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ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Thermodynamics > State Functions

State Functions

A state function is a property whose value does not depend on the path taken to reach that specific value. State functions are important as they are often encountered in thermodynamics.

Introduction

Whenever compounds or describe chemical reactions are discussed, one of the first things mentioned is the state of that specific molecule or compound. State is referring to temperature, pressure, and the amount and type of substance present. Once the substance's state has been established, one can define state functions. State functions are values that depend on the state of the substance, and not on how that state was reached. For example, density is a state function because a substance's density is not affected by how the substance is obtained. Lets say we have a certain amount H2O, it does not matter whether that H2O is obtained from one's tap, from a well, or from a bottle, because as long as all three have the same states, they will have the same density. When deciding whether a certain property is a state function or not, keep this rule in mind: is this property or value affected by the path or way taken to establish it? If the answer is no, then there is a state function, but if the answer is yes, then there is no state function.

Mathematics of State Functions

When talking about state functions, another way to think of them is as integrals. State functions can be thought of as integrals because integrals depend on only three things: the function, the lower limit and the upper limit. Similarly, state functions as well depend on three things: the property, the initial value, and the final value. In other words, integrals help illustrate how state functions depend only on the final and initial value and not on the object's history or the path taken to get from the initial to the final value.

Here, I have provided an example of the integral of enthalpy H, where t0 represents the initial state and t1 represents the final state.

\[ \displaystyle \int_{t_o}^{t_1} \; H(t) dt = H(t_1)-H(t_o) \]

In essence, we have devised enthalpy as we know it:

\[ \Delta H = H_{final} - H_{initial}\]

As represented by the solution to the integral, enthalpy is a state function because it only depends on the initial and final conditions and not on the path taken to establish these conditions. Thus in essence, you can take the integral of state functions using only 2 values: the final and initial value. On the other hand, you would need multiple integrals and multiple limits of integration just to be able to take the integral of one property that is a path function. So if you know that you can take an integral of a certain property using just the property and it's initial and final value, chances are that the property would be a state function.

State Functions vs. Path Functions

The main way we define state functions is by comparing it to path functions. As stated before, a state function is a property whose value does not depend on the path taken to reach that specific function or value. In essence, what we are saying is that if something is not a path function, it is probably a state function. In order to have a better understanding of state functions, lets first define path functions and then compare path and state functions.

Path functions are functions that depend on the path taken to reach that specific value. For example, lets imagine you have $1000 in your savings account. Let's say you want to deposit some money to this account. The amount you deposit is a path function because it is dependent upon the path taken to obtain that money. In other words, the amount of money you will deposit in your savings account is dependent upon the path or way taken to obtain that money. If you work as a CEO of a company for a week versus working at a gas station for a week, you would result in two different amounts of money at the end of the week. Thus, a path function is a property or value that is dependent on the path taken to establish that value.  

State functions on the other do not depend on the path taken at all. Let's use the same example that you had $1000 in your savings account. Now let's say you withdrew $500 from your savings account. It does not matter whether you withdraw the $500 in one shot or whether you do so at a rate of $50. At the end when you receive your monthly statement, you will notice a net withdrawal of $500 and will see your resulting balance as $500. Thus, the bank balance would be a state function because it does not depend on the path or way taken to withdraw or deposit money. In the end whether you do so in one lump or in multiple transactions, your bank balance will stay the same.

Below, I have made a figure that illustrates state functions in the form of enthalpy:

Enthalpy Diagram.png

In this figure, two different steps are shown to form NaCl(s). The first path takes only one step: Na+(g) + Cl-(g) = NaCl(s), which has an enthalpy of formation of -411 kJ/mol. The second path takes 5 steps to form NaCl(s) and when we add the enthalpy of formation of all these steps, the enthalpy of formation of NaCl(s) still ends up as -411 kJ/mol. This is the epitome of state functions as we have just stated that no matter what path we take to form NaCl(s), we still end up with the same enthalpy of formation of -411 kJ/mol.

