Lecture 21

 Hess's Law, State Functions, and Enthalpies of Reaction

A. Hess's Law of Heat Summations

Hess's Law combines the law of conservation of energy and the law of conservation of mass. It states: If a chemical reaction (or process) can be expressed as a sum of separate reactions, then the enthalpy change for the overall process is the sum of the enthalpy changes for the constituent reactions.

Let's see how this applies to a reaction for which the enthalpy change is experimentally unmeasureable: the production of gaseous methane by the reaction of carbon (as graphite) with hydrogen gas:

C (s, graphite) + 2 H₂ (g) ---> CH₄ (g)

(NOTE 1: the reasons that the enthalpy change for this reaction can't be measured experimentally are first, that the reaction is too slow, and second, millions of compounds--the class of organic compounds called the hydrocarbons--could be produced on the way to methane.)

(NOTE 2: remember that carbon has more than one form that is stable in the solid state. Diamond and graphite are the two best known, but buckminsterfullerenes and nanotubes add to the list. Different solid forms of a given element are called allotropes -- refer to Chapter 2 in Kotz and Treichel.)

There are other reactions whose enthalpy changes can be easily determined, and if summed properly, the chemical equations add to give the reaction above. A set of reactions which meet this specification are:

Chemical Equation
1. C(s) + O2 (g) ---> CO2 (g)	-393.5
2. H2 (g) + 1/2 O2 (g) ---> H2O (l)	-285.8
3. CH4 (g) + 2 O2 (g) ---> CO2 (g) + 2 H2O (l)	-890.3

In order to decide how to use these, it is important to remember that we want methane on the right-hand side, and graphite and hydrogen gas on the left-hand side. We also want all reactants and products other than C (s), H₂ (g), and CH₄ (g) to cancel. It is clear that we will have to reverse the direction of Equation 3 to get methane as a product, and that we will need to take double the amounts in Equation 2 to cancel the water in 1.

Chemical Equation
reverse 3: CO2 (g) + 2 H2O (l) ---> CH4 (g) + 2 O2 (g) 1: C(s) + O2 (g) ---> CO2 (g) double 2: 2 H2 (g) + O2 (g) ---> 2 H2O (l)	890.3 -393.5 -571.6
Overall: add the above reactions and cancel like terms on both sides of the chemical equation to get the desired equation. C(s) + 2 H2 (g) ---> CH4 (g)	Add the above enthalpy changes to get the overall enthalpy change for the desired reaction: -74.8 kJ

Now we have determined that the enthalpy change for the reaction, C(s) + 2 H₂ (g) ---> CH₄ (g), is -74.8 kJ

This process can be shown graphically. From equations 1 and 2 above, we find that a mole of carbon, two moles of hydrogen, and two moles of oxygen combine to form a mole of carbon dioxide and two moles of water to release 965.1 kJ of energy. Then we imagine that the carbon dioxide and water thus formed react to produce a mole of methane and two moles of oxygen, absorbing 890.3 kJ of energy when they do so. The net energy change must be the difference between these two processes, or 74.8 kJ of energy released when methane is formed from the elements carbon and hydrogen.

B. State Functions

A state function is a quantity whose value is determined only by the state of the system. State functions are path-independent: you can't tell how you got there by looking at where you are. Examples: You can't tell from looking at someone's bank balance how it was achieved. You can't tell by looking at a blown-up balloon whether it was blown up slowly or rapidly. For state functions, only the initial and final values are important, not the process used to get from the initial to the final state.

The enthalpy change (Delta H) and the overall energy change (Delta E) are examples of state functions.

A practical example of state functions. Remember that changes in the internal energy E are due to changes in the heat evolved (or absorbed) and to work done by (or on) the system.

Now consider the burning of gasoline, represented as the compound octane, C₈H₁₈:

2 C₈H₁₈ + 25 O₂ ---> 16 CO₂ + 18 H₂O

In the first case, we can burn the gasoline out in the open, in which case almost all the energy from the reaction is released as heat (there is a very small work term due to the volume change and the PV work). Or we can burn the gasoline in an automobile engine and harness a significant amount of the energy produced as mechanical energy, that is used to turn the wheels of the car and propel it from one destination to another.

The total energy change is divided differently in these two cases: the work terms w have different values, and the heat terms q have different values. Thus q and w are NOT state functions, even though the internal energy and enthalpy changes ARE state functions.

 

C. Standard Enthalpies of Formation

When a reaction occurs with all reactants and products in their standard states under standard conditions, the enthalpy change is called the standard enthalpy change for the reaction.

The standard state of an element or compound is the most stable form of the element or compound in the physical state in which it exists at 1 bar (100 kPa; 101.3 kPa = 1 atm) and a specific temperature, usually 25^o C.

The standard molar enthalpy of formation is the enthalpy change that results when one mole of the compound is produced directly from its elements, all in their standard states.

Some compounds and their standard molar enthalpies of formation are shown in the table below. The reactions may have fractional coefficients for the elements: the key is to take just enough to produce one mole of the product.

Compound name	Formation reaction
Gaseous water, H2O (g)	H2 (g) + 1/2 O2 (g) ---> H2O (g)	-241.8
Liquid water, H2O (l)	H2 (g) + 1/2 O2 (g) ---> H2O (l)	-285.8
Hydrogen sulfide, H2S (g)	1/8 S8 (s, rhombic) + H2 (g) ---> H2S (g)	-20.63
Hydrazine, N2H4 (l)	N2 (g) + 2 H2 (g) ---> N2H4 (l)	50.63

When most chemical compounds form from their elements, energy is released. The formation reaction is thus exothermic, and the sign of the standard enthalpy of formation is negative. However, there are some compounds whose formation is endothermic. Hydrazine is such an example: hydrazine is unstable with respect to decomposition back to its elements.

NOTE CAREFULLY: The units of kJ are understood to be "kJ/mol of product" because of the way the standard enthalpy of formation is defined.

 

D. Calculating Enthalpies of Reaction From Standard Enthalpies of Formation

Let's consider one of the early rocket fuels. Liquid hydrazine was stored in one set of fuel tanks, nitric acid in others. These two compounds react vigorously with each other, producing large volumes of gases (nitrogen and steam) along with much energy. The overall reaction is

5 N₂H₄ (l) + 4 HNO₃ (l) ---> 7 N₂ (g) + 12 H₂O (g)

We can imagine this reaction occurring in a series of 3 steps. These 3 steps sum to the overall reaction in the last line of the table, with an energy enthalpy that is equal to the sum of the enthalpy changes for the three steps.

Reactions
1. Hydrazine decomposes to its elements. 5 N2H4 ---> 5 N2 + 10 H2
2. Nitric acid decomposes to its elements. 4 HNO3 ---> 2 H2 + 2 N2 + 6 O2
3. The hydrogen and oxygen produced in the previous reactions combine to form water. 12 H2 + 6 O2 ---> 12 H2O
Sum of the above three equations: 5 N2H4 + 4 HNO3 ---> 7 N2 + 12 H2O

This set of steps can also be shown graphically:

If you followed this reasoning carefully, you will realize that the enthalpy change for the reaction was calculated from the standard enthalpies of formation for each of the compounds:

This leads to a general method for calculating the standard enthalpy change for any reaction:

 

 

This page is http://chemiris.labs.brocku.ca/~chemweb/courses/chem180/CHEM1P80_Lecture_21.html
Created October 30, 2000 by M. F. Richardson
© Brock University, 2000