Calculation of Enthalpy Values
Using
E = c m DT
Calculation of DH requires 3 steps
1. Use E = cmDT 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 DH is correct
Use E = cmDT 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)
Calculate
the
ofofwater
and use
for m
litreuse
weighs
1 kg
Do1not
thelitres
mass
chemical
thisthis
is used
in step 2
E = c m DT
The change in temperature
Specific heat capacity of water
be found
on the liquid
last page
of the
book
It This
does will
not matter
what
is, use
thedata
value
for water
When 1g of ethanol C2H5OH was burned the heat produced
warmed 5litres of water from 20.1 oC to 21.5 oC
Calculate the ethalpy of combustion of ethanol
E = c m DT
21.5 - 20.1 = 1.4
5 litre = 5kg
E = 4.18 x 5
4.18 from
databook
x 1.4
E = 29.26 kJ
Now do step 2
When 1g of ethanol C2H5OH was burned the heat produced
warmed 5litres of water from 20.1 oC to 21.5 oC
From step 1
E = 29.26 kJ
Scale up the value to obtain the energy change for 1 mole
2 x C =Mass
2 x of12ethanol
= 24 = 46 g
Gram Formula
6 xH = 6 x 1 = 6
1 x O = 1 x 16 = 16
1g
29.26 kJ
Gram Formula Mass = 46 g
46g
DH
29.26 x 46
1345.96
kJ kJ
=
1345.96 kJ
Now do step 3
When 1g of ethanol C2H5OH was burned the heat produced
warmed 5litres of water from 20.1 oC to 21.5 oC
DH
From step 2
=
1345.96 kJ
Check to make sure the sign of the DH is correct
Heat was produced in the reaction making it exothermic
DH will have a negative value
DH
=
- 1345.96 kJ mol-1
When 2g of a compound ( formula mass 40) is dissolved in
50 cm3 of water the temperature rises by 10 oC
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
2g
40g
Step 3
2.09 kJ
41.8 kJ
Temperature rise so exothermic
DH
=
- 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 100cm3 of hydrochloric acid concentration 0.8 mol l-1 is
neutralised by 100 cm3 of an alkali, both at 12 oC the temperature
of the salt solution rises to 16.6 oC
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
No. of moles = conc
0.8 x 0.1
x vol= in0.08
litres
0.08
3.8456 kJ
1 mole
Step 3
48.07 kJ
Temperature rise so exothermic
DH
=
- 48.07 kJ mol-1