Chapter 9. Thermochemistry: Chemical Energy

Energy and Its Conservation, First Law of Thermodynamics

Energy—ability to do work or to supply heat. Mathematically,

E = q + w

where q = heat and w = work.

Heat—Thermal energy transferred from one object to another.

Another way of defining energy is by

E = T + V

where T = kinetic energy and V is the potential energy. The kinetic energy is the energy of motion for a particle with mass m, and has $$T=\frac{1}{2}m{v}^2$$ where v = velocity of particle. The potential energy is related to position of that particle. The functional form of potential energy differs in different systems. For example, we have seen in Chapter 7 that interaction energies between particles in H₂ molecule is given by Coulomb interaction. The distances, hence position, between two charges give the strength of the interaction.

Energy conservation: The figure below explains the relationship between energy (total energy), kinetic energy, and potential energy. Let's say that you place a ball in a symmetrical bowl-like environment without friction. The ball has a mass, m. Since we are dealing with rolling a ball, we are dealing with the potential energy in gravity which has dependence on mass and the height of the ball. The potential energy is then given by

V = V(h) = mgh

where g is the gravitational force on earth, 9.81 m/s², and h is the height of the ball compared at the bottom of the well (h = 0). Energy unit is in Joules, J = 1 kg m²/s². When you let the ball go, the position of where you put the ball, labeled with h₀ = 1, the kinetic energy is zero at the instance of letting the ball go. The ball starts to accerelate until it passes the bottom of the well, labeled h₁ = 0, at which the potential energy is zero, but kinetic energy is maximum, because of the pull of the gravity. The ball then starts to slow down until it reaches at the same height as h₀ on the opposite side, labeled h₂.

At each point of passage, the total energy is exactly the same.

at h₀, T = 0, E = V(h₀),

at h₁, V = 0, E = T(h₁),

at h₀, T = 0, E = V(h₂),

at any other position h, T ≠ 0, V ≠ 0, E = T + V

Internal Energy and State Function

Thermodynamic system: An isolated system, i.e. the system of interest such as moleucules being surrounded by surrounding, and the surrounding is enclosed by an insulated wall so that no heat, no work, no molecules escape from the surrounding. The system can exchange heat and/or work to the surrounding.

The energy of the system, internal energy which consists of molecular motions and interaction between them, can change so that the final energy can be higher or lower than its initial energy. When the final energy is higher than the initial, then the energy flows into the system from the surrounding. On the other hand, if the final energy is lower, the energy flows out of the system to the surrounding. These are depicted above.

State Function—if the quantity is independent of path, we call it state function. Energy is a state function. So, when calculating the overall energy change in the system, all you need to do, is to calculate the difference between the final energy and initial energy,

Δ E = E_final - E_initial

Expansion Work

Work: In general work is force times distance. In this chapter, we only consider work involving expansion work. Under ordinary laboratory conditions, atmospheric pressure, p, is constant. So, when expansion work is performed by the system, work done by the system is given by the following equation,

w = -p ΔV

where ΔV is the change in volume, and p is the pressure. Pressure is measured in atmosphere, atm, and the volume is measured in L with if pΔV = 1 atm L = 101 J.

Since ΔV is defined as V_final - V_initial, if the gas is expanded, meaning that V_final > V_initial, the sign of the work is negative. In this case, it is said that the work is done by the system. If, however, the gas is compressed, then w is positive.

Energy and Enthalpy

We already saw that ΔE = q - pΔV. Since we can measure heat, by observing the temperatrue change, we can rearrange the equation and solve for q. Then, we get,

q = ΔE + pΔV

If there is no work, that is no change in volume, q = ΔE.

If the pressure is constant, then q = ΔE + pΔV. We define this heat measured at constant pressure to be enthalpy change, ΔH.

Enthalpy is state function, like energy. So, enthalpy change is not depended on the path. When the enthalpy of final state is higher than the initial state, it is said to be endothermic and one must inject heat into the system to achieve the final state. On the other hand, if the final state is lower in enthalpy, it is said to be exothermic where heat is generated (or liberated) by the system.

Thermodynamic Standard State

It is convenient for us to use standardized thermodynamic quantity, so that when reporting values can be compared directly.

Thermodynamic standard state—most stabel form of a substance at 1 atm pressure and at specified temperature, usually 25°C. If solution is involved, the concentration is 1 M.

We denote enthalpy change measure at the standard state to be ΔH° ← notice a knot!

