Every chemical and physical transformation involves some change in energy. These changes may involve heat, light, electricity, or some other form of energy. The branch of chemistry, which deals with the measurement of these changes, is called thermochemistry. Enthalpy is a word which means ‘heat content’. Every substance has some characteristic enthalpy because of its chemical makeup. It depends upon several factors: the total number of subatomic particles in each atom and how they are arranged; the total number of atoms in the substance and how these atoms are arranged; and the total number of molecules and how they are arranged (including any interactions between them). Because it is impossible to know all of these factors, we cannot calculate the absolute value for the heat content of a substance. However, we can measure the change in heat content, which accompany chemical or physical transformations.
The energy change of a reaction that occurs at constant pressure is termed the heat of reaction or the enthalpy change. The symbol Hrxn is used to denote the enthalpy change. If heat is evolved, the reaction is exothermic (Hrxn < 0) and if heat is absorbed the reaction is endothermic (Hrxn > 0).
If the transformation only involves a change in heat energy, it is a simple matter to monitor the energy change with a thermometer. The quantity of heat either evolved or absorbed may be calculated using the equation:
q = m · s · T
Where q is the quantity of heat, m is the mass of the substance undergoing the change, s is the specific heat of the substance, and T is the change in temperature in °C.
The quantity of heat is measured experimentally by allowing the reaction to take place in a thermally insulated vessel called a calorimeter. Ideally, the heat liberated in the reaction would only cause an increase in the temperature of the solution. However, because no calorimeter is perfect, some of the heat "leaks" out and we must calibrate it by determining its water equivalent, WE. Although it is a simple procedure, it is a bit messy and so in this example, we will just assume that the WE = 99 J/°C.OK, how do we determine the overall Hrxn for this reaction:
A + B C
We first need to determine the amount of
heat (qrxn) that
is evolved during this reaction. For this we will add 20.00 g of
M reactant 'A' and 15.00 g of 2.00 M reactant 'B' into the calorimeter
monitor the temperature. During the course of the reaction, the
g of solution increases in temperature by 6.5°C. From this
in temperature, we can calculated the true amount of heat, qrxn
, that was released:
|Heat gained by the solution = 35.00 g · 6.5°C · 4.18 J/g°C||
|Heat gained by the calorimeter = 99 J/°C · 6.5°C||
|True heat of the reaction, qrxn :||
Now that we know the true amount of heat that was released, we can calculate Hrxn by simply dividing qrxn by the moles of the limiting reagent (assume that the density of this solution is 1.000 g/mL):
moles of 'A' = 1.0 mole/liter
· 0.020 liters = 0.020 moles
moles of 'B' = 2.0 moles/liter · 0.015 liters = 0.030 moles
Therefore 'A' is the limiting reagent, and the Hrxn is:
-1595 J / 0.020 moles = -79,750 J/mol or -79.8 kJ/mol
In this experiment, you will determine the enthalpy of combustion, Hcomb for magnesium reacting with oxygen to form magnesium oxide:
Mg(s) + ½ O2(g) MgO(s)
The most straight forward method of determining this value would be to weigh a piece of magnesium, place it in a 'bomb' calorimeter with excess oxygen and initiate the reaction with a spark. Unfortunately, we do not have such a sophisticated piece of equipment and so must be a little bit cleverer.
Since we do not possess the technology to determine this Hcomb directly, we will use Hess' law to determine it indirectly. Basically, Hess' law states that it does not matter how many different steps a reaction goes through. As long as the overall reactants and products are the same, then the enthalpy of reaction will be the same. This of course is a very powerful tool since it allows us to calculate the H for reactions that would normally be difficult or impossible to measure directly.
For our study, we will need to determine
of each of the following three reactions:
As written, Reaction #1 provides one of the reactants we needed, Mg(s). However, it requires HCl(aq) as a reactant, and generates MgCl2(aq) and H2(g) as unwanted products. Likewise, Reaction #3 provides us with the other reactant we need, O2(g). Unfortunately, it introduces H2(g) as a reactant and generates H2O(l) as an unwanted product. Fortunately, all of these problems are solved if we reverse Reaction #2:
MgCl2(aq) + H2O(l) MgO(s) + 2 HCl(aq)
Now we can see that the MgCl2(aq) generated as a product in Reaction #1 is canceled by the reactant MgCl2(aq) in Reaction #2. Also, the two moles of HCl(aq) required as reactants in Reaction #1, are canceled by the two moles of HCl (aq) generated as products in Reaction #2. Finally, the H2(g) generated as a product in Reaction #1, is canceled by the H2(g) required as a reactant in Reaction #3.
Therefore, the overall Hcomb for the reaction of Mg(s) with O2(g) to form MgO(s) , can be calculated as:
Hcomb = HMg/HCl - HMgO/HCl + HH2/O2
In this experiment you will need to
determine the temperature change
or T for several reactions. Throughout this
discussion we have referred to T
as the change in temperature.
One might assume that this was simply the difference between the
beginning and final temperatures as determined with a thermometer.
Unfortunately, things are slightly more complicated.
For a variety of reasons, including the specific heat of the
inefficient mixing of the solution, and heat losses to the calorimeter,
temperature difference determined by the thermometer is less than the
temperature change of the solution. You may have to use some
'judgment' to determine the final temperature of the reaction mixture.
Procedure: (Note: we are now using good quality mini-calorimeters that have been calibrated to have a WE of 25 J/°C)
Preparing the Software
From your time/temperature graphs, please determine:
(Updated 9/30/13 by C.R. Snelling)