LECTURE 4
Variation of enthalpy with temperature
So far, we can only work at 25 °C.
Like cv we define a constant pressure heat capacity, cp, as the amount of heat energy needed
to raise the temperature of 1 mole of substance by 1 K at constant pressure.
 dq p 
 dH 
 =
cp = 


 p or
dT
dT

p
dH = cp dT
Thus, H must increase by (cp dT) if heated by dT.
T2
∫ dH = ∫ c
p
dT
the Kirchoff Equation
T1
First approximation: treat cp as independent of T for small temp. ranges or use the average
value over the range of T
H T − H T = c p (T2 − T1 )
2
1
Particularly useful: recall that standard enthalpies of formation are defined at 25 °C. Hence, we
can write
∆H°f, T = ∆H°f, 298 + c p ( T - 298.15 )
Enthalpies of Reaction
A model reaction:
R → P
The enthalpy of R at 25 °C is ∆H°f, 298 (R). The enthalpy at some other temperature T is [∆H°f,
298 (R) + c p (R) (T - 298.15)].
∆H°T (reaction) = [∆H°f, T (P) ] – [∆H°f, T (R)]
= [∆H°f, 298 (P) + c p (P) (T–298.15)]
– [∆H°f, 298 (R) + c p (R) (T–298.15)]
= [∆H°f, 298(P) - ∆H°f, 298(R)} + {[ c p (P) - c p (R)] (T–298.15)}
or
∆H°T (reaction) = ∆H°298 (reaction) + ∆ c p (T - 298.15)
where
∆cp =
∑ν
i
cp (products) -
∑ν
i
cp (reactants)
The complete combustion of ethane releases 1558.8 kJ mol-1 at 25 °C. Calculate
∆H°(combustion) at 100 °C.
cp / J K-1 mol-1: C2H6(g) 52.6; O2(g) 29.4; CO2(g) 37.1; H2O(l) 75.3
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C2H6(g) + 3½ O2(g) → 2 CO2(g) + 3 H2O(l)
∆H°298 = - 1558.8 kJ mol-1
∆cp = [2 cp (CO2) + 3 cp (H2O)] - [cp (C2H6) + 3½ cp (O2)]
= [2 (37.1) + 3 (75.3)] - [ 52.6 + 3½ (29.4)]
= 144.6 J K-1 mol-1
∆H°373 = ∆H°298 + ∆cp ∆T
= - 1558.8 x 103 (J mol-1) + (144.6 J K-1 mol-1 )(75 K)
= - 1547.9 kJ mol-1
.
Other types of enthalpy change
Enthalpy of combustion - often referred to as the ‘Heat of combustion’
the enthalpy change when 1 mol of a compound reacts completely with excess oxygen
gas.
e.g. the combustion of 1 mol of methane gas at 25 °C evolves 890.3 kJ.
CH4(g)+ 2O2(g) → CO2(g)+ 2H2O(l) ∆H°298(comb.) = -890.3 kJ mol-1
∆H°298(combust) are easy to measure and often used to determine ∆H°f,298 values.
e.g. combustion of methane
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
∆H° (comb) = [2∆H°f(H2O) + ∆H°f(CO2)]-[∆H°f(CH4) + 2∆H°f(O2)]
-890.3 = [2 (-285.8) + (-393.5] - [∆H°f (CH4) + 2 (0)]
-890.3 = -965.1 - ∆H°f (CH4)
∆H°°f (CH4) = 74.8 kJ mol-1
The bomb calorimeter - already seen for ∆U(combustion) at constant volume
∆H(combustion) = ∆U(combustion) + ∆(pV) (combustion)
If the gases behave ideally
∆(pV) = (pV)prod - (pV)react
=
RT {(ngas)prod - (ngas)react }
∆(pV) = ∆ngas R T
∆H(comb.) = ∆U(comb.) + ∆ngas R T
where ngas is the number of moles of gas involved in the reaction.
Why do we only consider gases?
Volume of 1 mol of ideal gas: 25 °C, 1 atm = 24470 cm3 .
At 25 °C: water occupies ~18 cm3 mol-1; hexane 132 cm3 mol-1, benzene 89 cm3 mol-1 and
naphthalene ∼120 cm3 mol-1.
∴ Volumes of solids and liquids are negligible.
