
Use the enthalpies for the reactions and processes given in the table at the left to answer the questions on the right.
