|Enthalpy and Internal Energy||Enthalpies of Reaction||Standard State Enthalpies|
|Hess's Law||Enthalpies of Formation||Bond Dissociation Enthalpies|
Enthalpy and Internal Energy
By definition, the enthalpy of the system is the sum of the internal energy of the system plus the product of the pressure of the gas in the system times its volume.
The change in the enthalpy of the system (H ) that occurs during a reaction is the enthalpy of the final state minus the initial state of the system.
H = Hfinal - Hinitial
When this equation is applied to a chemical reaction, the final state corresponds to the products of the reaction and the initial state of the system is the reactants. The change in the enthalpy of the system as the reactants are converted into the products of the reaction is therefore known as the enthalpy of reaction.
The relationships between heat, the internal energy of the system, and the enthalpy of the system during a chemical reaction can be summarized as follows.
E = qV
H = qp
H =E + (PV)
|Practice Problem 5:
For which of the following reactions is H about the same as E?
(a) CaCO3(s) CaO(s) + CO2(g)
(b) 2 NH3(g) N2(g) + 3 H2(g)
(c) Fe2O3(s) + 2 Al(s) Al2O3(s) + 2 Fe(l)
(d) CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
Enthalpies of Reaction
The enthalpy of reaction is the difference between the sum of the enthalpies of the products of the reaction and the sum of the enthalpies of the starting materials.
H = Hproducts - Hreactants
At constant pressure, when a reaction gives off heat to its surroundings, the enthalpy of the system decreases. Because the sum of the enthalpies of the products is smaller than the sum of the enthalpies of the reactants, exothermic reactions are characterized by negative values of H.
Exothermic Reactions: H is negative (H < 0)
Endothermic reactions, on the other hand, take in heat from their surroundings. As a result, the enthalpy of the system increases. Endothermic reactions are therefore characterized by positive values of H.
Endothermic Reactions: H is positive (H > 0)
There are several ways the enthalpy of reaction information can be added to the balanced equation for the reaction.
One approach assumes that the balanced equation is written in terms of moles. For example, the reaction between sodium and chlorine to form sodium chloride gives off 411.15 kilojoules of energy per mole of NaCl formed. When two moles of sodium react with a mole of chlorine two moles of sodium chloride are formed. Thus, a total of 822.30 kilojoules of energy is released.
2 Na(s) + Cl2(g) 2 NaCl(s)
H = -822.30 kJ
Another approach reports values of the enthalpy of reaction per mole of one of the reactants or products. This approach is indicated as follows.
2 Na(s) + Cl2(g)
H = -411.15 kJ/mol NaCl
Most endothermic reactions have to be driven by some external force, much as work has to be done to roll a boulder uphill. An example of this phenomenon is the electrolysis of molten sodium chloride.
|2 NaCl(l)||2 Na(s) + Cl2(g)|
A handful of endothermic reactions are spontaneous. One example of a spontaneous endothermic reaction is the basis of a commercial product, an ice pack that doesn't have to be kept in the freezer. These ice packs contain a small quantity of ammonium nitrate (NH4NO3) or ammonium chloride (NH4Cl), which is separated from a sample of water by a thin membrane. When the pack is struck with the palm of the hand, the membrane is broken, and the salt dissolves in the water.
|NH4NO3(s) + H2O||NH4+(aq) + NO3-(aq)||Hrxn = 25.7 kJ/mol|
Because the reaction is endothermic, it absorbs heat from its surroundings, and the ice pack can get cold enough to treat minor athletic injuries.
|Practice Problem 6:
Use your experience with ice, water, and steam to predict which of the following reactions are exothermic and which are endothermic.
(a) H2O(s) H2O(l)
(b) H2O(l) H2O(g)
(c) H2O(g) H2O(l)
Reversing the direction in which a reaction is written cannot change the magnitude of the enthalpy of reaction, only the sign of H.
H373 = 40.88 kJ/mol
Reversing the direction in which the above reaction is written changes the sign of H because the initial and final states of the system have been reversed.
H373 = -40.88 kJ/mol
Standard-State Enthalpies Of Reaction
The heat given off or absorbed by a chemical reaction depends on the conditions of the reaction. Three factors are particularly important: (1) the concentrations of the reactants and products involved in the reaction, (2) the temperature of the system, and (3) the partial pressure of any gases involved in the reaction.
The combustion of methane can be used to illustrate the magnitude of the problem.
Assume that we start with a mixture of CH4 and O2 in which the partial pressure of each gas is 1 atm and the temperature of the system is 25oC. Furthermore assume that we run the following reaction and then let the products cool to 25oC.
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
Under these conditions, the reaction gives off a total of 802.4 kilojoules of energy per mole of CH4 consumed. If we start with the reactants at 1000oC and 1 atm pressure, however, and return the products to the same conditions, the reaction gives off only 792.4 kJ/mol. The difference between these numbers is small (10.0 kJ/mol), but it is still 100 times larger than the experimental error (ħ0.1 kJ/mol) with which the measurements were made.
The effect of pressure and concentration on thermodynamic data is controlled by defining a set of standard conditions for thermodynamic experiments. By definition, the standard state for thermodynamic measurements fulfills the following requirements.