An example
Following the earlier calculation, calculate the enthalpy of combustion of naphthalene at 25 °C.
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From earlier, ∆U(comb.) for naphthalene at 25 °C = - 5142 kJ mol-1
C10H8 (s) + 12 O2 (g) → 10 CO2 (g) + 4 H2O (l)
To convert ∆H to ∆U using Eqn. 2.a, we need ∆ngas.
∆ngas = ngas (products) - ngas (reactants) = 10 - 12 = -2 mol
∆H(comb.) = - 5142000 J mol-1 + (-2 mol) (8.314 J K-1 mol-1) (298.2 K)
∆H(combustion) = - 5147000 J mol-1 = - 5147 kJ mol-1
Exercise: Take the values from Example 2.8 and calculate the error in neglecting the volumes of
naphthalene and water.
Note that the difference between the internal energy and enthalpy is relatively small in this case.
Enthalpies of phase change
Consider S, L, G for now.
solid to liquid: melting or fusion.
Liquid molecules have greater motion than solid - energy has to be added to the solid - fusion
is an endothermic process.
The enthalpy of fusion, ∆Hfus is the energy required at 1 bar pressure to melt 1 mole of a
pure component at its melting point, Tm .
An earlier name, still in common usage is that ∆Hfus is the latent heat of fusion. It essentially
represents the energy needed to overcome the intermolecular forces in the solid,
Also,
∆H(freezing) = - ∆H(fusion)
liquid → vapour - vaporization.
∆Hvap is the energy required at 1 bar pressure to vaporize 1 mole of a pure liquid at its
boiling point, Tb.
Vaporisation is also endothermic. ∆Hvap is considerable larger than ∆Hfus since the molecules
are completely separated in vaporization whereas a considerable degree of intermolecular
interaction remains in liquids.
Solid → vapour - sublimation.
Some compounds do not display a liquid phase an example is solid carbon dioxide which, at
atmospheric pressure, sublimates at -78 °C.
∆H(sublimation) = ∆H(fusion) + ∆H(vaporization)
Examples
Calculate the energy required to turn 100 g of ice at 0 °C into steam at 100 °C.
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The process required can be conveniently split into three parts.
1. 100 g ice → 100 g water at 0 °C
2. 100 g water, 0 °C → 100 g water, 100 °C
3. 100g water, 100 °C → 100g steam, 100 °C
∆H1 = ∆Hfus for 100 g
∆H2 = cp ∆T for 100 g
∆H3 = ∆Hvap for 100 g
100 g of water = 100 / 18 mol = 5.56 mol
∆Htotal = ∆H1 + ∆H2 + ∆H3
= 5.56 mol [ + 6.01 x 103 + 75.1 (373 - 273) + 40.7 x 103]
= 301.46 kJ (100g)-1
Note that the factor of 103 occurs in ∆H1 and ∆H3 since they are measured in kJ mol-1 while cp is in J K-1
mol-1.
A gas mixture consisting of 25% nitrogen and 75% hydrogen (by volume) was passed over a catalyst at a rate of 112 dm3
min-1 (Gas volumes corrected to 0 °C and 1 atm pressure). The reaction took place at 450 °C and 1 bar pressure and
complete conversion to ammonia was achieved. Given the following data, calculate the rate of heat evolution or
absorption.
∆H°f, 298 (NH3) - 46.0 kJ mol-1
Mean cp / J K-1 mol-1: N2 29.7; H2 29.3; NH3 39.7.
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N2(g) + 3 H2(g) → 2 NH3(g)
The enthalpy change at 25 °C can be calculated
∆H°298
= [ 2 x ∆H°f, 298 (NH3)] - [∆H°f, 298 (N2) + 3 ∆H°f, 298 (H2)]
= [ - 92.0 ] - [ 0 ] kJ mol-1
∆H°°298 = - 92.0 kJ mol-1
Correcting the enthalpy change to 450 °C makes use of the Kirchoff Eqn.
