Industrial Chemistry

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Industrial Chemistry

• Hesss law

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Index
Hesss Law
Hesss Law and its experimental verification
Hesss Law calculations, 4 examples.
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Hesss Law and calculations
Hesss law states that enthalpy change is
independent of the route taken
www 11.11
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Verification of Hesss Law
?H enthalpy change
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 Hesss Law applies, the enthalpy change for
route 1 must be the same as the overall change
for route 2.
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Verification of Hesss Law
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.
1. 2.50g 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 1 ?H 2 ?H 3 will verify Hesss Law
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Hesss Law Calculations
Hesss Law can be used to calculate enthalpy
changes that cannot be directly measured by
experiment.
Route 1
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
and
3C (s) 3O2 (g) ? 3CO2 (g)
3H2 (g) 1½O2 (g) ? 3H2O (l)
Route 2b involves the reverse combustion of
propane
3CO2 (g) 3H2O(l) ? C3H6 (g) 4½O2 (g)
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?H1
Example 1
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3
3
3
?H 1
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Alternative approach for example 1
C(s) O2 (g) ? CO2(g) ?Ho298 -394
kJ mol-1 H2(g) ½O2(g) ? H2O(g) ?Ho298
-286 kJ mol-1 C3H6(g) 4½O2(g) ? 3H2O(g)
3CO2(g)?Ho298 -2058.5 kJ mol-1
3C(s) 3H2 (g) ? C3H6(g)
?Hf ?
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)
?Hc 3 x -394 kJ
3H2(g) 1½O2(g) ? 3H2O(g)
?Hc 3 x -286 kJ
3H2O(g) 3CO2(g) ? C3H6(g) 4½O2(g)
?Hc 2058.5 kJ
Equation has been reversed (enthalpy now has
opposite sign)
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Now add the equations and also the corresponding
enthalpy values
3C(s) 3H2(g) ? C3H6(g)
?Hf (3 x -394) (3 x -286) (2058.5)
?Hf 18.5 kJ mol-1
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Calculate the enthalpy change for the reaction
Example 2
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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).
3
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Alternative approach for example 2
C6H12(l) 9O2 (g) ? 6H2O(l) 6CO2(g) ?H
-3924 kJmol-1 H2(g) ½O2(g) ? H2O(l)
?H -286 kJmol-1C6H6(l) 7½O2(g) ? 3H2O(l)
6CO2(g) ?H -3268 kJmol-1
C6H6(l) 3H2(g) ? C6H12(l)
?Hf ?
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)
?Hc -3268 kJ
3H2(g) 1½O2(g) ? 3H2O(l)
?Hc 3 x -286 kJ
6H2O(g) 6CO2(g) ? C6H12(g) 9O2(g)
?Hc 3924 kJ
Equation has been reversed (enthalpy now has
opposite sign)
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Now add the equations and also the corresponding
enthalpy values
C6H6(l) 3H2(g) ? C6H12(l)
?H f -3268 (3 x -286) 3924
?H f -202 kJ mol-1
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3. Use the enthalpy changes of combustion shown
in the table to work out the enthalpy change of
formation of ethyne, C2H2.
Using the Second method
Required equation, 2C(s) H2(g) ? C2H2(g)
?Hf ?

• 2C(s) 2O2(g) ? 2CO2(g) ?Hc
  2 x -394 kJ mol-1

• H2(g) ½O2(g) ? H2O(g) ?Hc -286
  kJ mol-1

C2H2(g) 2½O2(g) ? 2CO2(g) H2O(g) ?Hc
-1299 kJ mol-1

• 2CO2(g) H2O(g) ? C2H2(g) 2½O2(g) ?Hc
  1299 kJ mol-1

Adding bulleted equations gives us
2C(graphite) H2(g) ? C2H2(g)
?Hf (2 x -394) (-286) 1299
?Hf 225 kJ mol-1
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4. Using the following standard enthalpy changes
of formation, ?Hof / 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
2C(s) 3H2(g) ½O2(g) ? C2H5OH(l) ?Hf -278
kJ
? 2C(s) 2O2(g) ? 2CO2(g) ?Hf 2
x -394 kJ
? 3H2(g) 1½O2(g) ? 3H2O (g) ?Hf 3 x
-286 kJ
? C2H5OH(l) ? 2C(s) 3H2(g) ½O2(g)
?Hf 278 kJ
Add bulleted equations
C2H5OH(l) 3O2(g) ? 2CO2(g) 3H2O(l)
Solve equation for ?Hc
?Hc 278 (2 x -394) (3 x -286)
?Hc -1368 kJ mol-1