# Enthalpy

### Written by tutor Uneeb Q.

Enthalpy in thermodynamics is called a state function meaning it does not matter how it arrived at its final value (i.e. whether you took \$10 dollars at a time to pay a loan or \$100 at a time, the amount you paid at the end was still the same, assuming no interest charges). Such a distinction is more important in higher level mathematics, but on a basic level it shows you that everything contains a certain amount of enthalpy, no matter how you use it. Enthalpy on a basic level however can be described as the transfer of potential energy, in the form of heat, in a system (chemical bonds as used in chemistry).

Mathematically the thermodynamic term used to describe enthalpy is H, as seen in formulas like Gibb’s Free energy where G = H – TS . In these cases it is important to distinguish the units used so that it is easier to obtain via dimensional analysis. The units of molar enthalpy are kiloJoules per mole (kJ/mol), recall that kilo is the metric prefix for 1000 and Joules is the SI Unit of energy. We need to use thousands of Joules in our unit because chemical reactions can yield a lot of energy sometimes. Also recall that moles are a unit used to describe a quantity of molecules.

Sometimes you might see a degree (°) symbol next to H, this implies Standard heat of formation. Standard heat of formation is defined as the enthalpy when one mole of a compound is formed from free elements at 1 atm and 25°C. These conditions are typical for reactions, thus 1 atm and 25°C is called standard thermodynamic conditions.

Finally it is important to remember phase designations in balanced chemical equations. For instance in the equation below, letters are written in parentheses.

2 Na (s) + 2 H2O (l) –> 2 NaOH (aq) + H2 (g)

In the equation, lowercase s in parentheses implies a solid while lowercase g implies a gas. Liquids are written with the lowercase letter of L either in cursive or in italics (this is done to not confuse the readers with the letter uppercase of i). Last but not least aq implies aqueous (dissolved in water) this is of no concern in thermodynamics as we do not measure aqueous values of thermodynamics. With this knowledge in mind one can now decipher a typical set of values for thermodynamics, the tables below show reference data that every chemistry student should be able to read and interpret. Keep in mind that there are distinctions even between values of H, for instance Hf is defined as the heat of formation of something. This will be important later. For example in the equation above H2 is a gas that forms in the reaction so one would look up its heat of formation to find its enthalpy.

For Liquids

 Molecular formula Name ΔfH° kJ/mol ΔfG° kJ/mol S° J/mol K Cp J/mol K C2H6Zn Dimethyl Zinc 23.4 201.6 129.2 C2H7N Dimethylamine -43.9 70.0 182.3 137.7 C2H6S2 Dimethyl disulfide -62.6 235.4 146.1 C2H6S Dimethyl sulfide -65.3 196.4 118.1 C2H6O4S Dimethyl sulfate -735.5

For Gases

 Molecular formula Name ΔfH° kJ/mol ΔfG° kJ/mol S° J/mol K Cp J/mol K C2H6Zn Dimethyl Zinc 53.0 C2H7N Dimethylamine -18.8 68.5 273.1 70.7 C2H6S2 Dimethyl disulfide -27.4 C2H6S Dimethyl sulfide -37.4 286.0 74.1 C2H6O4S Dimethyl sulfate -687.0

As you can see different thermodynamic values for these chemicals, different enthalpies are needed for different phases of matter. For now we are only focusing on the column containing our ΔH values. Recall that capital delta (Δ) is used to define the overall or net change. While the values may appear small, the amount of energy released by some reactions is quite large and could possibly burn you. It is sometimes important to understand the conversion between kJ and J, as some questions may ask for an answer in one unit or the other.

Question: What are the enthalpy values for C2H6O4S (l) and for C2H6O4S (g)? How much difference is there between both states in terms of J? In kJ?