Enthalpy of Physical and Chemical Change

Enthalpy change associated with physical change Melting process (fusion), and boiling (vaporization) needs extra heat to melt or vaporize. For H₂),

H₂O (s) → H₂O (l)       ΔH_fus = 6.01 kJ/mol @ 0°C

and

H₂O (l) → H₂O (g)       ΔH_vap = 40.7 kJ/mol @ 0°C

Then, by the virtue of enthalpy being state function, sublimation process is

H₂O (s) → H₂O (g)       ΔH_sub = ΔH_fus + ΔH_vap = 46.7 kJ/mol.

Enthalpy change associated with chemical change Of course, we are now talking about heat generated or absorbed by the reaction. We call the heat, heat of reaction. We write down the heat of reaction at the end of chemical equation, for example, combustion of propane (C₃H₈)

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l)           ΔH° = -2220 kJ

The chemical equation that includes thermodynamic data, such as heat of reaction, is called thermochemical equation. In the above example, we have standard heat of reaction, and the reaction is exothermic.

If you reverse the reaction, that is,

3 CO₂(g) + 4 H₂O(l) → C₃H₈(g) + 5 O₂(g)           ΔH° = +2220 kJ

the reaction becomes endothermic.

If on the other hand, the reaction is doubled the above,

6 CO₂(g) + 8 H₂O(l) → 2 C₃H₈(g) + 10 O₂(g)           ΔH° = +4440 kJ

If halved, the value of the ΔH° must be multiplied by ½.

Calorimetry and Heat Capacity

Heat Capacity Since heat, q, has to do with temperature change, heat and temperature change must be proportional to each other, i.e.

q ∝ ΔT

where ΔT = T_final - T_initial. If you want an equation, instead of proportionality, we can put a proportionality constant, C, such that,

q = C ΔT

and C is called heat capacity. Heat capacity tells how much energy required by a substance to raise its temperature by 1°C. It has a unit of J/°C.

We can make heat capacity dependent on mass, it means that,

q = m c_sp ΔT

where m is the mass, and c_sp is specific heat. Specific heat is defined as energy required by 1 g of substance to raise its temperature by 1°C. Its unit is J/(°C g).

Furthermore, we can make depended on number of moles, as well. In such a case,

q = n c_m ΔT

where n is the number of moles, and c_m is the molar heat capacity, which is defined as energy required to by 1 mol of substance to raise its temperature by 1°C. Its unit is J/(°C mol).

Using these, we can measure heat.

Calorimeter—a device to measure heat. Some devices are rather elaborate, but the priciples are the same for any calorimeter. So, we illustrate a calorimetry experiment using a styrofoam coffee cup, as follows.

The styrofoam cup acts as insulating walls in the description of isolated system at the beginning of the chapter. We need to obtain temperature readings, so a thermometer is necessary, and in order to keep the temperature of water to be consistent throughout the cup, a stirring rod is provided. A calorimeter usually use water as the surrounding to the sample chamber. For simple experiments, such as the ones we perform in our labs, it doesn't have to have the sample chamber.

Specific heat of water is given as 4.182 J/(°C g).

In elaborate experiments, we can not use specific heat of water as the specific heat of surrounding, since the wall might absorb some of the heat and it might change the value. The specific heat of the calorimeter must be independently determined. For us in this course, we use the specific heat of water as the specific heat of calorimeter.

If the system releases heat to the surrounding, the surrounding receives the same amount of heat. So the amount of heat transferred is exactly the same magnitude. However, in this case, system is "pushing" the heat, and the surrounding is "pulling" the heat. This can be considered as opposite effect. In terms of mathematics, we denote this "push-pull" system to be,

q_sys = -q_cal

where q_sys is the heat from/to the system, and q_cal is the heat to/from the calorimeter (surrounding).

If we use the specific heat equation for q, then one can expand the equation above as,

m_sysc_sys ΔT_sys = -m_calc_cal ΔT_cal

The ΔT_sys on the left-hand side of the equation, is given by T_final - T_initial of the system, while ΔT_sys on the right-hand side of the equation, is given by T_final - T_initial of the calorimeter or the surrounding. In either case, T_final is common to both the system and the surrounding.

If the system is exothermic, then the heat is pushed into the surrounding. Hence the temperature of the surrounding becomes higher, compared to the initial temperature. On the other hand, if the system is endothermic, the heat is extracted from the surrounding and provided to the system. It means that the final temperature of the surrounding must become lower than the initial temperature. These situations are shown in the figure to the right.

