Calculation of Enthalpy Values Using E = c m

D

T

Calculation of D H requires 3 steps 1. Use E = cm D 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 D H is correct

Use E = cm

D

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 The mass of water in kg (litres)

E = c m

D

T

The change in temperature Specific heat capacity of water

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

D

T

5 litre = 5kg 4.18 from databook E = 4.18 x 5 x 1.4

E = 29.26 kJ 21.5 - 20.1 = 1.4

Now do step 2

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 1 E = 29.26 kJ

Scale up the value to obtain the energy change for 1 mole

6 x H = 6 x 1 = 6 1 x O = 1 x 16 = 16 1g 29.26 kJ Gram Formula Mass = 46 g D H = 1345.96 kJ

Now do step 3

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 D H = 1345.96 kJ Check to make sure the sign of the D H is correct

Heat was produced in the reaction making it exothermic

D H will have a negative value D H = - 1345.96 kJ mol -1

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 1 E = c m DT E = 4.18 x 0.05 x 10

E = 2.09 kJ

Step 2 Step 3 2g 2.09 kJ 40g

41.8 kJ

Temperature rise so exothermic D

H = - 41.8 kJ mol -1

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.

No. of moles = concentration x volume in litres

c) We scale up (or down) the number of moles to 1

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 1 E = 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

Step 3 Temperature rise so exothermic D

H = - 48.07 kJ mol -1