∆H°T (reaction) = ∆H°298 (reaction) + ∆ c p (T - 298.15)
∆cp = [2 x c p (NH3)] - [ c p (N2) + 3 x c p (H2)} = [2 x 39.7] - [ 29.7 + (3 x 29.3)]
= - 38.2 J K-1 mol-1
∆H°723 = - 92000 Jmol-1 + (- 38.2 J K-1 mol-1 ) (723.15 - 298.15)K
= - 92000 Jmol-1 + ( - 16235 Jmol-1 )
∆H°°723 = - 108.24 kJ mol-1
Hence, at 450 °C, 108.24 kJ is evolved for each mole of the equation as written. From the stoichiometry, this means that
108.24 kJ is evolved per mole of nitrogen which reacts. Since, for a gas, the volume is proportional to the number of
moles, the 1:3 mixture is in the stoichiometric amount. We know that at 0 °C and 1 atm pressure, 1 mole of an ideal gas
occupies 22.4 dm3. Thus, the total amount of gas passing over the catalyst amounted to 5 moles per minute, of which
1.25 moles were N2.
Rate of heat evolution = (moles of N2 min-1 ) (heat evolved per mole N2)
= ( 1.25 mol min-1 ) (- 108.24 kJ (mol N2-1 )
= - 135.3 kJ min-1
The negative sign indicates an exothermic reaction so that heat is evolved.
Using the data from a text, estimate the standard enthalpies of formation for (a) gaseous cyclohexane, C6H12 and (b)
gaseous benzene, C6H6. Compare the true values in a data table.
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(a) The formation of cyclohexane can be written
6 C(graph) + 6 H2(g) → C6H12(g) ∆H°1 = ∆H°f(C6H12(g))
To form cyclohexane, we need to “make” 6 C–C bonds and 12 C–H bonds. From the definition of bond energies, this
refers to the reaction
6 C(g) + 12 H(g) → C6H12(g) ∆H°2 = 6 (-344) + 12 (-415) = -7044 kJ mol-1
the values being negative since this is the reverse of bond dissociation.
Prior to this step, we need
6 C(graph) + 6 H2(g) → 6 C(g) + 12 H(g) ∆H°3 = 6 ∆H(graph.) + 6 ∆H(H–H)
∆H°3 = 6 (716.7) + 6 (436) = 6916.2 kJ mol-1.
It is clear that ∆H°°1 = ∆H°°2 +∆
∆H°°3 = -7030 + 6916.2 = -127.8 kJ mol-1
The actual value is -123.2 kJ mol-1. The agreement is therefore very good considering that average bond energies
have been used throughout.
(b) This calculation is performed in an analogous manner to part (a) except that benzene contains 6 C–H, 3 C–C and 3
C=C bonds.
6 C(graph) + 3 H2(g) → C6H6(g)
∆H°1 = ∆H°f(C6H12(g))
6 C(g) + 6 H(g) → C6H6(g) ∆H°2 = 3 (- 615) + 3 (-344) + 6 (-415) = -5367 kJ mol-1
6 C(graph) + 3 H2(g) → 6 C(g) + 6 H(g) ∆H°3 = 6 ∆H(graph.) + 6 ∆H(H–H)
∆H°3 = 6 (716.7) + 3 (436) = 5608.2 kJ mol-1.
It is clear that ∆H°°1 = ∆H°°2 +∆
∆H°°3 = -7030 + 6916.2 = 241.2 kJ mol-1
The actual value is 83 kJ mol-1. Clearly, there is poor agreement for benzene. This is due to the assumption of the
alternating single and double bonds in benzene whereas the ‘actual’ structure is a resonance stabilised hybrid of these.
Note: This illustrates that care must be taken to consider the chemical properties of the molecules involved when using
average bond enthalpy data. At best approximate results can be expected.
The complete combustion of 2.0026 g of sucrose (C12H22O11) in a bomb calorimeter with a heat capacity of 11140 JK-1 at
25 °C resulted in a temperature increase of 2.966 °C. Calculate ∆Hcombustion for sucrose.
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The energy released on combustion is given by (11140 JK-1 × 2.966 K) = 33.041 kJ
Since the bomb calorimeter is a constant volume device, this gives ∆Ucombustion. The experiment used 5.85 x 10-3 moles.
Hence,
∆Ucombustion = - 33.041 / 5.85 x 10-3 = - 5647.5 kJ mol-1
The chemical equation for the reaction is C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(l)
so the number of moles of gas is the same before and after the reaction. Hence, ∆ngas = 0 and ∆Hcombustion = ∆Ucombustion
∴ ∆Hcombustion = - 5647.5 kJ mol-1