Hess's Law

When multiple stpes are involved in the reaction, the sum of all enthalpy changes in different stpes give the overall enthalpy of reaction.

For example, we will look at the heat of reaction of the following.

Mg(s) + ½O₂(g) → MgO(s)         ΔH° = ? Manesium burns at very high temperature and give off ultraviolet rays. So, we can not make you to do the experiment to measure the heat of reaction directly by burning it.

Instead, we can do two separate experiments that are amenable to our course. We carry out the heat measurements of the following two reactions. The experiment and how to obtain ΔH_rxn is illustrated in the example below. Before we move to the example, few tips when combining chemical equations.

When adding two equations We want the following equation.

2A + 2B → 3D       ΔH = ?

The two steps involved in this reaction, and they are:

A + B → 3C       ΔH₁

D → 2C           ΔH₂

In order to have D on the right-hand side in the overall equation, we must flip the equation around, and make 2C → D. When equation is flipped, you're reversing the flow of energy (heat), so the sign of the ΔH must also be flipped. Therefore, the enthalpy of reaction 2 is now -ΔH₂.

In addition, there is no C in the overall equation, therefore, we must have the same number of C in both reactions. It means that we need to multiply 2 to the first reaction, and 3 to the second. Then, we have,

2A + 2B → 6C       2ΔH₁

6C → 3D           -3ΔH₂

---

overall     2A + 2B → 3D     ΔH = 2ΔH₁ - 3ΔH₂

Here is an example of experiment we actually do in our lab.

Hess's law can be illustrated in terms of energy diagram. For a reaction,

Overall     CH₃CH₂OH(l) + O₂(g) → CH₃CO₂H(l) + H₂O(l)         ΔH°_rxn = ???

We have the reaction in two steps:`

Rxn 1     CH₃CH₂OH(l) + ½O₂(g) → CH₃CHO(g) + H₂O         ΔH° = -174.2 kJ

Rxn 2     CH₃CHO(g) + ½O₂(g) → CH₃CO₂H(l)             ΔH° = -318.4 kJ

Standard Heats of Formation

Formation Reaction—reaction to make a compound from its stable elements. The reactants are the elements and the product is the moleucle of interest. For example, the formation reaction for water is,

H₂(g) + ½O₂ → H₂O(l)

as making liquid water from its elements, hydrogen molecule and oxygen molecule. The enthalpy associated with the formation reaction is called, heat of formation, ΔH_f . As we have seen already,

H₂(g) + ½O₂ → H₂O(l)         ΔH_f(H₂O) = -285.8 kJ/mol

If we measure the heat of formation at standard state, then the heat of formation is standard heat of formtion, ΔH°_f.

Heats of Formation of Elements are zero! So, H₂, O₂, C(graphite), Cu, Fe, Cl₂ have all zero for its formation enthalpy.

Standard Reaction Enthalpy. What's useful about the heats of formation, is the fact that one can obtain heat of reaction, by the following for the reaction,

aA + bB → cC + dD

where small letters, a, b, c, and d are the stoichiometry of the balanced chemical equation. Then,

$$\Delta H_{rxn}^{\circ }=\; c\Delta H_f^{\circ }(C) \; +\; d\Delta H_f^{\circ }(D) \; -\; a\Delta H_f^{\circ }(A) \; -\; b\Delta H_f^{\circ }(B)$$ This is quite useful because you don't have to know anything about the reaction to know the enthalpy of the reaction, simply by looking up on a table of ΔH°_f (In our case, look at Appendix B in our text).

Bond Dissociation Energies

We have looked at the energy profile of a diatomic molecule, the last picture of this section. We discussed that when making a chemical bond from two atoms, energy is released. When a chemical bond is to break, one must input energy into the molecule. The energy required to break a bond into two fragments of molecule is called bond energy.

The bond energies can be useful in estimating the reaction enthalpy. Afterall, we are only rearranging chemical bonds during chemical reaction. So, for example, two ethane (C₂H₆) react to make butane, C₄H₁₀, and H₂, as shown below.

We can estimate the reaction enthalpy by adding the average bond energies of bonds being broken and formed. Remember that bond breaking is endothermic and bond formation is exothermic. In this example, the bonds in red color are being broken or formed. Two C—H bonds are broken and one C—C and one H—H bonds are formed. The average bond energies are given in Table 1 of Chapter 7, but here I reproduce it below.

