The process of measuring energy changes is known as calorimetry, after the old unit for heat, a calorie. Any vessel that is used to measure enthalpy changes is known as a calorimeter.
ENTHALPY CHANGE OF COMBUSTION
The following are the standard enthalpy changes of combustion of some fuel; -286 kJ mol -1 for hydrogen, -890 kJ mol -1 for compressed natural gas (CNG), -2219 kJ mol -1 for liquid petroleum gas (LPG), -726 kJ mol -1 for methanol and -1367 kJ mol -1 for ethanol. I also mentioned this in my previous post here. Being combustion reactions, they are all exothermic. We can find these values by measuring the increase in temperature of a certain mass of water to which energy from the combustion reaction is transferred, making its temperature rise. One gram of water requires 4.2 joules of energy to raise its temperature by 1°C; this is water’s specific heat capacity. Specific heat capacity (c) is the energy required to raise the temperature of 1 gram of substance by 1 K. It is measured in Jg-1K-1. A temperature rise of 1 K is the same as a temperature rise of 1 °C.
So, the specific heat capacity of water is 4.2 J/g/K (joules per gram per kelvin). We can use a simple equation to work out the energy transferred to the water:
Energy transferred, q = mass of water, m (g) × specific heat capacity of water, c (Jkg-1K-1) × temperature change, ΔT (K).
The specific heat capacity of water is unusually high. It takes energy to break the hydrogen bonds between clusters of water molecules. This means that the energy does not go towards increasing the kinetic energy of molecules and hence the temperature of the water.
SIMPLE DETERMINATION OF THE ENTHALPY CHANGE OF COMBUSTION OF A LIQUID FUEL
A simple experimental set-up to measure the enthalpy change of combustion of ethanol is shown below. The experiment is carried out as follows.
- Step 1. Weigh the spirit burner with ethanol in it at the beginning of experiment and take the water temperature.
- Step 2. Allow the spirit burner to heat up a known mass of water by about 20°C. Stop heating and take the temperature of the water. This gives the temperature rise in Kelvin.
- Step 3. Weigh the spirit burner again. This gives the mass of the ethanol used.
The next Example explains how this experiment may be used to calculate the enthalpy change of combustion for ethanol.
EXAMPLE
0.40g ethanol raises the temperature of 100.00g water in the metal calorimeter by 21.0 °C. Calculate the enthalpy change of combustion of ethanol.
ANSWER
Step 1. Work out the energy transferred to the water:
Energy transferred to the water, q = mcΔT
= 100.00 × 4.2 × 21.0 = 8820 J = 8.82 kJ (ignoring significant figures)
Step 2. Calculate the amount of fuel used in moles:
Mass of 1 mole of ethanol (C2H5OH)
= (12 × 2) + (1 × 5) + 16 + 1 = 46 g
Moles of ethanol in 0.40 g = mass in grams/mass of one mole (in grams)
= 0.40/46
= 0.0087 mol (correct to 2 sig. Figs)
Step 3. Work out the enthalpy change of combustion, that is, the energy transferred when 1 mole of ethanol burns:
0.0087 mol ethanol releases 8.82 kJ
1 mole of ethanol releases 8.82 × 1/0.0087 = 1014 kJ
So the enthalpy change of combustion of ethanol = -1000 kJ mol-1 (correct to 2 sig. figs)
The bomb calorimeter
The bomb calorimeter measures enthalpy changes of combustion. The weighed sample is inside a stainless steel container – the bomb – filled with oxygen under pressure, and the fuel is ignited electrically.
The principle is the same as for the simplified experiment that’s explained above. Energy is transferred from the combusted fuel to the surrounding water, and the temperature rise is measured. However, this apparatus gives a much more accurate value than the simple version, provided the readings are taken quickly, because energy losses to the surroundings are reduced to almost zero. In the simplified experiment above, another potential error arises when the spirit burner is weighed, as some of the ethanol may evaporate between weighings.
Enthalpy change is transfer of energy into or out of the system at constant pressure. As the bomb has a constant volume, it does not measure enthalpy change exactly. However, because the difference is small, we need not be concerned with the mathematical correction used to adjust the value to constant pressure.
THE ENERGY VALUES OF FOOD
The bomb calorimeter is used to measure the enthalpy changes of combustion of different foods. Although our bodies don’t burn foods in quite the same way as we burn natural gas in a cooker, the outcome is the same. Oxygen is still required and the energy we obtain from compounds in foodstuffs is the samne as if they were burnt in a bomb calorimeter. The crucial difference is that when we burn fuels in cooking or in the combustion engine of a car, most of the released energy is wasted and goes to heat up the surroundings.
When we ‘burn’ glucose in our bodies, there is not just one high-temperature reaction, but several small steps with small transfers of energy, each step catalyzed by complex molecules called enzymes. The human machine is nowhere near 100 per cent efficient, and we do lose energy to our surroundings (hence, our body warmth), but much of the energy of the step-wise reactions is either used to maintain electrical and chemical body functions or stored, and so does not emerge as heat. Of course, when we store energy it is usually as fat, so taking in more energy than we require will make us obese.
ENTHALPY CHANGE OF NEUTRALIZATION
Many reactions take place in aqueous solution and the enthalpy changes of these reactions can be measured using an expanded polystyrene cup fitted with a lid. Expanded polystyrene is a very good insulator and also absorbs little energy from the reaction itself.
