Introduction

In studying chemical processes, much of our attention is generally focused upon the properties of the substances involved. However, such processes also encompass distinct changes in the state of the system we are investigating, changes which are found to affect the overall energy of a system. In order to clearly understand what was just stated, we need to familiarize ourselves with a few terms. One of the first terms mentioned was ‘system’, and this is specifically defined as the region of the universe we are investigating. On the other hand, everything else, other than the system itself, is defined as the surroundings. In its entirety, the state of the system is specified by a number of variables including its temperature, pressure, volume, and even the chemical composition. A more extensive discussion of these energy relationships can be found in Chapter 6 of your textbook.

Energy and a little bit of Work

When a system undergoes any chemical or physical change, the internal energy (ΔE) of the system is equal to the heat (q) gained, or released, plus the work (w) done on, or by, the system. Shown as a mathematical relationship below, this is the basis for the definition of the First Law of Thermodynamics which basically states that the energy of the universe is constant.

Therefore, referring to the equation above, we can see that the energy of a system can increase either when heat is added (+q) to the system, or when work is done upon the system (+w).

Continuing with our initial equation, systems are sometimes studied under constant pressure. When this is the case, our equation can be manipulated to give rise to the relation now shown. NOTE: The subscript p denotes constant pressure.

The value of q_P in this equation is important. Specifically, this value is the amount of heat absorbed or given off by a system at constant pressure and is referred to as the heat of reaction, or more commonly, the enthalpy of reaction (ΔH). By substituting our new variable ΔH for our old variableq_Pand rearranging the second equation we finally arrive at a relationship you will possibly see time and time again.

Let’s quickly look a little deeper into this equation. As is the case with this lab, a majority of chemistry involves reactions in solution where no gaseous products are formed. In other words, under these conditions the pressure remains constant (we’re working in solution) and there is no change in volume (no gas expelled). Thus, when this is the case, ΔH is quite a useful quantity as it is equal to the change in energy of the system at constant pressure.
  

Heat

The amount of heat (q) required to raise the temperature of a particular system is an extensive property. It depends on the amount of material present, as well as the chemical make-up of the substance and how much the temperature changes. Mathematically, this can be expressed in one of two ways. The first expression, shown below on the left, deals with the number of moles of material (n) while the second expression, below on the right, incorporates the mass of the material (m).

The variables C and c are in fact different, there is not a typo! In the first expression, C is the molar heat capacity generally given in units of J/mol•K, and c in the latter expression is the specific heat capacity of the material under investigation which has units of J/g•K. In both equations, ΔT is the change in temperature which can be in Kelvin or Celsius, it does not matter. The question we present to you however is why not? Finally, in order to determine the enthalpy (ΔH), the heat (q) of a particular reaction is divided by the number of moles (n) of material involved in the reaction:

The formal definitions of the two constants described above include:

1. Molar Heat Capacity (C): the amount of heat required to raise the temperature of one mole of material by one degree Kelvin or Celsius.
   
   
2. Specific Heat Capacity (c): the amount of heat required to raise the temperature of one gram of material by one degree Kelvin or Celsius.

In particular, the values for these constants depend on the identity of the material, the state of the material, and of course its temperature. Below is a table of the values these constants take on for various states and temperatures of water. It is important to note that for the range of liquid water temperatures, that the value does remain fairly constant at two significant figures.

Direct your attention to the last column of the table. Another unit used to express heat is the calorie, which is a member of the older English System. Similar to the definitions given above, it is defined as the amount of heat necessary to raise one gram of water from 14.5ºC to 15.5ºC at one atmosphere of pressure. Since a majority of work is done in aqueous solutions, this is still a useful unit even though it is not in the SI System. On a side note, the ‘calorie’, abbreviated Cal, many of you see on the back of your Coke® can or Nestle® wrapper is actually 1,000 calories or 1 kilocalorie.

The Calorimeter Cup

The term calorimetry refers to the measurement of heat released, or absorbed, during a chemical or physical process. Well insulated so that its contents do not gain or lose heat to the surroundings, the ideal calorimeter is constructed of a material of low heat capacity so that only a small amount of heat is exchanged between the contents and the calorimeter itself. For many processes, an unsealed insulated cup can be used because of its low heat capacity and excellent insulating properties. Furthermore, constant pressure during the reaction is maintained simply by our atmosphere.

In reality however, no calorimeter is ideal. Therefore in order to obtain reliable, and most importantly reproducible, results, the calorimeter must be calibrated to determine just how much heat is exchanged with the surroundings. This correction factor is termed the calorimeter constant (C_Cal), and is simply the heat capacity of the calorimeter given in units of J/K. In addition, this variable fits into an equation similar to the ones above…

and since the ΔT of the cup will change with each experiment, it is useful to have a calorimeter constant which can be used in all experiments. In these types of experiments, the temperature change of the cup can be assumed to be equal to the temperature change of the solution contained within it.

The heat capacity of the calorimeter cup (C_Cal) is determined by performing a separate experiment in which no chemical reaction takes place. Hot water is added to some cold water contained within the calorimeter. The amount of heat lost by the hot water (q_HW) must be equal to the heat gained by both the cold water (q_CW) and the calorimeter (qCal)—remember the 1st Law of Thermodynamics! Mathematically this is expressed as the following:

Noting that the signs of these amounts of heat are opposite in sign due to the hot water losing energy and the cold water and cup gaining energy, we have to use the absolute value of ΔT  ( |T_F  – T_I| ) to solve the problem.

