1.Problems that give you an equation and D H value and ask you to find the heat absorbed or released for a given amount of substance.

The thermochemical equation for the combustion of ethyne is:
C₂H₂(l) + 5O₂(g) --> 4CO₂(g)+2H₂O(g) D H=-2600kJmol^-1
(a)How much energy is produced when 13.0g of ethyne are burnt?

Working through moles

The DH value tells us that for every mole of the equation as written there will be 2600 kJ of energy released. That means 2 moles of ethyne will produce 2600kJ when burnt (so 1 mol will produce 2600/2 = 1300 kJ).
It is vital that you work our how much energy is involved for one mole of the substance given, so if the equation involves 2 moles then divide through, etc.
Now that you have found out how much energy is produced per mole of your substance, you need to work out how many moles you actually have,(an n=m/M calculation), then multiply by this.
m=13.0g means n= 13.0/26.0 = 0.500 mol.

Energy released, E = 1300 x 0.500 = 650kJ.

The steps of the calculation were:
*Work out the energy involved for 1 mol of the substance you are asked about. (Divide DH by coefficient in equation
* Work out no. of moles of substance given (often n=m/M but can be other formulae)
* Multiply Energy(J/mol or kJ/mol) by moles to get the Energy released/absorbed.

Note that this is not a DH value, but an Energy value. DH values only go with equations.

For an alternative method - working through mass instead of moles see the box below.

There are variations on this type of problem where you are given DH , the Energy involved in an experiment and you need to calculate the amount of substance used.
In this case the steps are:
*Work out the energy involved for 1 mol of the substance you are asked about. (Divide DH by coefficient in equation
* Divide the Energy involved for your experiment by the amount of Energy expected per mol (from the step before).
*This gives you the number of mole present which you can then convert to other quantities like mass, volume, etc.

2.The 2nd type of problem involves using a reaction to heat up some water ( or a calorimeter).

The combustion of butane is given by: 2C₄H₁₀(g) + 13O₂(g) --> 8CO₂(g) + 10H₂O(g) DH = -5760kJmol^-1

What mass of butane would be needed to heat 100g of water from 20°C to 100°C?
Work out the energy involved for 1 mole of butane. DH/2 since equation deals with 2 moles of butane. 5760/2 = 2880kJ released per mole of butane.
Next, calculate the amount of energy involved in heating the water, given that the heat capacity of water is 4.184 J °C^-1 g^-1. This means it takes 4.184 J to heat 1 g of water by 1°C.
E=4.184 x m(H₂O) x DT
E=4.184 x 100 x 80° = 33472 J = 33.5kJ
Work out how many mole of butane involved by dividing actual energy involved by the amount of energy expected to be used, i.e. 33.5kJ/2880kJ mol^-1 = 0.0116mol
Finally, convert moles to grams.
m=nxM = 0.0116 x 58 = 0.673g

2½. The same procedure is used when you are given a calorimeter with a calibration factor which is heated or cooled during the experiment.
A calorimeter and its contents has a calibration factor,C_F= 318J °C^-1. When sulfur trioxide was added to water, the temperature of the calorimeter and its contents rose by 15.0°C. How many grams of SO₃ were added, given the following equation:
SO₃(g) + H₂O(l) --> H₂SO₄(aq) DH = -129.6 kJ mol^-1

Use E = C_F x DT so E = 318 x 15.0 = 4770 J or 4.77kJ
Now, according to the equation, 1 mole of SO₃ should produce 129.6 kJ of energy, so to find out how much 4.77kJ represents:-
4.77/129.6 = 0.0368 mol
and it's a straightforward step to convert this to grams
m(SO₃) = 0.0368 x 80 g = 2.94g

3. The most challenging of all thermochemistry problems requires you to calculate the calibration factor first (usually by electrical calibration) and then go on to use this calibration factor to work out the energy involved in a subsequent reaction.

1.50g of a fossil fuel was burnt in a bomb calorimeter in the presence of excess oxygen,immediately after a calibration run. The following results were obtained:
Initial temp. of calorimeter and sample = 21.6°C
Temperature after applying a voltage of 4.80V and a current of 4.34A for 2.00 minutes = 22.9°C
Temperature after complete combustion of sample = 41.6°C
Calculate the heat of combustion of the fuel in kJ/g.

First, you will need to work out the calibration factor, C_F for the calorimeter based on the electrical calibration data and the formula E = VIt
, then C_F = E/DT
E = 4.80 x 4.34 x 120.00 (don't forget t must be in seconds)
= 2500J
C_F =E/DT = 2500/(22.9-21.6°)=1923J/°C
Now, use this calibration factor for the second part
For the combustion reaction,DT = 41.6-22.9°C = 18.7°C
The energy, E, produced by this combustion = C_Fx DT = 1923 x 18.7 = 35960 J = 36.0 kJ
To work out the heat of combustion in kJ/g, we need to divide by the no. of grams of fuel burnt in this reaction, ie. 1.50 g
i.e the heat of combustion = 36.0/1.50 kJ/g = 24.0kJ/g
Note: I have seen problems where the chemical reaction has been carried out before the calibration. Do not be afeared!! The order makes no difference. Just don't get your DT values mixed up!! One is for the calibration, the other is for the reaction.

Of course, there can be variations on these themes, such as finding T or finding an unkown calibration factor. Try to think up as many different ways to ask questions involving these variables (and then you won't be so rattled should the exam have one of these deviations...I mean variations!)