The enthalpy value for liquid Dimethyl sulfate (C2H6O4S) is -735.5 kJ/mol
The enthalpy value for gaseous Dimethyl sulfate (C2H6O4S) is -687.0 kJ/mol
The difference in enthalpy between both states is |(-735.5)-(-687.0)| = 48.5 kJ/mol
In Joules this difference is 48.5 kJ/mol * 1000 J/kJ = 48,500 J/mol

Now that we understand how to use the information given to us, now we can apply it to sample problems. The form of the equation below follows for any thermodynamics problem involving chemical reactions;

ΔHrxn=ΣΔHproducts – ΣΔHreactants

Recall that sigma (Σ) is used for the mathematical operation of sums, and that products are found in the right side of the chemical equation (or the head of the arrow) and reactants are found at the left (or the tail of the arrow). For example

Reactants –> Products

A hypothetical double replacement (displacement) reaction can be written as follows

CuO (s) + 2 HCl (aq) –> CuCl2 (aq) + H2O (l)

In the example above we can obtain enthalpy values for these chemicals using a reference table similar to the one above. I have found the following; ΔHf (H2O) = -241.8 kJ/mol, ΔHf (HCl) = -92.3 kJ/mol, ΔHf (CuO) = -155.2 kJ/mol, ΔHf (CuCl2) = 220.1 kJ/mol. Therefore the enthalpy of our reaction above can be found by

ΔHrxn=ΣΔHproducts – ΣΔHreactants

ΔHrxn=(ΔHH2O + ΔHCuCl2) – (ΔHCuO + (2*ΔHHCl)

(Not that there is a 2 in the balanced equation in front of HCl, this must be accounted)

ΔHrxn=(-21.7) – (29.4) = -51.1 kJ/mol

Not bad huh? That really is all you have to do mathematically, add your product and reactant enthalpies and subtract the initial reactant enthalpies from final products of the reaction. Now let’s investigate a case where the math won’t make complete sense.

2 H2 (g) + O2 (g) –> 2 H2O (l)

In the equation above we can calculate the enthalpy of this reaction similarly to before. Again let’s assume that the number of moles are the same as written in the balanced equation above. We begin our solution with

ΔHrxn=ΣΔHproducts – ΣΔHreactants

ΔHrxn=ΣΔHH2O – Σ(ΔHO2 + ΔHH2)

Now this is where things get hairy. Using values obtained from a textbook we find the enthalpy of formation of water is approximately -241.8 kJ/mol, but the molar enthalpies of oxygen and hydrogen gas are 0 kJ/mol. Now wait, why are some things zero? Well in thermodynamics a naturally occurring (free) element always has an enthalpy of zero. It’s in its natural state, a state in which nothing has been done to it, so since we can’t exactly “create” or “form” a naturally occurring element (besides nuclear reactions) it has no enthalpy of formation (Hf).

Recall the diatomic elements (which are the naturally occurring forms of some gases). In particular they are hydrogen (H2), nitrogen (N2), oxygen (O2), and group 7 halogens (F2, Cl2, Br2, etc). All of which have enthalpies of zero.

Despite these facts lets go back to our problem above and solve it.

ΔHrxn=ΣΔHH2O – Σ(ΔHO2 + ΔHH2)

ΔHrxn= (-241.8) – (0 + 0) = -241.8

In fact it is these types of experiments, combining free elements to create a compound, that allow us to determine the enthalpy values in the first place. So don’t freak out if your solutions have no enthalpies in places, this can and will happen. Often these types of enthalpies are used as trick questions on exams or homework.

Question: Determine the enthalpy of the following hypothetical reaction

Cu (s) + CO (g) –> CuO (s) + C (s)

Constants: ΔHCuO = -155.2 kJ/mol, ΔHCO = -110.5 kJ/mol

Answer: First set up your equations, remember free elements have zero enthalpy!

ΔHrxn=Σ(ΔHC + ΔHCuO) – Σ(ΔHCO + ΔHCu)

ΔHrxn= (0 + (-155.2)) – ((-110.5) + 0) = -44.7 kJ/mol

So now we have some examples of enthalpies of reactions. So what do these negative and positives have to do with anything? Enthalpies that are negative are considered exothermic (exo – outside, thermic – heat) meaning heat is LEAVING the system. Enthalpies that are considered endothermic (endo – inner) meaning that heat is ENTERING the system (adding to it).

More generally, reactions that give off heat are considered exothermic (negative enthalpy, -H), and reactions that get cold and need heat to run are endothermic (positive enthalpy, +H). Therefore a negative enthalpy or heat of reaction (-H) does not mean that the reaction is cold, it is telling you that the chemicals are LOSING their internal heat as the reaction continues to the environment (i.e. you). Conversely, a positive enthalpy (+H) is considered a cold reaction because the chemicals NEED TO GAIN HEAT energy from the environment (i.e. a bunsen burner) in order to continue and will feel cold to the touch.