Continuing, the amount of heat absorbed by the calorimeter (C_Cal) can be obtained from the values of q_HW and q_CW, which can be obtained from previous equations. Specifically, the two latter values are ascertained by plugging in the corresponding masses, specific heats, and temperature changes into the formula:

With these two values now known, we can find q_Cal by using the following formula:

and then finally solve for C_Cal using:

An Example

To clarify the principles of calorimetry, consider the following hypothetical experiment where we can use calorimetry to determine the enthalpy change (ΔH) of the reaction of NaOH in HCl.

In this experiment, 75-mL (75-g) of a 0.6-M solution of HCl was placed in a calorimeter and allowed to equilibrate. Once at room temperature, an equal volume, mass, and concentration of NaOH is added to the calorimeter while stirring. Temperature readings, taken for 6-min before and after the addition of NaOH, were recorded (see table above) and a graph was constructed as shown above.

Realizing that the plot is in fact a cooling curve, we conclude that the data of interest is constricted to the left hand portion of the graph. Continuing, the initial temperature (T_I) of the acid in the cup was found to be 22.1°C and upon addition of NaOH at 60-sec, the temperature of the solution slowly increased to a maximum of 27.8°C and then steadily began to decline. The precise determination of the final temperature (T_F) is slightly more complicated since the heat exchange is occurring between the contents and the cup during and after the reaction.

Evaluation of T_F is accomplished, as shown above, through extrapolation. Mathematically, extrapolation is the process of estimating new data points that lie outside a set of known points. In other words, this method gives a theoretical temperature representing the temperature obtained if instantaneous equilibrium was achieved within the system. Thus, we arrive at a final temperature of 29.3°C.

With the initial and final temperatures at last known, we can begin traveling the rather tedious road of calculations that will end with the heat of reaction. However, before we begin we have to discuss one last thing. You were questioned earlier as to why it does not matter which unit of temperature you use in calorimetry. We will now answer that very question. Recalling that ΔT is defined as the final temperature minus the initial temperature, we can see as shown below that it is truly unnecessary to convert Celsius to Kelvin:

Now, for our particular example, the heat of reaction is equal to heat gained by the acid plus the heat gained by the calorimeter as shown below.

Thus, our first step is to calculate the heat gained by HCl and the calorimeter. Assuming that the heat capacity for a dilute solution of acid in water is very similar to that of water, we can calculate the heat absorbed by the acid. The same general concept can be used to calculate the heat absorbed by the calorimeter. All that is different is the equation used, and the fact that the heat capacity for the calorimeter is given to be 6.80 J/K.

Realizing that the calorimeter constant is very small compared to that of the acid, we can just sum the two numbers to ascertain a final value of ~2.35-kJ for the heat of this particular reaction.

Hess’s Law

One of the last calculations you will be held responsible for in this experiment involves Hess’s Law. Hess’s Law states that if a reaction is carried out in a series of steps, than the overall change in enthalpy (ΔH) for the reaction is equal to the sum of the enthalpy changes for each individual step. In fact, the overall enthalpy change for the process is independent of the number of steps taken as well as the particular nature of the path by which the reaction is carried out. Therefore, we can use information tabulated for a relatively small number of reactions to calculate the overall change in enthalpy for a large number of different reactions.

In this particular experiment, we want to know the heat of reaction for burning magnesium metal in oxygen to generate magnesium oxide (MgO).

Overall, this is a very exothermic reaction thus making it extremely difficult to calculate ΔH using our relatively elementary calorimetry setup. However, by experimentally measuring the heat of reaction for two other reactions…

and using the heat of formation reaction for water…

we can use Hess’s Law to calculate the heat of reaction. If you are unfamiliar with the process, or just need to refresh your memory on just how to solve such a problem, consult Chapter 6 of your textbook pages 261-265, and then attempt the following problem.

Practice

Using the equations provided below, see if you can determine the overall change in enthalpy for the reaction provided below:

The following reactions are given with their corresponding values of ΔH:

The first step to successfully solving a Hess’s Law problem is to understand just exactly how you are allowed to manipulate the giving reactions. Overall there are two relatively simple rules:

1. If you need to reverse a reaction, you must reverse the sign corresponding to the value of ΔH. In other words if it’s positive and you reverse it, it has to become negative and vice-versa.
   
   
2. If you need to get more of one species on one side, or reduce, you can multiply, or divide, the ENTIRE reaction by some number. If this is done, the corresponding value of ΔH must also be treated accordingly.

For the practice problem, the first thing we need to do is get all of the species being reacted on their correct side with respect to the final equation. Thus, we need to reverse the first reaction as shown below:

In the final equation above, we see that there is no PCl₃, thus we must somehow eliminate it. Species can be eliminated during Hess’s Law when they exist on both sides of two different equations. Thus, if we multiply the first equation by four, we are left with 4 mols of PCl₃ on both sides.

Now if we treat this as one big addition problem, we can cancel out common species as well as add species existing on the same side of the equation which results in the following expression:

Finally, adding the values of ΔH, we are left with the overall change in enthalpy for this particular reaction!