Constant-Pressure Calorimetry
Atmospheric pressure is constant
ΔH = qP
qsystem = -qsurroundings
System = reaction or process conducted in the water
Surroundings = water in the calorimeter + calorimeter.
Assume insulator prevents heat transfer to the room.
qsystem =-qsurroundings = - (qwater + qcalorimeter)
For most calculations, the qcalorimeter can be ignored
qsystem = - qwater
csystemmsystem ΔTsystem = - cwatermwater ΔTwater
Example
3.88 g of ammonium nitrate is dissolved in 60.0 g of water
in a coffee cup calorimeter, that is initially at 23.0 oC. As
the ammonium nitrate dissolves the temperature goes to
18.4 oC. What is the molar ΔH for dissolving ammonium
nitrate?
NH4NO3(s)  NH4+(aq) + NO3-(aq)
ΔTsolution = 18.4oC – 23.0oC = -4.6oC
mwater = 60.0g
msolution = 60.0 g + 3.88 g = 63.88 g
cwater = 4.184J/goC = cfinal solution (by assumption)
qsolution = csolutionmsolution ΔTsolution
qsolution = 4.184J/goC)(63.88g)(-4.6oC) = -1229J
mNH4NO3 = 3.88g
molesNH4NO3 = 3.88g/80.032g/mol = 0.04848mol
molar enthalpy for dissolving = ΔH = qNH4NO3/molesNH4NO3
q NH4NO3 = -qsolution
ΔH = +1229J/0.04848mol
ΔH = 25.4 kJ/mol
Bomb Calorimetry (Constant-Volume Calorimetry)
- For combustion reactions
- Substance placed in a sealed “bomb” and filled to a
high pressure of O2
- Substance ignited by a spark
- Heat from the combustion is transferred to the water
- Must account for heat capacity of the calorimeter
(includes heat capacity of water). Value of Ccalorimeter is
found by calibration (usually by manufacturer)
qrxn = -Ccalorimeter(ΔT)
A 1.800g sample of octane, C8H18, was burned in a bomb
calorimeter whose total heat capacity is 11.66 kJ/oC. The
temperature of the calorimeter plus contents increases from
21.36oC to 28.78oC. What is the enthalpy of combustion
per gram of octane? Per mole of octane?
2 C8H18 + 25O2  16 CO2 + 18 H2O
ΔTwater = 28.78oC – 21.36oC = 7.42oC
Ccal = 11.66kJ/oC
msample = 1.80g
qrxn = -Ccal (ΔTwater)
qrxn = -11.66kJ/oC(7.42oC) = -86.52kJ
ΔHcombustion(in kJ/g)
ΔHcombustion = -86.52kJ/1.80g = -48.1 kJ/g
ΔHcombustion(in kJ/mol)
molesC8H18 = 1.80g(1mol/114.2302g)=0.01575 mol
ΔHcombustion = -86.52kJ/0.01575mol = -5492 kJ/mol
Hess’s law - If a reaction is carried out in a series of steps,
H for the overall reaction is the sum of H’s for each
individual step.
Example:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
2H2O(g)  2H2O(l)
H = -802 kJ
H = -88 kJ
Sum:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H = -890 kJ
To sum reactions: Sum things separately on each side of
the arrows. Cancel out things that appear equally left and
right of the arrow. It's just like summing math equations.
Remember........enthalpies are extensive, so they refer to
the reactions as written.
H2O(g)  H2O(l)
H = -44 kJ
compare to above
Standard Enthalpy of Formation (ΔHf0 )
The enthalpy change for the formation of one mole of a
substance from the most stable forms of its component
elements at 298 Kelvin, 1 bar pressure, 1 mol/L
The standard enthalpy of formation of the most stable
form on any element is zero
Many ΔHf0 values have been tabulated.
Using ΔHf0’s to Calculate Enthalpies of Reaction
For a reaction:
o
H rxn
  nH of products    mH of reactants 
Where n and m are the stoichiometric coefficients of
each reactant or product from the balanced chemical
equation.
Example
Calculate the enthalpy change for the following reaction:
N2O4(g) + 4 H2(g)  N2(g) + 4 H2O(g)
From a table of standard enthalpies of formation.
N2O4(g)
H2(g)
N2(g)
H2O(g)
9.66 kJ/mol
0 kJ/mol
0 kJ/mol
-241.82 kJ/mol
element
element
notice value for
the gas is used
ΔH = [(1 x ΔHN2)+ (4 x ΔHH2O)] –
[(1 x ΔHN2O4) + (4 x ΔHH2)]
ΔH = [(1x 0 kJ/mol) + (4 x -241.82kJ/mol)] –
[(1x l9.66kJ/mol) + (4 x 0kJ/mol)]
= -976 kJ
PAY ATTENTION TO THE PARENTHESES AND
BRACKETS TO KEEP THE ORDER OF ADDITIONS
AND SUBTRACTIONS CORRECT!
Example
Calculate the enthalpy change for the following reaction:
2 KOH(s) + CO2(g)  K2CO3(s) + H2O(g)
From a table of standard enthalpies of formation.
KOH(s)
CO2(g)
K2CO3(s)
H2O(g)
-424.7 kJ/mol
-393.5 kJ/mol
-1150.18 kJ/mol
-241.82 kJ/mol
ΔH = [(1 x ΔHK2CO3) + (1 x ΔHH2O)] –
[(2 x ΔHKOH) + (1x ΔHCO2)]
ΔH = [(1x (-1150.18kJ/mol)) + (1x(-241.82kJ/mol))] –
[(2 x (-424.7kJ/mol)) + (1 x (-93.5kJ/mol))]
= -149.1 kJ