Enthalpy

### Bringing heat and work together

We've seen how James Prescott Joule showed the equivalence of heat and work. He showed that the simple act of stirring a solution is enough to heat it, and that if we carefully measure the kinetic energy of that stirring, it will show up as a change in temperature of the solution. All of the energy of stirring winds up as heat added to the system. (Actually, there is always some unavoidable loss, but we'll learn later to account for that.)

Heat is just an expression of the kinetic energy of the atoms and molecules that make up a system. Jiggling of atoms and molecules is heat, and it doesn't matter how we accomplish faster jiggling – by heating or by stirring, it's all the same.

When we speak of heat, we really have to speak of work, too. The way we do this in chemistry is to invent a new kind of function, a state function, called the internal energy (U). Once we have the internal energy in hand, we'll derive a more useful state function, Enthalpy (enth' · uhl · pee), and use it to answer some important questions about chemistry. With his apparatus, Joule showed that dropping weights to stir water imparts a known amount of kinetic energy to the stirring paddles, and thus to the water, to heat the water by that amounf of KE. The KE of a weight drop is equal to the amount of work required to raise the weight and to the amount of PE the weight has at the top.

### Internal energy (U)

Internal energy will be a function that will depend only on the present state of our system, whatever that is. It doesn't depend on how the system got that way. Thus we call it a state function.

We define internal energy as heat added to, plus work done on a system: U = q + w. Remember our sign conventions from the thermodynamics intro section:

positive heat and work are heat and work added to our system (see the box below).

Now we have a function that tracks both heat and work together. Notice that with internal energy, we allow for a system to be endothermic – to absorb heat, but to also do so much work that its total energy content, measured as internal energy, might not change much or at all.

We also allow for a system to be exothermic – to lose heat to the surroundings – but to have so much work done on it that its internal energy stays more or less the same. Internal energy is a much finer instrument for us to monitor what happens in a chemical process because it tracks both heat and work.

#### Internal energy

Internal energy is heat (q) added to the system plus work (w) done on the system.

#### Sign convention in thermodynamics

Positive heat is heat added to the system.

Positive work is work done on the system.

### Pressure-volume (PV) work and working at constant pressure

Now work done by a chemical system is almost always pressure-volume work. A system heats up and expands, thus doing work by compressing the surroundings (negative work – it's work done by the system), or it cools down and is compressed by the surrounding atmosphere, or it evolves gas so that it might be able to do the work needed to expand a balloon, and so on.

The units of work are Joules, and 1J = 1 Kgm2/s2. If we multiply pressure units (N/m2 = Kg·m/s2) by volume units (m3), we get Joules, so the product PV is work in joules.

So we can write

$$U = q + w$$

and

$$U = q + PV$$

What we really want when we're talking about internal energy is not the amount present in a system at any given time, but the change in internal enery, ΔU. We write that like this:

$$\Delta U = \Delta q + \Delta(PV)$$

Now most of the work we do in the laboratory occurs at constat pressure, which means ΔP = 0. And when we add heat to a system (Δq > 0), it generally expands, which means that Δw < 0, so we can rewrite ΔU for the constant pressure situation like this:

$$\Delta U = \Delta q - P \Delta V$$

Here we've made sure to get the signs right: Heat is absorbed so Δq is positive and work is done by the system, so PΔV is negative.

#### The pressure-volume product (PV) has units of energy (usually Joules).

The change in internal energy, ΔU, of a system going from state 1 to state 2 at constant pressure is

$$\Delta U = \Delta q - P \Delta V$$

or →

$$\Delta U = (q_2 - q_1) - P(V_2 - V_1)$$

So that's our first state function. We'll come back to it in time as our study of thermodynamics unfolds, but for now, we'll make a slight modification to find a new state function that will prove to be more useful.

### A new state function: Enthalpy (H)

Let's do something that will seem a little contrived and first, and add the term PΔV back to the internal energy. Obviously that just cancels the -PΔV that was already there:

We call this new state function change the Enthalpy (H), and the change in enthalpy is ΔH:

$$\Delta H = \Delta U + P \Delta V$$

Plugging in for ΔU and PΔV, we can show that this is just the change in heat when the pressure is constant.

\require{cancel} \begin{align} \Delta U + P \Delta V &= \Delta q - \cancel{P \Delta V} + \cancel{P\Delta V} \\[5pt] &= \Delta q \end{align}

So ΔH is just the heat added to a system at constant pressure:

$$\Delta H = \Delta q$$

← heat added to the system at constant pressure.

Notice that again, we don't really care how much heat a system has, but rather how much is added to it (Δq > 0) in some change like a reaction, or how much is liberated from it (Δq < 0). We're interested in ΔH.

The change in Enthalpy (ΔH) is the amount of heat added to a system at constant pressure, for example, to an open beaker.

### Enthalpy is a state function

Enthalpy is one of a number of quantities you'll encounter in chemistry and thermodynamics which are called state functions. Others are entropy and free energy.