Question: Is the reaction in the previous equation endothermic or exothermic?

Answer: It is exothermic, because the enthalpy of reaction is negative.

Using what we know about enthalpy, in terms of endo and exothermicity, we can then extend our analysis of chemical nature by asserting that creating bonds stores chemical potential energy as heat (endothermicity). And therefore breaking bonds releases energy as heat (exothermicity). This does not change our understanding of the mathematical nature but it can extend it to the bonds in a chemical itself. Bond Bond-Enthalpy kJ/mol Cl-Cl 242 C-Br 285 C-C 347-356 C-H 410 H-Cl 431 H-H 436 C-F 490

As you can see, this molecule has 2 types of bonds. It has a total of three C-H bonds and one C-Cl bonds. Let’s calculate the enthalpy that would result from breaking all of these bonds, say in a combustion reaction using the data in the table.

ΔH=Σ(3(C-H)) + 1(C-Cl))

ΔH=(3(410)) + 1(331))

ΔH=(1230 + 331) = 1561 kJ/mol

Because we are talking about breaking bonds (releasing stored energy ΔH is negative)

ΔH = -1561 kJ/mol

This reaction is EXOTHERMIC. Just for the sake of example, let’s look at a more complex question you may be asked to solve, first let’s utilize a balanced chemical equation

2 HCl (aq) –> Cl2 (g) + H2 (g)

Break this reaction into products and reactants based on bonds formed and broken
Products = Formed Cl-Cl bond + Formed H-H bond (both positive enthalpy)
Products = 2 H-Cl bonds broken (negative enthalpy)

ΔHrxn=ΔHproducts – ΔHreactants

ΔH= Σ[(Cl-Cl) + (H-H)] – (-2(H-Cl))

ΔH= [(242 + 436) – (-2(431))]

ΔHrxn= -183 kJ/mol

### Relation to Hess’ Law

The last enthalpy topic that deserves mention is a mathematical system described by Hess’ law. Recall that enthalpy is a state function, it does not matter how it obtains its final value. Because of this, knowing a series of reactions allows you to determine the enthalpy of a new reaction. For example,

CS2(l) + 2 O2(g) –> CO2 (g) + 2 SO2(g)

Let’s assume we want to solve for the above equation and it’s enthalpy value. But we are only given the following information.

C(s) + O2(g) –> CO2(g); ΔHf = -393.5 kJ/mol
S(s) + O2(g) –> SO2(g); ΔHf = -296.8 kJ/mol
C(s) + 2 S(s) –> CS2(l); ΔHf = 87.9 kJ/mol

While at first this appears intimidating, it is in fact just a matter of algebra. First try to rearrange the equation to mimic the one above, let’s first reverse the last equation

CS2(l) –> C(s) + 2 S(s); ΔHf = -87.9 kJ/mol

CS2(l) + 2 O2(g) –> CO2(g) + 2 SO2(g)

By doing this we have aligned CS2 as a reactant in both equations. By reversing a reaction we have reversed its enthalpy value as well (from 87.9 to -87.9 kJ/mol). Next let’s try to align the CO2 and SO2

C(s) + O2(g) –> CO2 (g); ΔHf= -393.5 kJ/mol
S(s) + O2(g) –> SO2(g); ΔHf= -296.8 kJ/mol
CS2(l) + 2 O2(g) –> CO2 + 2 SO2(g)

Notice how these products are already on the proper side of the equation we can leave them as is. Also notice how oxygens also align in this manner.