Using this table, we have,

Type	#affected bonds	Energy	Total
Reactants
C—H	2	410 x 2	820
total of reactants			820
Product
C—C	1	350 x 1	-350
H—H	1	436 x 1	-436
total of product			-786
ΔHestimated = 34 kJ/mol(8 kcal/mol) Actual = 42 kJ/mol(10 kcal/mol)

In this example, the estimated enthalpy of reaction is 34 kJ/mol, while calculating it from the heats of formation yields 42 kJ/mol. The difference between the estimated by the average bond dissociation energy and by heats of formation is quite small. We can do the similar series of hydrocarbons by breaking two C—H bonds and forming one C—C bond and one H—H bond. With our estimation, all should come out to be 34 kJ/mol, is this value correct for different reactions? Here is the answer,

Ethane (C₂H₆) + Propane (C₃H₈) → Pentane (C₅H₁₂) + H₂         ΔH° = 42 kJ/mol

Propane (C₃H₈) + Propane (C₃H₈) → Hexane (C₆H₁₄) + H₂         ΔH° = 43 kJ/mol

Propane (C₃H₈) + Butane (C₄H₁₀) → Heptane (C₇H₁₄) + H₂         ΔH° = 43 kJ/mol

Butane (C₄H₁₀) + Butane (C₄H₁₀) → Heptane (C₈H₁₈) + H₂         ΔH° = 43 kJ/mol

So, as long as the reaction of this type or similar, one can estimate the reaction enthalpy by using the average bond energies.

If the reaction is more ivolved, this estimate just doesn't work. For example, in the following example the estimation breaks down.

In this case we break two C—Hbonds and one H—Cl bond, and form one C—H bond and one C—Cl bond. In this case the estimation fails.

Using the average bond energy, we obtain ΔH_estimate = +70 kJ/mol, while calculating it from the heats of formation, it is -69 kJ/mol. Therefore, it is off by nearly 140 kJ/mol and has wrong sign for the reaction enthalpy! In this case, the reaction is not just simply breaking bonds and forming the similar bonds; we have change in hybridization of both carbons from sp² to sp³.

Entropy

Spontaneity. Our experience tells us that heat never seems to flow from cold to hot object in a natural way; it is always one direction, from hot to cold. Heat can not be coveyed from a cold object to hot object spontaneously. For example, if you have two pieces of metal with one piece heated to high temperature (metal 1 in the figure), and the second piece originally at room temperature (metal 2). It means that temperature of metal 1 is higher than that of metal 2. In this case, direction of the heat flow is to the right. This natural tendency for heat to flow in one direction is the statement of 2^nd law of thermodynamics .

Some chemical reactions are spontaneous. For example, ice melts at room temperature, despite it must overcome the heat of fusion, ΔH_fus > 0. Or when you turn your iPod on, electricity flows through the system immediately. This is due to that fact that red-ox reaction takes place in the battery to supply the electrons needed by the device.

Also from our experience that for a gas in a balloon the situation on right is much more likely than the spontaneous organization of gas molecules into a corner of the balloon at the same temperature! (Of course, if you lower the temperature, the gas molecules would condense into a liquid.

As far as energy is concerned, we have talked about energy has tendencey to flow toward lower energy side, since lower energy means the state is more stable. ....

Free Energy

Spontaneous process occurs when the Gibbs free energy is lower in the product, compared to the reactant. Gibbs free energy combines the enthalpy and entropy contributions in the spontaneous processes. For a chemical reaction, the Gibbs free energy change is given as,

ΔG = ΔH - TΔS

Just as enthalpy and entropy are state functions, Gibbs free energy is also a state function. As shown in the figure below, spontaneity only comes when the ΔG < 0.

If the ΔG > 0, the reaction is non-spontaneous. If the Δ G = 0, the reaction is said to be in equilibrium where the rates of forward reaction is the same as that of the reverse direction. In such a reaction we use a double-arrow sign, ⇄ in chemical equation to indicate there is an equilibrium. Do you remember how we represent weak acids? For acetic acid, we wrote,

H₃CCO₂H ⇄ H⁺ + H₃CCO₂^-

In order to determine the spontaneity of chemical process, there are three factors: 1) ΔH, 2) &DeltaS, and 3) T. Following table summarizes.

Signs of
ΔH	ΔS	T	Spontaneity
+	-	high	spontaneous
+	-	low	non-spontaneous
+	+	high	non-spontaneous
+	+	low	non-spontaneous
-	-	high	non-spontaneous
-	-	low	spontaneous
-	+	high	spontaneous
-	+	low	spontaneous