Measurements done under standard-state conditions are indicated by adding a superscript "o" to the symbol of the quantity being reported. The standard-state enthalpy of reaction for the combustion of natural gas at 25oC, for example, would be reported as follows: Ho = -802.4 kJ/mol CH4.
Measurements taken at other temperatures are identified by adding a subscript specifying the temperature in kelvin. The data collected for the combustion of methane at 1000oC, for example, would be reported as follows: H1273 = 792.4 kJ/mol.
The enthalpy of a system can be defined in terms in terms of the internal energy, pressure, and volume of the gas in the system.
H = E + PV
Because the internal energy, pressure, and volume of a gas are all state functions, the enthalpy of a system is also a state function. As a result, the difference between the initial and final values of the enthalpy of a system does not depend on the path used to go from one of these states to the other.
Hess's law states that the enthalpy of reaction (H) is the same regardless of whether a reaction occurs in one step or in several steps. We can therefore calculate the enthalpy of reaction by adding the enthalpies associated with a series of hypothetical steps into which the reaction can be broken.
|Practice Problem 7:
The standard-state molar enthalpies of reaction for the formation of water as both a liquid and a gas have been measured.
Use these data and Hess's law to calculate Ho for the following reaction.
|Practice Problem 8:
Before pipelines were built to deliver natural gas, individual towns and cities contained plants that produced a fuel known as town gas by passing steam over red-hot charcoal.
C(s) + H2O(g) CO(g) + H2(g)
Calculate Ho for this reaction from the following information.
Enthalpies of Formation
By definition, Hof is the enthalpy associated with the reaction that forms a compound from its elements in their most thermodynamically stable states.
|Practice Problem 9:
Which of the following equations describes a reaction for which Ho is equal to the enthalpy of formation of a compound, Hof?
(a) 2 Mg(s) 2 MgO(s) s) + O2(g)
(b) MgO(s) + CO2(g) MgCO3(s)
(c) Mg(s) + C(s) + 3/2 O2(g) MgCO3(s)
Hess's law can be used to calculate the enthalpy of reaction for a chemical reaction from the enthalpies of formation of the reactants and products of the reaction.
|Practice Problem 10:
Use Hess's law to calculate Ho for the reaction
from the following enthalpy of formation data.
No matter how complex the reaction, the procedure used in the above example works. All we have to do as the reaction becomes more complex is add more intermediate steps.
We obtained the answer to this exercise by adding the enthalpy of formation of each of the products and subtracting the enthalpy of formation of each of the reactants. In general, the enthalpy of reaction for any chemical reaction is equal to the difference between the sum of the enthalpies of formation of the products and the sum of the enthalpies of formation of the reactants.
Ho = Hof products - Hof reactants
This formula works because enthalpy is a state function. Thus, Ho is the same regardless of the path used to get from the starting materials to the products of the reaction.In the second step, these elements combine to form the products of the reaction.
|Practice Problem 11:
Which of the following substances should have a standard-state enthalpy of formation equal to zero?
|Practice Problem 12:
Pentaborane(9), B5H9, was once studied as a potential rocket fuel. Calculate the heat given off when a mole of B5H9 reacts with excess oxygen according to the following equation.
2 B5H9(g) + 12 O2(g) 5 B2O3(s) + 9 H2O(g)
As we have seen, Ho can be calculated with the following formula when the enthalpy of formation is known for all of the reactants and products of a chemical reaction.
Ho = Hof products - Hof reactants
Only a limited number of enthalpies of formation have been measured, and there are many reactions for which Hof data is not available for one or more reagent. When this happens, Ho for the reaction can not be predicted. The enthalpy of reaction can be estimated using bond-dissociation enthalpies. By definition, the bond-dissociation enthalpy for an X-Y bond is the enthalpy of the gas-phase reaction in which this bond is broken to give isolated X and Y atoms.
XY(g) X(g) + Y(g)
The bond-dissociation enthalpy for a C-H bond can be calculated by combining Hof data to give a net equation in which the only thing that happens is the breaking of C-H bonds in the gas phase.
|CH4(g) C(s) + 2 H2(g)||Ho = 1 mol x 74.81 kJ/mol CH4|
|C(s) C(g)||Ho = 1 mol x 716.68 kJ/mol C|
|2 H2(g) 4 H(g)||Ho = 4 mol x 217.65 kJ/mol H|
|CH4(g) C(g) + 4 H(g)||Ho = 1662.09 kJ|
If it takes 1662 kJ/mol to break the four moles of C-H bonds in a mole of CH4, the average bond-dissociation enthalpy for a single C-H bond is about 415 kJ/mol.
Bond-dissociation enthalpies are always positive numbers because it takes energy to break a bond. When a table of bond energies is used to estimate the enthalpy associated with the formation of a bond, the sign becomes negative because energy is released when bonds are formed.
|Practice Problem 13:
Use bond-dissociation enthalpies to estimate Ho for the gas-phase reaction between hydrogen and nitrogen to form ammonia.
N2(g) + 3 H2(g) 2 NH3(g)
Assume that N2 molecules are held together by bonds.
Bond-dissociation enthalpies can only provide an estimate of the value of Ho because they are based on estimates of the strength of an average bond.