## Lecture 3

Today we'll figure out why you need salt to make ice cream and what the USDA is doing with so many bombs. But first a

### Review

Endothermic reactions:

Heat flows from the surroundings into the system, making q > 0 This heat flow occurs in response to a lowering of the Temperature of the system

K.E.products < K.E.reactants

P.E.products > P.E.reactants

The potential energy stored in the chemical bonds of the products is greater than the potential energy stored in the bonds of the reactants. The products have weaker bonds and are less stable than the reactants.

Exothermic reactions:

Heat flows from the system to the surroundings, making q < 0 . This heat flow occurs in response to an elevation of the Temperature of the system

K.E.products > K.E.reactants

P.E.products < P.E.reactants

The potential energy stored in the chemical bonds of the products is less than the potential energy stored in the bonds of the reactants. The products are more stable than the reactants (stronger bonds).

We can not measure heat flow, but we can measure Temperature differences (T2 - T1 = DT )

Temperature changes can be related to heat flow using the concept of heat capacity.

Heat Capacity, C, is the heat required to raise the Temperature of an object 1oC.

Adding heat to a compound increases the average Translational kinetic energy of the molecules in the system. For substances other than monatomic gases, adding heat also increases the internal vibrational and rotational energy of the molecules. The larger and more complex the molecule, the more possibilities there are for internal vibrations and rotation, and the more Energy is required to raise the Temperature of 1 mol of the substance 1oC.

Example 1.

How can we determine the specific heat capacity of a 50 g piece of unknown metal? We can place it in a container of boiling water (100oC), and leave it there until the temperature reaches 100o again, i.e. when they have reached thermal equilibrium. The piece of metal (100oC) is then placed in 1 L (1000 g) of water at room temperature (27o C). The temperature of the water rises until it reaches 27.8oC, the final temperature. Heat flowed between the metal and the water:

qmetal = - qwater

qwater = Cwater x DT x masswater = (4.184 J/oC g) (27.8 - 27.0 oC) x (1000 g)

qmetal = Cmetal x DT x massmetal = Cmetal( J/oC g) x (27.8 - 100.0 oC) x (50 g)

Setting qmetal = - qwater, we can solve for Cmetal, the specific heat capacity of the unknown metal. It is 0.93 J/o g. Looking back at the list of specific heat capacities tells us that the unknown metal must be Aluminum. Even without the list, we could have determined this. We know the specific heat capacity ( 0.93 J/oC g) of the metal and we also know that all metals have a molar heat capacity of about 25 J/oC mol. These two are related by the atomic mass of the metal.

(25J/oC mol) / (0.93 J/oC g) = 27.8 g / mol, the approximate atomic mass of Al

Now we can measure T and, using heat capacities, convert this to a measure of q, heat flow. How does this help us understand or quantify the thermodynamic properties of chemical reactions? Remember that q is a path function, which changes with experimental conditions. It would be much more useful to have a State function to describe heat flow.

Remember:

DE = q + w = q - PDV

which rearranges to give:

q = DE + PDV

There are only State Functions on the right, so a new State Function can be defined to replace q. It is given the symbol H, is called Enthalpy and is defined as:

H = E + PV

Expanding this expression gives:

DH = DE + DPV = DE + PDV + VDP

this isn't exactly like

q = DE + PDV

where q is a path function.

But, if we require that the experiment be carried out under conditions of constant P, DP = 0 and the two expressions become the same:

qp = DH = DE + PDV

and qp is now a State Function, DH. The subscript, P, indicates that this is heat measured under conditions of constant Pressure

The Enthalpy of a reaction is also called the Heat of Reaction and is measured under conditions of constant Pressure, i.e., in an open container. The most simple device for doing this is called a "coffee cup" calorimeter (calorie meter). This is illustrated on p.249 of Chemistry, by Zumdahl.. The following problem also comes from that text, problem 45, p. 281.

