Energy changes

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Hi I'm quite confused on how to work out the enthalpy change of neutralisation of a reaction. Please help me, and possibly give me a typical example of this to use as a set example and template to follow in further questions based on this.

Ulex writes
It would be most unwise of you to try to follow a template for doing this sort of question. You really must try to understand what to do and then apply your understanding to the particular circumstances.
 
There are usually two parts to the calculation:
 
Part 1 Work out the energy change occurring in the reaction vessel (calorimeter)
 
Part 2 Take your answer to part 1 and scale it up to give the energy change which would have occurred if you used 1 mole of reactant.
 
Example: 10 cm³ of hydrochloric acid and 10 cm³ of sodium hydroxide solution (both of 1 mol dm^-3 concentration) are mixed and the temperature went up by 5 °C. Calculate the enthalpy change of neutralisation.
 
Part 1
Energy given out = mass x specific heat capacity x temperature rise
 
Standard (usual) assumptions – remember these:
 
- Assume that all the energy given out goes into the water present. In this part of the calculation ignore the mass of dissolved substances.
 
- Assume that the density of water is 1 g per cm³. The mass of the solutions is therefore taken as 20 g and the specific heat capacity is 4.2 J g^-1.
 
Energy given out = 20 x 4.2 x 5 J
= 420 J
 
Part 2
10 cm³ of HCl contains 0.01 moles
0.01 moles of HCl produce 420 J when neutralised (notice that you base this part of the calculation on only one of the reactants, whereas in Part 1 you used the mass water in both reactants)
 
1 mole of HCl produces 42000 J (i.e. 100 times as much)
The enthalpy change of neutralisation is therefore 42 kJ mol^-1

Risk assessment
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updated: 11 May 2006