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4. Heat of NeutralisationExperiment: To measure heat of neutralisation.1. Take a known volume of standard acid (e.g. 100 ![]() 2. Take a known volume of base (e.g. 100 ![]() 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 ![]() ![]() 6. Calculate the number of moles of acid neutralised: ![]() 8. 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![]() ![]() ![]() ![]() ![]() ![]() 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![]() ![]() 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 ![]() 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 ![]() ![]() ![]() ![]() ![]() ![]() 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 calculationsThe 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 ![]() ![]() ![]() Now balance the equation but 1 mole must remain on the right hand side. ![]() Example 3Write equations representing heats of formation of![]() ![]() Solution Equations for heat of combustion: If you are told that the heat of combustion of carbon is -393 kJ ![]() ![]() When an organic compound is burned the carbon produces carbon dioxide and the hydrogen produces water. Example 4Write equations representing heats of combustion of![]() ![]() 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:![]() ![]() ![]() ![]() Solution Example 6 (L.C.H. 1995)Calculate the heat of reaction for![]() Given that the heats of formation of ![]() ![]() ![]() ![]() ![]() ![]() Solution 6. Bond Energy calculationsWhen 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:![]() ![]() ![]() ![]() 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![]() E (H-H) = 436 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. |
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