A state function is a quantity that depends only on the initial and final states of the system of interest, but not on how it got from one to the other.

The figure below shows two hypothetical states of a system, the gray boxes labeled reactants and products. Enthalpy is measured on the vertical scale, so the enthalpy of the products, in this case, is lower than that of the reactants. The lines represent pathways from one state to another. Getting from reactant to product can take a straight-forward or path or a tortured one, but the resulting ΔH is the same. ### How enthalpy is used: Thermochemistry

What we really need is a way to determine the enthalpy change for any given reaction. Take the model double-displacement reaction, .

(we're just using fake elements for now). We can write synthesis reactions for each of the components, AB, CD, AD and BC, like this:

#### A   +   B   →   AB,

and so on.

We can write the synthesis equationes for AB and CD as decomposition reactions to obtain these four elementary reactions: Now we can just add up all of the components on both sides of these equations, and cancel – just as we would do in an algebraic equation – to recover our original equation (below).

It turns out that we can write any reaction as a sum of these so-called elementary steps, usually synthesis or decomposition reactions. The general idea is called Hess's law (see box below). #### Hess's Law

Hess's law of constant heat summation says that the enthalpy change, ΔH, during a chemical reaction is not dependent on how that reaction occurs – that it is not dependent on the path of the reaction.

The conseqence of Hess's law is that we can write any reaction as a summation of synthesis and decomposition steps, and add up the enthalpy changes of those steps to calculate the enthalpy change of the reaction.

Standard enthalpies of formation (synthesis), ΔHof are tabulated for many compounds.

### Example 1

Consider the combustion of solid sulfur to sulfur-trioxide (SO3):

$$2 \, S_{(s)} + 3 \, O_{2 \: (g)} \: \longrightarrow \: 2 \, SO_{3 \: (g)}$$

We'd like to know the total enthalpy change for this reaction, but it's difficult to measure. However, we do have measurements of the enthalpy changes for combustion of solid sulfur to form SO2 and for the combustion of SO2 to form the trioxide. Those equations are

\begin{align} S_{(s)} + O_{2 \: (g)} \: &\longrightarrow \: SO_{2 \: (g)} \\[5pt] 2 \, SO_{2 \: (g)} + O_{2 \: (g)} \: &\longrightarrow \: 2 \, SO_{3 \: (g)} \end{align}

and the enthalpy changes are ΔHf = -296.0 KJ and ΔHf = -198.2 KJ, respectively. Note that the negative signs mean that these reactions are exothermic – they give off heat.

Now we can prepare to add these two steps to get the overall reaction:

\begin{align} S_{(s)} + O_{2 \: (g)} \: &\longrightarrow \: SO_{2 \: (g)} \phantom{00000} \Delta H_o^f = -296.0 \, KJ \\[5pt] 2 \, SO_{2 \: (g)} + O_{2 \: (g)} \: &\longrightarrow \: 2 \, SO_{3 \: (g)} \phantom{0000} \Delta H_o^f = -198.2 \, KJ \end{align}

But notice that SO2 does not appear in our overall equation, so it must cancel when we add the elementary steps. In order for that to happen, we'll have to multiply the first equation through by two, including the enthalpy.

Here's how that looks; the multiplication and cancellations are done: Finally, notice that this enthalpy change is for the formation of two moles of SO3. It is customary to give enthalpy changes in terms of energy (usually Joules) per mole of product.

All we have to do is divide by 2 to get our final enthalpy, ΔHf = 391.5 KJ/mol or 391.5 KJ·mol-1.

### A step further: Standard enthalpies of formation

#### Standard enthalpy of formation, ΔHf

The standard enthalpy of formation of a compound is the heat absorbed (positive) or emitted (negative) when one mole of the compound is made from its elements in their standard states.

The ΔHf of elements in their standard states is arbitrarily set to zero. This is OK because we're only interested in changes in enthalpy, ΔH.

### Example 2

Calculate the enthalpy change of combusting 1 mole of ethanol, C2H5OH.

Solution: First, we have to recall that combustio means combination with oxygen to produce only carbon dioxide (CO2) and water H2O. That's only the case when we combust hydrocarbons or oxy-hydrocarbons. The balanced reaction for the reaction is: We will use Hess's law and the reactions of formation of each of the reactants and products to calculate the total enthalpy change for this reaction.

The formation reaction, with enthalpy change, for ethanol from elements in their standard states (solid carbon, gaseous hydrogen and oxygen) is: CO2 is formed from solid carbon and gaseous oxygen in this reaction and water is formed from H2 and O2 like this: O2 is a diatomic gas at standard temperature and pressure (STP), therefore its standard heat of formation is zero.

These standard heats of formation can be looked up in a table. You can download one here.

