1. Calculation of Enthalpy Values Using E = c m  T

2. Calculation of  H requires 3 steps 1.Use E = cm  T to calculate the energy change from the experiment 2.A scaling up of this value to obtain the energy change for 1 mole 3. Checking to make sure the sign of the  H is correct

3. Use E = cm  T to calculate the energy change On some occasions this may already be done for you There will be no temperature but there will be a value in kJ E = c m  T Specific heat capacity of water This will be found on the last page of the data book It does not matter what the liquid is, use the value for water The mass of water in kg (litres) 1 litre weighs 1 kg Calculate the litres of water and use this for m Do not use the mass of chemical this is used in step 2 The change in temperature

4. When 1g of ethanol C 2 H 5 OH was burned the heat produced warmed 5litres of water from 20.1 o C to 21.5 o C Calculate the ethalpy of combustion of ethanol E = c m  T E = 29.26 kJ 4.18 from databook 5 litre = 5kg 21.5 - 20.1 = 1.4 Now do step 2 E =4.18xx51.4

5. Scale up the value to obtain the energy change for 1 mole When 1g of ethanol C 2 H 5 OH was burned the heat produced warmed 5litres of water from 20.1 o C to 21.5 o C E = 29.26 kJFrom step 1 2 x C = 2 x 12 = 24 6 x H = 6 x 1 = 6 1 x O = 1 x 16 = 16 Gram Formula Mass = 46 g Gram Formula Mass of ethanol = 46 g 1g 29.26 kJ 46g 29.26 x 46 kJ46g 1345.96 kJ Now do step 3  H = 1345.96 kJ

6. Check to make sure the sign of the  H is correct When 1g of ethanol C 2 H 5 OH was burned the heat produced warmed 5litres of water from 20.1 o C to 21.5 o C From step 2  H = 1345.96 kJ Heat was produced in the reaction making it exothermic  H will have a negative value  H = - 1345.96 kJ mol -1

7. When 2g of a compound ( formula mass 40) is dissolved in 50 cm 3 of water the temperature rises by 10 o C Calculate the enthalpy of solution Step 1E = c m DT E = 4.18 x 0.05 x 10 E = 2.09 kJ Step 22g 2.09 kJ 40g 41.8 kJ Step 3Temperature rise so exothermic  H = - 41.8 kJ mol -1

8. When asked to calculate the enthalpy of neutralisation some minor changes must be made to the method. a)All liquids, both the acid and alkali volumes, are heated and so are included when calculating the mass of water b)We will not be give a mass of acid We will be given a concentration and a volume we use these to calculate the number of moles of acid used. c) We scale up (or down) the number of moles to 1 No. of moles = concentration x volume in litres

9. When 100cm 3 of hydrochloric acid concentration 0.8 mol l -1 is neutralised by 100 cm 3 of an alkali, both at 12 o C the temperature of the salt solution rises to 16.6 o C Calculate the enthalpy of neutralisation of hydrochloric acid Step 1E = c m DT E = 4.18 x 0.2 x 4.6 E = 3.8456 kJ Step 2 0.08 3.8456 kJ 1 mole 48.07 kJ No. of moles = conc x vol in litresNo. of moles = 0.8 x 0.1 = 0.08 Step 3Temperature rise so exothermic  H = - 48.07 kJ mol -1