ΔH^{Ψ}_{f} (methane) =
??? kJmol^{1} 
C_{(s)}
+ 2H_{2(g) } CH_{4(g)} 
ΔH^{θ}_{c}(C_{(s)})
+ 2 x ΔH^{θ}_{c}(H_{2(g)})
= (393)
+ (2 x 286)
C => CO_{2}
and 2H_{2} => 2H_{2}O 
+2O_{2(g) } 
+2O_{2(g)} 
ΔH^{θ}_{c}(CH_{4(g)})
= 890
OR change the arrow round and change the signs as below,
remember, change direction, you change the sign BUT not the
numerical energy value! 
OR
ΔH^{θ}_{f}(CO_{2(g)})
+ 2 x ΔH^{θ}_{f}(H_{2}O_{(l)})
same numbers due to coincidence
of enthalpy names 
+2O_{2(g) } 
2O_{2(g) } 
= ΔH^{θ}_{c}(CH_{4(g)})
= (890) = +890 
CO_{2(g)}
+ 2H_{2}O_{(l)}
Note that you double the enthalpy of
combustion of hydrogen to make the cycle balance in molar terms 
The cycle for the standard enthalpy of formation of
methane, ΔH^{θ}_{f}
From Hess's Law, add
up the sequence of enthalpy changes via the lower 'staged route'
to get the overall enthalpy change for the enthalpy of formation
of methane from its elements in their normal stable states.
ΔH^{θ}_{f} (methane) = (393) + (2 x 286)
+ (+890) = 75 kJmol^{1}

2nd example of Hess's Law
Cycle calculation
From the
following thermochemical data
(1)
^{1}/_{2}H_{2(g)} + ^{1}/_{2}Cl_{2(g)}
==> HCl_{(g)} ΔH^{θ}_{f}(hydrogen
chloride) = 92.3 kJmol^{1}
(2)
2C_{(s)}
+ 3H_{2(g)} + ^{1}/_{2}O_{2(g)} ==>
CH_{3}CH_{2}OH_{(l)} ΔH^{θ}_{f}(ethanol)
= 278.0 kJmol^{1}
(3)
2C_{(s)}
+ 1^{1}/_{2}H_{2(g)} + ^{1}/_{2}O_{2(g)}
+ ^{1}/_{2}Cl_{2(g)} ==> CH_{3}COCl_{(l)} ΔH^{θ}_{f}(ethanoyl
chloride) = 275.0 kJmol^{1}
(4)
4C_{(s)}
+ 4H_{2(g)} + O_{2(g)} ==> CH_{3}COOCH_{2}CH_{3(l)} ΔH^{θ}_{f}(ethanol)
= 481.0 kJmol^{1}
Calculate the
enthalpy change for the reaction
(5)
CH_{3}COCl_{(l)}
+ CH_{3}CH_{2}OH_{(l)} ==> CH_{3}COOCH_{2}CH_{3(l)}
+ HCl_{(g)} ΔH^{θ}_{r@}(esterification)
= ??? kJmol^{1}
ΔH^{Ψ}_{reaction}(esterification)
= ??? kJmol^{1} 
CH_{3}COCl_{(l)}
+ CH_{3}CH_{2}OH_{(l)
} CH_{3}COOCH_{2}CH_{3(l)}
+ HCl_{(g)} 
ΔH^{θ}_{f}(CH_{3}COCl_{(l)})
+ {ΔH^{θ}_{f}(CH_{3}CH_{2}OH_{(l)})}
= (275.0) + {(278.0)}




ΔH^{θ}_{f}(HCl_{(g)})
+ ΔH^{θ}_{f}(CH_{3}COOCH_{2}CH_{3(l)})
= (92.3) + (481.0) 
4C_{(s)}
+ 4^{1}/_{2}H_{2(g)} + O_{2(g)} +
^{1}/_{2}Cl_{2(g)}
from {2C_{(s)}
+ 1^{1}/_{2}H_{2(g)} +
^{1}/_{2}O_{2(g)} +
^{1}/_{2}Cl_{2(g)}} + {2C_{(s)}
+ 3H_{2(g)} +
^{1}/_{2}O_{2(g)}}
Note that you didn't have to double
or treble etc any enthalpy value because the molar ratio is 1 :
1 ==> 1 : 1 
Adding up all the ΔH^{θ}'s
in the lower route of the cycle
ΔH^{θ}_{298}(esterification) = (+275.0) + (+278.0)
+ (92.3) + (481.0) = 20.3 kJmol^{1}
Note that the sign of enthalpy of
formation of ethanoyl chloride and ethanol is reversed to fit in
with the direction of change.