Now our goal will be to get these reactions to add to our combustion reaction using Hess's law. In order to do that, we'll need to flip one of them around the → sign, and we'll have to multiply some of the ΔH's by the coefficients in the balanced combustion reaction to account for the fact that we're working with more than one mole of some compounds. Here is a reminder of the combustion reaction and those coefficients. Here are the adjustments to the formation reactions. First, we'll have to flip the ethanol formation reaction, which means that the enthalpy of formation (now an enthalpy of decomposition) will switch signs. Then we'll multiply the CO2 formation reaction by the coefficient of CO2 in the combustion reaction and the water formation reaction by 3 for the same reason.

Here is the result, showing how all three reactions add to the combustion reaction. Look carefully at the cancellations and convince yourself that this procedure works. The last step is just to add all of the ΔH's to get the enthalpy change of the reaction. The burning of ethyl alcohol (ethanol) is quite exothermic. For that reason, ethanol is often used as a fuel for heating and cooking. The cans of Sterno, for example, which are used for heating trays of food at buffet tables, can contain thickened or "jellied" ethanol.

Ethanol is also the main ingredient in alcohols for consumption, like vodka and rum. When it's found in products like Sterno, it is denatured – other alcohols are added which cause temporary stomach sickness when consumed. Image: Sterno

### Example 3

Calculate the enthalpy change for the reaction
2NH3 + 2CH4   →   2HCN + 6H2O

Solution: The overall reaction is The formation reactions and standard enthalpies of formation of each compound are: Here again, we don't need to worry about O2; the ΔHfo of pure elements in their standard-state forms is zero.

Because ammonia (NH3) and methane (CH4) appear on the reactant side of the equation, we'll have to reverse those formation equations to use Hess's law to add the elementary equations and arrive at our target equation.

Also, most standard enthalpies of formation are scaled for one mole of product, as these are, so don't forget to multiply by the coefficient(s) occuring in the balanced equation. In this case that's 2 ammonias, 2 methanes and two moles of hydrogen cyanide (HCN).

Writing the equations and cancelling what's common to both sides, as in the previous example and in this one, can help you to do that bookkeeping. The result below shows that this reaction is exothermic.

Once you get some practice using Hess's law and standard enthalpies of formation, you probably won't need to write the equations any more. I recommend doing so until that comes naturally, though. You can download a table of standard enthalpies of formation of a select group of compounds here. If you need others, they can usually be found in on-line sources. Wikipedia is a particularly good source. Be aware that different tables will have slightly different enthalpies, mainly due to the way the enthalpies are measured, but they'll be close enough.

### Practice problems

Using Hess's law and the elementary reactions and enthalpies given, calculate the enthalpy change for these reactions:

 1.   2 C(s) + H2 (g) → C2H2 (g) C2H2 (g) + (5/2)O2 (g) → 2 CO2 (g) + H2O (l) ΔH = -1299.5 KJ/mol C (s) + O2 (g) → CO2 (g) ΔHfo = -393.5 KJ/mol H2 (g) + ½O2 (g) → H2O (l) ΔHfo = -285.8 KJ/mol 2. C2H4 (g) + 6 F2 (g) → 2 CF4 (g) + 4 HF(g) H2 (g) + F2 (g) → 2 HF(g) ΔH = -537 KJ/mol C(s) + 2 F2 (g) → CF4 (g) ΔH = -680 KJ/mol 2 C(s) + 2 H2 (g) → C2H4 (g) ΔH = 52 KJ/mol 3. HO(g) + Cl2 (g) → HOCl(g) + Cl(g) Cl2 (g) → 2 Cl(g) ΔH = 242 KJ/mol H2O2 (g) → 2 HO(g) ΔH = 134 KJ/mol H2O2 (g) + 2 Cl(g) → 2 HOCl(g) ΔH = -210 KJ/mol

Using standard enthalpies of formation (download the table for those), calculate the enthalpy change for these reactions:

 4 C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O(g) 5 SO2 (g) + 2 HCl(g) → H2O(g) + SOCl2 (g) 6 H2SO4 (aq) + BaI2 (aq) → BaSO4 (s) + 2 HI (aq) 7 C12H22O11 (s) + 12 O2 (g) → 12 CO2 (g) + 11 H2O(l)

### Video examples

1. Finding the enthalpy of a neutralization reaction using standard enthalpies of formation and Hess's law

Minutes of your life: 2:28

2. Finding the enthalpy of a combustion reaction using standard enthalpies of formation and Hess's law

Minutes of your life: 2:32

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### State

To know the "state" of a system means to know enough about it that we then can at least calculate all of the other properties that could be measured. It's having all the information there is. For thermodynamic systems, we usually mean pressure, temperature, volume and number of moles of the component(s).

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### Endothermic

A system that is endothermic absorbs heat fromt he surroundings. The melting of ice is endothermic: Heat flows from the surroungings into the ice.

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### Exothermic

A system that is exothermic releases heat to the surroundings. For example, when sodium hydroxide (NaOH) is dissolved in water, the solution heats up.

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