1. Industrial Chemistry
Hess’s law

2. Index Hess’s Law Hess’s Law and its experimental verification
Hess’s Law calculations, 4 examples.

3. “enthalpy change is independent of the route taken”
Hess’s Law and calculations
Hess’s law states that
“enthalpy change is independent of the route taken”

4. Verification of Hess’s Law
H = enthalpy change
NaOH (s)
NaCl (aq)
Route 1
H 1
NaOH (s)
NaOH (aq)
NaCl (aq)
Route 2
H 2
H 3
The conversion of solid NaOH to NaCl solution can be achieved by two
possible routes. Route 1 is a single-step process, (adding HCl (aq) directly
to the solid NaOH) and Route 2 is a two-step process (dissolve the solid NaOH in
water, then adding the HCl(aq)) All steps are exothermic.
If Hess’s Law applies, the enthalpy change for route 1 must be the same as the
overall change for route 2.
H 1 =
H 2
H 3 +

5. Verification of Hess’s Law
Route 1 H 1
50 ml 1mol l-1 HCl
Route 2 H 2 + H 3
50 ml H2O
50 ml HCl
then
2.50g of NaOH added to a dry, insulated
beaker.
Before adding the acid, its temperature
is recorded. The final temperature
after adding the acid is also recorded.
g of NaOH added to a dry, insulated
beaker.
2. Before adding the water, its temperature
is recorded. The final temperature rise
after adding the water is also recorded.
3. Now add the acid, again, recording the final
temperature.
Use the equation below to calculate H2 and H 3
H 2
Knowing the specific heat capacity for
water, it is then possible to calculate the
Enthaply change for this reaction.
H 3
H = c m T
H 1 = c m T
H 1 = H 2 + H 3 will verify Hess’s Law

6. Hess’s Law Calculations
Hess’s Law can be used to calculate enthalpy changes that cannot be directly measured by experiment.
Route 1
3C (s) + 3H2 (g)
C3H6 (g)
3CO2 (g) + 3H2O (l)
Route 2a
Route 2b
Route 1 cannot be carried out
in a lab, as carbon and hydrogen
will not combine directly.
The enthalpy of combustion reactions can act as a stepping stone which
enables a link with carbon and hydrogen (the reactants) with propene
(the product)
Route 2a involves the combustion of both carbon and hydrogen
3C (s) + 3O2 (g)  3CO2 (g)
and
3H2 (g) + 1½O2 (g)  3H2O (l)
Route 2b involves the reverse combustion of propane
3CO2 (g) + 3H2O(l)  C3H6 (g) + 4½O2 (g)

7. Example 1 + H1 H 1 = H 2a H 2b + H 1 = + = + 18.5 kJ mol -1
3C (s) + 3H2 (g)
C3H6 (g)
Route 1
3CO2 (g) + 3H2O(l)
Route 2a
Route 2b 3 3
H 1 =
H 2a
H 2b +
Route 2a
H c C= -394 kJ mol –1
H c H = -286 kJ mol –1
H 2a = kJ
H 2a
= -( x 394) = kJ mol -1
-( x 286) = -858 kJ mol -1 + 3 3
Route 2b
H 2b
= kJ (note the reverse sign)
H c Propene = kJ mol –1
H 1 =
-2040 kJ +
( )
= kJ mol -1

8. Alternative approach for example 1
C(s) + O2 (g)  CO2(g) ΔHo298 = -394 kJ mol-1
H2(g) + ½O2(g)  H2O(g) ΔHo298 = -286 kJ mol-1
C3H6(g) + 4½O2(g)  3H2O(g) + 3CO2(g)ΔHo298 = kJ mol-1
3C(s) H2 (g)  C3H6(g)
ΔHf = ?
Re-write the equations so that the reactants and products are on the same side of the “arrow” as the equation you are interested in. Multiply each equation so that there are the same number of moles of each constituent also.
3C(s) + 3O2 (g)  3CO2(g)
ΔHc = 3 x -394 kJ
3H2(g) + 1½O2(g)  3H2O(g)
ΔHc = 3 x -286 kJ
3H2O(g) + 3CO2(g)  C3H6(g) + 4½O2(g)
ΔHc = kJ
Equation has been reversed; (enthalpy now has opposite sign)

9. 3C(graphite) + 3O2 (g)  3CO2(l) ΔHc = 3 x -394
3H2(g) + 1½O2(g)  3H2O(l)
ΔHc = 3 x -286
3H2O(g) + 3CO2(g)  C3H6(g) + 4½O2(l)
ΔHc =
Now add the equations and also the corresponding enthalpy values
3C(s) + 3H2(g)  C3H6(g)
ΔHf = (3 x -394) + (3 x -286) + ( )
ΔHf = kJ mol-1