TOP OF PAGE
3rd example of Hess's Law
Cycle calculation
Problem solving from the
following thermochemical data:
1.2b(ii) Solving enthalpy problems using an
'algebraicequation style' method
BUT, remember, the question might specify
solving the problem via a Hess's Law cycle!
Given the following data
(1)
C_{(s)} + O_{2(g)} ==> CO_{2(g)
} ΔH^{θ} = 393 kJmol^{1}
(2)
H_{2(g)} + ^{1}/_{2}O_{2(g)} ==> H_{2}O_{(l)}
ΔH^{θ} = 286 kJmol^{1}
(3)
3C_{(s)} + 4H_{2(g)}
==> C_{3}H_{8(g)} ΔH^{θ} =
104 kJmol^{1}
Calculate the standard
enthalpy of combustion of propane
(4)
C_{3}H_{8(g)} + 5O_{2(g)} ==> 3CO_{2(g)} +
4H_{2}O_{(l)} ΔH^{θ}_{c,298}(propane)
= ??? kJ mol^{1}
What you do is rearrange, if
necessary, the data equations and add up the results both equation
components and delta H values, cancelling out the equation
components should leave you with the correct equation whose enthalpy
value you require.
3C_{(s)}
+ 3O_{2(g)} ==> 3CO_{2(g)} 
(1) 
add ΔH^{θ} = 3 x
393 kJmol^{1} 
4H_{2(g)}
+ 2O_{2(g)} ==> 4H_{2}O_{(l)} 
(2) 
plus ΔH^{θ} = 4 x
286 kJmol^{1} 
C_{3}H_{8(g)} ==>
3C_{(s)} + 4H_{2(g)}

(3) 
plus ΔH^{θ}
= +104 kJmol^{1} (equation and sign reversed) 
C_{3}H_{8(g)} + 5O_{2(g)} ==> 3CO_{2(g)} +
4H_{2}O_{(l)} 
(4) 
= ΔH^{θ}_{c,298}(propane)
= 2219 kJ mol^{1} 
adding up
(1) + (2) + (3) gives (4),
cancelling out
unwanted equation components, ==> treated as an = sign 
Note that here you are using factors
of 3 and 4 to make the cycle balance in molar terms
2nd example
of 'algebraic' calculation style
Given the
following thermochemical data:
(1)
^{1}/_{2}H_{2(g)} + ^{1}/_{2}Cl_{2(g)}
==> HCl_{(g)} ΔH^{θ} = 92.3
kJmol^{1}
(2)
2C_{(s)}
+ 3H_{2(g)} ==> CH_{3}CH_{3(g)} ΔH^{θ}
= 84.7 kJmol^{1}
(3)
2C_{(s)}
+ 2H_{2(g)} + Cl_{2(g)} ==> ClCH_{2}CH_{2}Cl_{(l)} ΔH^{θ}
= 166.0 kJmol^{1}
Calculate the
enthalpy change for the reaction
(4)
2Cl_{2(g)}
+ CH_{3}CH_{3(g)} ==> ClCH_{2}CH_{2}Cl_{(l)}
+ 2HCl_{(g)} ΔH^{θ} = ???
kJmol^{1}
H_{2(g)}
+ Cl_{2(g)} ==> 2HCl_{(g)} 
(1) 
add ΔH^{θ} = 2 x
92.3 kJmol^{1} 
CH_{3}CH_{3(g)}
==> 2C_{(s)} +
3H_{2(g)} 
(2) 
plus ΔH^{θ} =
+84.7 kJmol^{1} (equation and sign reversed) 
2C_{(s)} + 2H_{2(g)}
+ Cl_{2(g)} ==> ClCH_{2}CH_{2}Cl_{(l)} 
(3) 
plus ΔH^{θ}
= 166 kJmol^{1} 
2Cl_{2(g)}
+ CH_{3}CH_{3(g)} ==> ClCH_{2}CH_{2}Cl_{(l)}
+ 2HCl_{(g)} 
(4) 
= ΔH^{θ}_{298}(chlorination
reaction)
= 265.9 kJ mol^{1} 
adding up
(1) + (2) + (3) gives (4),
cancelling out
unwanted equation components, ==> treated as an = sign 
Note that you double the enthalpy of
formation hydrogen chloride to make the cycle balance in molar terms
1.2b(iii)
A method using the summation of enthalpies of reactants and products
A very simple example of
an enthalpy
summation method
molecule 
ethanoic acid 
+ ethanol 
==> 
ethyl ethanoate 
+ water 
equation 
CH_{3}COOH_{(l)} 
+ CH_{3}CH_{2}OH_{(l)} 
==> 
CH_{3}COOCH_{2}CH_{3(l)} 
+ H_{2}O_{(l)} 
ΔH^{θ}_{f,298}/kJmol^{1} 
487 
278 