A calorimeter contains 125 g of water at 24oC. To this is added 10.5g of KBr (24oC). After the salt has dissolved, the temperature of the solution is 21oC. Calculate the DH of this reaction in J g-1 and kJ mol-1, assuming that the heat capacity of the solution is 4.18 J oC-1g-1.

The net reaction is:

KBr(s) → K+ (aq) + Br-(aq)

qreaction = -qsurroundings

The solution is the surroundings, and the dissolution of the salt is the reaction.

qsurroundings = Csurroundings x DT x masssurroundigs = (4.18 J KoC -1g-1)(21.1 - 24.2o C) (10.5 + 125 g) = -1756 J

In an insulated container,

qsurroundings = - qreaction

qreaction = + 1756 J

The "system" has gained heat, so it is an Endothermic Reaction. To calculate the heat of reaction per gram of KBr,

(1756 J) / (10.5 g KBr) = 167 J g-1 KBr

And to calculate the molar heat of reaction:

(167 J g-1) (39.1 + 79.9 g mol-1) = 19873 Jmol-1 = 1.9 x 101 kJ mol-1

The solution loses energy since heat flows out, lowering the Temperature of the solution.

This energy is absorbed by the KBr and is used to dissolve the salt.

Since this was carried under conditions of constant Pressure

qrxn = DHrxn

The dissolution of KBr is endothermic

This is why you need salt in when making ice cream. The DH for the process of dissolving NaCl is also positive. As the salt dissolves, heat flows from the water/ice/salt mixture, lowering its T. This, in turn, lowers the temperature of the cream in the mixer.

The DHsolution of the salt CaCl2 is negative, making it an exothermic process. This would make it a poor choice for freezing ice cream, but an excellent method of melting ice on sidewalks.

H, like E, is an extensive property, so if we had dissolved twice the mass of KBr, the Enthalpy change DHsolution would have been twice as large.

Let's see what else we can derive from

DE = q + w = q - PextDV

rearrange it to give:

q = DE + PextDV

If we could measure q under conditions of constant volume (DV = 0),

qv = DE

In the term qV, the subscript, V, indicates that the heat flow was measured under conditions of constant volume.

This type of experiment is carried out using a second type of calorimeter, a bomb calorimeter, which is a rigid steel container of known heat capacity. Its volume does not change during the reaction. This is illustrated on p. 251 of Chemistry.

The fuel value, or caloric content, of foods is obtained by combusting the food in a bomb calorimeter (hence the USDA's possession of "bombs".)

Since foods are not pure substances, the Heat of Combustion is called the calorie content and is reported on a per gram basis.

Example 2.

A slice of banana weighing 2.5 g was burned in a bomb calorimeter and produced a T rise of 3.05oC. That particular calorimeter has a Cv = 0.61 kcaloC-1. An average banana weighs 125 g. How many calories are there in an average banana?

qv = Ccalorimeter x DTcalorimeter

qv = (0.616 kcaloC-1) x ( + 3.05oC)

qv = 1.88 kcal for the calorimeter

This is the heat flow into the calorimeter from the reaction. This heat came from the banana

qv = -1.88 kcal for the banana slice

(-1.88 kcal)/(2.5 g banana) x (125 g banana/average banana) = -94 kcal (Cal) per average banana

DEcombustion of an average banana is -94 Cal.

Let's look now at the difference between DH and DE.

By definition, H = E + PV.

In reactions involving only solids and liquids, very little change in Volume occurs during a reaction. In such reactions,

DH DE

A greater change in volume occurs when gases are involved, but only if the number of mole of gases changes in going from reactants to products:

DH = DE + DngRT

Since PV = nRT, PDV = DnRT, where ng = nprod - nreact

For example, in the combustion of naphthalene:

C10H8(s) + 12 O2(g) → 10 CO2(g) + 4 H2O

DE = -5152 kJ mol-1 and Dn = -2 so,

DH = -5156 kJ mol-1

The difference between them is about 0.1%, important theoretically, but not practically.