So once again, lets make a table to review the differences between a state and path function:

State Function Path Function
Independent of path taken to establish property or value. Dependent on path taken to establish property or value.
Can integrate using final and initial values. Need multiple integrals and limits of integration in order to integrate.
Multiple steps result in same value. Multiple steps result in different value.
Based on established state of system (temperature, pressure, amount, and identity of system). Based on how state of system was established.
Normally represented by an uppercase letter.1 Normally represented by a lowercase letter.1

1The last comparison made on the table is one that I have observed through the calculations I have done so far. It does not necessarily hold for all aspects and calculations involved in chemistry.

Analogy

The main point to remember when trying to determine a state function is asking yourself whether the path taken to reach the function affects the value. Here I present an analogy that will hopefully make it clear how to tell whether a certain property is a state function or not.

Every morning, thousands and millions of people are faced with the dilemma of how to reach their office. Some opt for taking the stairs as others take the elevator. In this situation, ∆y, or change in vertical position is the same whether a person take the stairs or the elevator. The distance from the office lobby to one's office stays the same, irrespective of the path taken to get to your office. As a result, ∆y is a state function because it's value is independent of the path taken to establish its value. 

In the same situation described above, time, or ∆t, would not be a state function. If someone takes the longer way of getting to your office (climbing the stairs), ∆t would be greater, while ∆t would be smaller if you take the quicker way of getting to your office (taking the elevator). In this analogy, ∆t is not a state function because it is dependent on the path taken to establish its value. Once again, the golden rule is to ask if this value affected by the path or way taken to establish it? If the answer is no, then there is a state function, and if the answer is yes, there is no state function.

 

Example 1

Given the following data, prove that \(w\) is not a state function:

A gas has a pressure of 2.4 atm at a volume of 1.36 L. This gas is expanded, due to which its pressure decreases from 2.4 atm to 1.8 atm to 1.2 atm.

SOLUTION

U and H are state functions, but w and q are not. It is fairly easy for one to list what quantities are state functions and which are not. Another way of telling whether a certain property is a state function or not is calculating its value in two given situations where the only difference between the two is the path taken to establish the value. If the value stays the same in both situations, then one knows they have a state function. If it does not, then one know that the property is not a state function.

The way we can prove that w is not a state function is by showing that the amount of work is different when a different path is taken. In other words, if we can show that the work is different when the gas is expanded in 2 steps from 2.4 atm to 1.2 atm than when it is expanded in 3 steps from 2.4 atm to 1.8 atm to 1.2 atm, we have proven that work is not a state function.

In the first step, calculate the work when expanding the gas in two steps from 2.4 atm to 1.2 atm:

To calculate the volume of the gas at 1.2 atm, I will use Boyle's equation (P1V1 = P2V2)

\[P_1V_1 = P_2V_2\]

\[(2.4)(1.36) = (1.2)(x)\]

\[3.264 = (1.2)(x)\]

\[x = \dfrac{3.264}{1.2} = 2.72\, L \]  

Now calculate the work for this 2 step expansion as both volumes are given:

\[w = -p\Delta{V}\]

\[w = -(1.2 L)(2.72\, atm - 1.36\, atm) = -1.632\,L atm\]

Now calculate the work when the gas is expanded in 3 steps:

\[w_{total}= w_{1.8} + w_{1.2}\]

Use Boyle's equation again to calculate the volume of the gas when the \(P_1=2.4\, atm\), \(V_1=1.36\, L\), and \(P_2=1.8 atm\)

Once finished with all the calculations, one should end up with V=1.81 L.

w1.8= -p∆V = -1.8(1.81-1.36) = -0.81 L atm

w1.2= -p∆V = -1.2 (2.72-1.81) = -1.092 L atm

wtotal= w1.8 + w1.2  = -0.81 - 1.092 = -1.902 L atm

When a 2 step expansion is used, the work = -1.632 L atm. When a 3 step expansion is used the work = -1.902 L atm. Thus, we have just established that work is a property whose value is affected by the path taken to calculate work and as a result, we just proved that work is not a state function.