481 
286 
ΔH_{reaction} = ∑H_{products}
∑H_{reactants}
ΔH^{θ}_{reaction,298} =
∑ΔH^{θ}_{f,298}(products)
∑ΔH^{θ}_{f,298}(reactants)
ΔH^{θ}_{esterification}
= {ΔH^{θ}_{f}(ethyl ethanoate) + ΔH^{θ}_{f}(water)}
{ΔH^{θ}_{f}(ethanoic
acid)
+ ΔH^{θ}_{f}(ethanol)}
ΔH^{θ}_{esterification}
= {481 + 286}
{487
+ 278}
ΔH^{θ}(esterification
reaction)
= (767) (765) = 2 kJmol^{1}
What would be ΔH^{θ}(hydrolysis)?
Answer! just reverse the sign! i.e. +2 kJmol^{1}
2nd example of enthalpy
summation method
A more complicated example where you
need to think more about mole ratios in the equation.
Given the following data from
laboratory measurements
(1)
C_{(s)} + O_{2(g)} ==> CO_{2(g)
} ΔH^{θ}_{f,298}(carbon dioxide) =
393 kJmol^{1}
(2)
H_{2(g)}
+ ^{1}/_{2}O_{2(g)} ==> H_{2}O_{(l)}
ΔH^{θ}_{f,298}(water) = 286
kJmol^{1}
(3)
C_{4}H_{10(g)} +
6^{1}/_{2}O_{2(g)} ==> 4CO_{2(g)} +
5H_{2}O_{(l)} ΔH^{θ}_{c,298}(butane)
= 2877 kJ mol^{1}
Calculate the standard enthalpy
of formation of butane gas, which cannot be determined by
experiment.
(4)
4C_{(s)}
+ 5H_{2(g)}
==> C_{4}H_{10(g)} ΔH^{θ}
= ??? kJmol^{1}
To solve this you can use
equation (3) and the data from equations (1) and (2) to obtain a
value for equation (4)
ΔH_{reaction} = ∑H_{products}
∑H_{reactants}
for equation (3)
ΔH_{combustion}(butane) =
∑ΔH^{θ}_{f}(products)
∑ΔH^{θ}_{f}(reactants)
ΔH^{θ}_{c,298}(butane)
= {4 x ΔH^{θ}_{f}(carbon dioxide) + 5 x ΔH^{θ}_{f,298}(water)}
{ΔH^{θ}_{f}(butane)
+ ΔH^{θ}_{f}(oxygen)}
Since oxygen is an element, ΔH^{θ}_{f}(oxygen)
= 0, therefore after rearranging we get
ΔH^{θ}_{c,298}(butane)
= {4 x ΔH^{θ}_{f}(carbon dioxide) + 5 x ΔH^{θ}_{f,298}(water)}
{ΔH^{θ}_{f}(butane)}
2877 = {(4 x
393) + (5 x
286)} ΔH^{θ}_{f}(butane)
ΔH^{θ}_{f}(butane)
= 2877 1572 1430 = 125 kJmol^{1}
Enthalpy calculation problems with worked out answers based on
enthalpies of reaction,
formation, combustion
EnergeticsThermochemistryThermodynamics Notes INDEX
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QUICK INDEX for
Energetics:
GCSE Notes on the basics of chemical energy changes
important to study and know before tackling any of the three Advanced Level
Chemistry pages Parts 13 here
* Part 1ab
ΔH Enthalpy Changes
1.1 Advanced Introduction to enthalpy changes
of reaction,
formation, combustion etc. : 1.2a & 1.2b(i)(iii)
Thermochemistry Hess's Law and Enthalpy
Calculations reaction, combustion, formation etc. : 1.2b(iv)
Enthalpy of reaction from bond enthalpy
calculations : 1.3ab
Experimental methods
for determining enthalpy changes and treatment of results and
calculations :
1.4
Some enthalpy data patterns : 1.4a
The combustion of linear alkanes and linear
aliphatic alcohols
:
1.4b Some patterns in Bond
Enthalpies and Bond Length : 1.4c
Enthalpies of
Neutralisation : 1.4d Enthalpies of
Hydrogenation of unsaturated hydrocarbons and evidence of aromatic
ring structure in benzene
:
Extra Q page
A set of practice enthalpy
calculations with worked out answers **
Part 2 ΔH Enthalpies of
ion hydration, solution, atomisation, lattice energy, electron affinity
and the BornHaber cycle : 2.1ac What happens when a
salt dissolves in water and why? :
2.1de Enthalpy
cycles involving a salt dissolving : 2.2ac
The
BornHaber Cycle *** Part 3
ΔS Entropy and ΔG Free Energy Changes
: 3.1ag Introduction to Entropy
: 3.2
Examples of
entropy values and comments * 3.3a ΔS, Entropy
and change of state : 3.3b ΔS, Entropy changes and the
feasibility of a chemical change : 3.4ad
More on ΔG,
free energy changes, feasibility and
applications : 3.5
Calculating Equilibrium
Constants from ΔG the free energy change : 3.6
Kinetic stability versus thermodynamic
feasibility  can a chemical reaction happen? and will it happen?
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