10. Example 2 Calculate the enthalpy change for the reaction:
C6H6(l) + 3H2 (g)  C6H12(l)
Route 1 ?
6H2O(l) + 6CO2(g)
Route 2a
Route 2b 3
The products of combustion act as a stepping stone which enables a link to be
made with benzene and hydrogen (the reactants) with cyclohexane (the product).
Route 2a involves the combustion of both benzene and hydrogen
C6H6(g) + 7½O2(g)  3H2O(l) + 6CO2(g)
H c benzene = kJ mol –1
H2(g) + ½O2(g)  H2O(l)
H c hydrogen = -286 kJ mol –1
Route 2b involves the reverse combustion of cyclohexane
6CO2 (g) + 6H2O(l) => C6H12 (g) + 7½O2 (g)
H c cyclohexane = kJ mol –1
H 1 = H 2a + H 2b
= ( ( x - 286)) = 202 kJ mol -1 3

11. Alternative approach for example 2
C6H12(l) + 9O2 (g)  6H2O(l) + 6CO2(g) ΔH = kJmol-1
H2(g) + ½O2(g)  H2O(l) ΔH = -286 kJmol-1 C6H6(l) + 7½O2(g)  3H2O(l) + 6CO2(g) ΔH = kJmol-1
C6H6(l) H2(g)  C6H12(l)
ΔHf = ?
Re-write the equations so that the reactants and products are on the same side of the “arrow” as the equation you are interested in. Multiply each equation so that there are the same number of moles of each constituent also.
C6H6(l) + 7½O2 (g)  6CO2(g) + 3H2O(g)
ΔHc = kJ
3H2(g) + 1½O2(g)  3H2O(l)
ΔHc = 3 x -286 kJ
6H2O(g) + 6CO2(g)  C6H12(g) + 9O2(g)
ΔHc = kJ
Equation has been reversed; (enthalpy now has opposite sign)

12. C6H6(l) + 7½O2 (g)  6CO2(g) + 3H2O(g) ΔHc = -3268
3H2(g) + 1½O2(g)  3H2O(g)
ΔHc = 3 x -286
6H2O(g) + 6CO2(g)  C6H12(l) + 9O2(g)
ΔHc =
Now add the equations and also the corresponding enthalpy values
C6H6(l) + 3H2(g)  C6H12(l)
ΔH f = (3 x -286)
ΔH f = -202 kJ mol-1

13. 3. Use the enthalpy changes of combustion shown in the table to work out the enthalpy change of formation of ethyne, C2H2.
Substance
C(graphite)
H2(g)
C2H2(g)
ΔHo(combustion)
-394 kJmol-1
-286 kJmol-1
-1299 kJmol-1
“Using the Second method”
“Required” equation, 2C(s) + H2(g)  C2H2(g) ΔHf = ?
2C(s) + 2O2(g)  2CO2(g) ΔHc = 2 x -394 kJ mol-1
H2(g) + ½O2(g)  H2O(g) ΔHc = kJ mol-1
C2H2(g) + 2½O2(g)  2CO2(g) + H2O(g) ΔHc = kJ mol-1
2CO2(g) + H2O(g)  C2H2(g) + 2½O2(g) ΔHc = kJ mol-1
Adding “bulleted” equations gives us 2C(graphite) + H2(g)  C2H2(g)
ΔHf = (2 x -394) + (-286)
ΔHf = kJ mol-1

14. ΔHc = + 278 + (2 x -394) + (3 x -286) ΔHc = -1368 kJ mol-1
4. Using the following standard enthalpy changes of formation,
ΔHof / kJmol-1 : CO2(g), -394; H2O(g), -286; C2H5OH(l), -278 calculate the standard enthalpy of combustion of ethanol i.e. the enthalpy change for the reaction C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
“Using the Second method”
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l) ΔHc = ?
2C(s) + 3H2(g) + ½O2(g)  C2H5OH(l) ΔHf = -278 kJ
● 2C(s) + 2O2(g)  2CO2(g) ΔHf = 2 x -394 kJ
● 3H2(g) ½O2(g)  3H2O (g) ΔHf = 3 x -286 kJ
● C2H5OH(l)  2C(s) + 3H2(g) + ½O2(g) ΔHf = +278 kJ
Add bulleted equations
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
Solve equation for ΔHc
ΔHc = (2 x -394) + (3 x -286)
ΔHc = kJ mol-1