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4. Heat of Neutralisation
Experiment: To measure heat of neutralisation.1. Take a known volume of standard acid (e.g. 100
of 1 mol HCl) in a calorimeter (e.g. a polystyrene cup) and take its temperature.2. Take a known volume of base (e.g. 100
of 1 mol NaOH) in a graduated cylinder and take its temperature.3. Get the average of both temperatures (this is the initial temperature).
4. Pour the base into the acid, stir and record the highest temperature reached.
5. Use the formula
where m = total mass of the liquid (acid + base), c = Specific heat capacity of the solution and
= Rise in temperature.6. Calculate the number of moles of acid neutralised:
molarity8. Calculate the amount of heat that would be produced if 1 mole of had been neutralised.
Example 1 (L.C.H. 1991)
Define heat of neutralisation. In what units is it measured? When a 200 of a sodium hydroxide solution was added to a 200 of 0.4mol
solution of sulphuric acid in a plastic container, a neutral solution was produced and the temperature rose by 5.5
. The density and the specific heat capacity of the neutral solution (assumed equal to those of water) are 1.0 
and 4.2 
respectively.i) Why was a plastic container used?
ii) What steps would you have taken to ensure an accurate measurement of the temperature rise?
iii) Calculate the heat of neutralisation of 1 mole of sulphuric acid by sodium hydroxide.
Solution
Example 2 (L.C.H. 1992)
When 200 of 1 mol hydrochloric acid solution was neutralised by a solution of sodium hydroxide 11.4 kJ of heat was produced. Calculate the heat of neutralisation of hydrochloric acid by the sodium hydroxide solution.Solution
Heat of neutralisation values
The heat of neutralisation of any strong acid e.g. HCl by any strong base like NaOH is the same - about 57.3 kJ
. Why? Strong acids and strong bases are fully dissociated (ionised) in solution. No energy is required for their dissociation. So, regardless of what strong acid and what strong base are reacting the common reaction taking place in all cases is
+ 
, for which 
is - 57.3 kJ .However if the acid or the base is weak the value of the heat of neutralisation is less. Why?
For complete neutralisation to take place the weak acid (or weak base) must first become fully dissociated and this requires energy. The overall heat change is the result of the two reactions, dissociation into ions (endothermic) and neutralisation (exothermic).
5. Hess's Law calculations
The definitions for heat of formation, heat of combustion and heat of reaction are extremely important.Equations for heat of formation:
If you are told that the heat of formation of
is -70 kJ , then, to write a balanced equation for this, you must first write down one mole of the product on the right of the equation. On the left of the equation write the formulae for the elements in their standard state:
Now balance the equation but 1 mole must remain on the right hand side.
Example 3
Write equations representing heats of formation of and 
.Solution
Equations for heat of combustion:
If you are told that the heat of combustion of carbon is -393 kJ
, you must be able to write an equation between one mole of carbon and excess oxygen.
When an organic compound is burned the carbon produces carbon dioxide and the hydrogen produces water.
Example 4
Write equations representing heats of combustion of and 
.Solution
Steps used to solve Hess's law problems:
1. Write out the equation for the heat change required.
2. Write out the given information in equation form.
3. Rearrange the equations from step two so that when all are added together, it gives the heat change for the required equation.
Example 5 (L.C.H. 1993)
Calculate the heat of formation of carbon monoxide from the following data:
(= -393 kJ)
(= -570 kJ)Solution
Example 6 (L.C.H. 1995)
Calculate the heat of reaction for
Given that the heats of formation of
, 
, 
are 233 kJ , 114.5 kJ and 194.5 kJ respectively.Solution
6. Bond Energy calculations
When a chemical reaction takes place, all the bonds in reactant molecules are broken, and all the bonds between product molecules are made.Energy is supplied to break bonds and energy is released when bonds are made.
Bond energy values are only average values therefore calculations done using these values will not be as accurate as calculations done using Hess's law, which uses accurately measured experimental values.
To solve bond energy problems:
(i) Write out the equation showing all the individual bonds.
(ii) Calculate the total energy required to break bonds in the reactant molecules.
(iii) Calculate the total energy released when the product molecules are formed.
(iv) Equate the sum of (ii) and (iii) with the heat of reaction.
Example 7 (L.C.H. 1996)
Chloroethane and bromoethane can be produced from ethane by the following reactions:
( = -55 kJ)
(= -58 kJ)Use the following data (where E stands for molar bond energy), to work out a value for E (C-Cl) in chloroethane, and a value for E (C-Br) in bromoethane.
E (C-C) = 348 kJ
E (H-Cl) = 431 kJ
E (C-H) = 412 kJ
E (C=C) = 612 kJ
E (H-Br) = 366 kJ
Suggest why the values of E (C-Br) is less than that of E (C-Cl).
Solution
Example 8 (L.C.H. 1990)
The heat of combustion of methane is - 890 kJ. Use this together with the following thermochemical data, (where E stands for the molar bond energy) to work out a value for E (C-H) in methane.E (H-H) = 436 kJ
= 718 kJ 
= -394 kJ 
= -286 kJ Solution
Leaving Certificate Higher Level Examination Questions
1987 Q10(b), 1988 Q4, 1989 Q4, 1990 Q5, 1991 Q10(c), 1992 Q9, 1993 Q10(c), 1994 Q4, 1995 Q5, 1996 Q10(b), 1997 Q4, 1998 Q4.
