Introduction:
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 can not 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. This can be readily determined by measuring the temperature change that occurs when a known amount of hot water is added to a known amount of cold water in the calorimeter. The heat lost by the warm water is equal to the heat gained by the cold water and the calorimeter (we will assume no heat is lost to the laboratory). For example, if:
T1 equals the temperature of 30 g of room temperature
water, and
T2 equals the temperature of 50 g of warmer water just
before it is added to the room temperature water, and
Tf equals the temperature after mixing,
then the heat lost by the warmer water is
qlost = (T2 – Tf) · 50g · 4.18 J/g°C
The specific heat of water is 4.18 J/g°C, and the density of water is assumed to be 1.00 g/mL. The heat gained by the room temperature water is
qgained = (Tf – T1) · 30g · 4.18 J/g°C
The heat lost to the calorimeter is the difference between the heat lost by the warmer water and that gained by the cooler water:
qlost – qgained = (Tf – T1) · WE
Thus, by measuring T1, T2, and Tf, the water equivalent (WE) of the calorimeter can be calculated:
Now lets look at some real numbers and
see how the (WE) is
determined and
used. Assume that we mix 50 g of warm water with 30 g of cool
water:
| Temperature of 50 g of warm water |
37.9°C
|
| Temperature of 30 g of room temperature water |
20.9°C
|
| Temperature after mixing |
29.1°C
|
| qlost = 50 g · 8.8°C · 4.18 J/g°C |
1841 J
|
| qgained = 30 g · 8.2°C · 4.18 J/g°C |
1028 J
|
| Heat lost to calorimeter: |
813 J
|
Using this data, we can calculated the WE as:
WE = 813 J / 8.2°C = 99 J/°C
So, now that we have the (WE), what do we
do with it? Well, lets say
that we want to determine the
Hrxn for the
following 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
1.00
M reactant 'A' and 15.00 g of 2.00 M reactant 'B' into the calorimeter
and
monitor the temperature. During the course of the reaction, the
35.0
g of solution increases in temperature by 6.5°C. From this
increase
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 |
951 J
|
| Heat gained by the calorimeter = 99 J/°C · 6.5°C |
644 J
|
| True heat of the reaction, qrxn : |
1595 J
|
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
Purpose:
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 more clever.
Method:
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
the
Hrxn
of each of the following three reactions:
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As written, the first reaction 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, the third reaction 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 consider reversing of the Reaction #2:
MgCl2(aq) + H2O(l)
MgO(s)
+ 2 HCl(aq)
Now we can see that the MgCl2(aq) generated as a product in the first reaction is canceled by the reactant MgCl2(aq) in the second reaction. Also, the two moles of HCl(aq) required as a reactant in the first reaction, are canceled by the two moles of HCl (aq) generated as a product in the second reaction. Finally, the H2(g) generated as a product in the first reaction, is canceled by the H2(g) required as a reactant in the third reaction.
Therefore, the overall
Hcomb
for the
reaction of Mg(s) with O2(g) to form MgO(s)
, can be calculated as:
Hcomb =
HReaction#1
-
HReaction#2 +
HReaction#3
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 heat capacity of the
thermometer,
inefficient mixing of the solution, and heat losses to the calorimeter,
the
temperature difference determined by the thermometer is less than the
actual
temperature change of the solution (see Figure below).
Procedure:
Construction of Calorimeters
Use the following table as a guide for your data collection:
| |
|
|
| Initial temperature of room temperature water | ||
| Initial temperature of warm water | ||
| Initial temperature of 2.00 M HCl | ||
| Weight of MgO, g | ||
| Initial temperature of 2.00 M HCl | ||
| Weight of Mg strip, g |
From your time/temperature curves, calculate:
(Updated 6/5/07 by C.R. Snelling)