When an acid reacts completely with an alkali the reaction is called a neutralization. The reaction between aqueous sodium hydroxide and aqueous hydrochloric acid can be represented by the equation:
HCl(aq) + NaOH(aq) → NaCI(aq) + H2O(l)
In this reaction all the reactants are present as ions, so:
H+(aq) + Cl-(aq) + Na+ (aq) + OH- (aq) → Na+(aq) + Cl-(aq) + H2O(l)
You will notice that the ions reacting are only H+(aq) and OH-(aq), while Na+ (aq) and Cl-(aq) take no part in the reaction and are called spectator ions. So the neutralization reaction is more simply written as:
H+(aq) + OH-(aq) → H2O(l)
So, standard enthalpy change of neutralization is defined as: The enthalpy change when an acid and a base react to form one mole of water under standard conditions (298 K, 100 kPa).
The experiment to measure this enthalpy change is carried out as follows:
Step 1. Measure out accurately a known volume and amount of aqueous acid and take the temperature of the solution.
Step 2. Mix the acid from Step 1 with a known volume and amount of alkali in a polystyrene cup and stir the mixture thoroughly.
Step 3. Calculate the temperature rise. This is usually done by taking the temperature of the alkali in the polystyrene cup each minute for four minutes. On the fifth minute the acid is added, the mixture stirred and then the temperature measured again every minute from the sixth minute until the 12th minute. A graph is drawn and the temperature rise as the acid is added to the alkali is extrapolated.
EXAMPLE
An experiment to measure the enthalpy change of neutralisation is carried out. When 50.0 cm3 HCI(aq) is added to 50.0 cm3 NaOH(aq), both of 1.0 mol dm-3 concentration, the temperature rise is 6.0°C. Calculate the enthalpy change of neutralization.
ANSWER
Step 1. Work out the energy transferred to the water. The acid and alkali are very dilute so the volume of water is assumed to be:
50.0 cm3 HCl(aq) + 50.0 cm3 NaOH(aq) = 100.0 cm3
The mass of 100 cm3 water = 100 g (1.0 cm3 water has a mass of 1.0 g)
Energy transferred to the water, q = mcΔT
= 100.0 × 4.2 × 6.0 = 2520 J = 2.52 kJ (ignoring significant figures)
Step 2. Calculate the amount of acid and alkali used in moles:
Number of moles NaOH in 1000 cm3 = 1.0 mol
So number of moles NaOH (and also HCl) in 50.0 cm3 = 1.0 × 50.0/1000 = 0.050 mol
Step 3. Work out the enthalpy change of neutralization, that is, the energy transferred when 1 mol of water is formed from 1.0 mol NaOH and 1.0 mol HCI.
When 0.050 mol of water is formed the energy released 2.52 kJ
1 mole of water being formed releases 2.52 × 1/0.050 = 50.4 kJ
So enthalpy change of neutralisation = -50 kJ mol-1 (correct to 2 sig. Figs)
The data book value ΔHꝋneut = 57.2 kJ mol-1, so heat energy was lost to the surroundings in this experiment.
RELEASING ENERGY FROM FUEL MOLECULES
We know that fuel molecules need oxygen to combust and so release energy. Where does this energy come from? To answer this question we need to look at the energies of the bonds that hold atoms together, and what happens when these bonds are broken and made.
BREAKING BONDS: BOND ENTHALPY
Bond enthalpy or bond energy is the energy required to break a bond between two atoms in a gaseous molecule. However, because the energy required to break one bond is so small, we define bond enthalpy as the energy required to break one mole of bonds. Thus: Bond enthalpy, E, is the energy required to break one mole of bonds of the same type in gaseous molecules under standard conditions (298 K, 100 kPa).
Consider hydrogen, which has an H-H bond:
H – H(g) → H(g) + H(g) ΔHꝋ = +436 kJ mol-1
Bond enthalpy, E, is often written as:
E (H – H) = +436 kJ mol-1
Notice that bond breaking is an endothermic process. This is because we are putting in energy to overcome the force of electrostatic attraction in the bond.
The bond energy in oxygen molecules is for breaking a double bond:
O=O(g) → O(g) + O(g) ΔHꝋ= +498 kJ mol-1
So E(O=O) = +498 kJ mol-1. This is larger than for the H-H bond, because double bonds are stronger than single bonds and so require more energy to be broken.
Up to now, we have considered bond energies only in diatomic molecules (molecules that contain two atoms). There are two O – H bonds in the triatomic water molecule. The energy required to break the first O – H bond is not the same as the energy required to break the second:
H – O – H(g) → H(g) + O – H(g) ΔHꝋ = +502 kJ mol-1
O – H(g) → O(g) + H(g) ΔHꝋ = +427 kJ mol-1
The first bond is in a molecular environment in which two O – H bonds exist, and more energy is required to break this bond. The second O – H bond is in a changed environment, and clearly this has an effect on its bond energy. A total of 929 kJ mol-1 is required to break both the O – H bonds in water, so we say the average or mean bond enthalpy for the O – H bond in this molecule is +464 kJ mol-1
The differences in the bond enthalpy of the O – H bond in other molecules can also be found because of the different molecular environments of the bond. For example, the bond enthalpy of the O – H bond in methanol is +437 kJ mol-1. Usually, the differences in bond enthalpies are not very great and they are average to give a good approximation that can be used.
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