C(s) + O2(g) –> CO2(g); ΔHf= – 393.5 kJ/mol
S(s) + O2(g) –> SO2(g); ΔHf= -296.8 kJ/mol
CS2(l) + 2 O2(g) –> CO2 + 2 SO2(g)

In this manner, when we add these two reactions together we create this new equation

C(s) + O2(g) –> CO2(g); ΔHf= – 393.5 kJ/mol
+ S(s) + O2(g) –> SO2(g); ΔHf= -296.8 kJ/mol
——————————————————————
S(s) + C(s) + 2 O2(g) –> SO2(g) + CO2(g); ΔHf= -690.3 kJ/mol

Just like in algebra, when you added systems of equations you sum up similar values as written. However this is not our final answer, let’s compare what we want, with what we just created

S(s) + C(s) + 2 O2(g) –> SO2 + CO2(g); ΔHf= -690.3 kJ/mol
CS2(l) + 2 O2(g) –> CO2(g) + 2 SO2(g)

As you can see only a few parts are adding up correctly at this time, we need to balance this out or add/subtract a new system to get our answer. Hey…remember that one equation we reverse in the beginning to align variables? What happens if we combine them?

S(s) + C(s) + 2 O2(g) –> SO2(g) + CO2g; ΔHf= – 690.3 kJ/mol
+ CS2(l) –> C(s) + 2 S(s); ΔHf= -87.9 kJ/mol
————————————————————————-
S(s) + C(s) + 2 O2(g) + CS2(l) –> SO2(g) + CO2(g) + C(s) + 2 S(s); ΔHf= -778.2 kJ/mol

Bolded above are similar variables found on both sides of the equation. Recall from algebra that you can subtract a variable from one side so long as it had a similar variable on the other side. “What you do to one side, you must do to the other.” For example in the equation

3x – 4 = 2x

When you subtract 3x from both sides you get:

-4 = -x
x = 4

Apply this same math to this system of equations, except this time subtract chemicals.

S(s) + C(s) + 2 O2(g) + CS2(l) –> SO2(g) + CO2(g) + C(s) + 2 S(s); ΔHf= -778.2 kJ/mol

Note that this operation does not change the value of H, because it is equivalent whether you have it written like this or if you subtract similar variables (a la algebra). In the end you obtain your final equation (the one you were looking for)

CS2(l) + 2 O2(g) –> CO2(g) + 2 SO2(g)

And you determine that it has an enthalpy of -778.2 kJ/mol!

## Enthalpy Practice Quiz

Identify the true statement(s).

An endothermic reaction:
i. Releases heat to the surroundings
ii. Absorbs heat from the surroundings
iii. Has a negative value of DH
iv. Has a positive value of DH

A.
i only
B.
iii only
C.
ii and iv only
D.
All of the above are true
The correct answer here would be C.

Identify the true statement(s).

An exothermic reaction:
i. Has products higher than reactants in coordinate diagrams
ii. Has a negative value of DH
iii. Has products lower than reactants in coordinate diagrams
iv. Has a positive value of DH

A.
i only
B.
ii and iii only
C.
ii and iv only
D.
All of the above are true
The correct answer here would be B.

Identify the true statement(s).

An exothermic reaction:
i. Heat can flow into the system
ii. Heat can flow out of the system
iii. The system can do work or release energy to the surroundings
iv. The surroundings can do work or add energy to the system

A.
i only
B.
ii and iii only
C.
ii only
D.
All of the above are true
The correct answer here would be D.

A certain reaction has a positive value of H. Which of the following is true?

A.
Heat will be released by the reaction
B.
Heat will be absorbed by the reaction
The correct answer here would be B.

For the reaction 2 SO2(g) + O2(g) –> 2 SO3(g), which of the following is the correct expression for evaluating the reaction enthalpy?

A.
Hrxn= H(SO2(g)) + H(O2(g)) + H(SO3(g))
B.
Hrxn= H(SO2(g)) + H(O2(g)) + 2H(SO3(g))
C.
Hrxn= 2H(SO3(g)) – H(O2(g)) + 2H(SO2(g))
D.
Hrxn= 2H(SO3(g)) – H(O2(g)) – 2H(SO2(g))
The correct answer here would be D.

The enthalpy of formation of Na(s) is:

A.
Negative
B.
Positive
C.
Zero
D.
Unknown
The correct answer here would be C.

The superscript o in the term Ho means:

A.
The reaction is carried out at zero degrees Celsius
B.
The reaction is carried out at zero pressure
C.
The enthalpy of the reaction is zero
D.
The reaction is carried out at standard conditions
The correct answer here would be D.

The subscript f in the term Hf means:

A.
Fluorine
B.
Fairly big
C.
Fairly small
D.
Formation of the compound from the elements
The correct answer here would be D.
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