Applications

State functions are commonly encountered in thermodynamics as many of the equations involved with thermodynamics, such as \(\Delta U\) and \(\Delta H\), are state functions. Additionally, state functions are crucial in thermodynamics because they make calculations simple and easy and allow one to calculate data that could otherwise only obtain through experiments.

More specifically, state functions really help with Hess's Law, which allows us to manipulate (add, subtract, multiply etc.) the enthalpies of half reactions when we are adding multiple half reactions to form a full reaction. Hess's Law is dependent upon the fact that enthalpy is a state function, which allows us to manipulate the enthalpies of half reactions. If enthalpy was not a state function, Hess's Law would be a lot more complicated as we would not simply be able to add the enthalpies of half reactions. Instead, we would have to go through several additional calculations, that too if even possible, before we took the sum of the enthalpies of the half reactions. Furthermore, state functions and Hess's Law helps one calculate the enthalpy of complex reactions without having to actually replicate these reactions in a laboratory. All we have to do is write out and sum the enthalpy of the half reactions or of the hypothetical steps leading to the chemical reaction. State functions are also encountered in many other equations involved with Thermodynamics such as internal energy (∆U), Gibb's free energy, enthalpy, and entropy.

Problems

  1. In terms of what we have discussed in this module, is going from the 1st floor of Sproul hall, to the 9th floor of Sproul hall, the same thing as going from the 1st floor of Sproul hall, to the 3rd floor, to the 5th floor, to the 9th floor of Sproul hall?
  2. Is ∆U a state function?
  3. Is temperature a state function?
  4. Is volume a state function? (prove with an example)
  5. Although pressure and volume are state functions, why is work (which is often expressed as -P∆V) not a state function?

Solutions

  1. Yes it is because the question describes a state function. Your position is dependent only on the final and initial position, which are respectively 9th floor of Sproul and 1st floor of Sproul, and not on the path or way taken to get there.
  2. The formula for ∆U is, ∆U = Ufinal- Uinitial . The formula of ∆U itself proves that it is a state function because ∆U is only dependent on Ufinal and Uinitial . In other words, ∆U is not affected by the path taken to establish its values. This is the definition of a state function and as a result, ∆U is a state function.
  3. Temperature is a state function as it is one of the values used to define the state of an object. Furthermore, temperature is dependent on the final and initial values, not on the path taken to establish the values.
  4. Volume is a state function because volume is only dependent on the final and initial values and not on the path taken to establish those values. Any example that shows this statement in function is acceptable. Here's an acceptable answer: Imagine a balloon is inflated until a certain volume. If inflated in multiple steps or in a single step, it will still attain the same volume at the end. As a result, volume is a state function because it is not dependent on the object's path or history.
  5. The reason work is a state function depends on the definition of work rather than the formula of work. The definition of work is moving an object against a force. Thus is essence, the definition of work states that work depends on its history or path it takes because the movement of an object is dependent upon the path taken to execute that movement (i.e. running vs. walking). Therefore, if an object is dependent on its history or on the path it takes, the resulting value or property is not a state function. Even though pressure and volume are state functions, the definition of work illustrates why work is not a state function.

References

  1. Petrucci, Ralph H., Harwood, William S., Herring, F. G., and Madura Jeffrey D. General Chemistry: Principles & Modern Applications. 9th Ed. New Jersey: Pearson Education, Inc., 2007. Print.
  2. Kotz, John C., Treichel, Paul M., and Townsend, John. Chemistry & Chemical Reactivity. 7th Ed. Belmont: Thomson Higher Education, 2006. Print.

Contributors

  • Allison Billings (UCD), Rachel Morris (UCD), Ryan Starr (UCD), Angad Oberoi (UCD)

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Last Modified
21:34, 29 Jun 2014

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