### Set 2

```Calorimetry
1
The specific heat (s) of a substance is the amount of heat (q)
required to raise the temperature of one gram of the
substance by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat
(q) required to raise the temperature of a given quantity (m)
of the substance by one degree Celsius.
C=mxs
Heat (q) absorbed or released:
q = m x s x Dt
q = C x Dt
Dt = tfinal - tinitial
2
How much heat is given off when an 869 g iron bar cools
from 94oC to 5oC?
s of Fe = 0.444 J/g • oC
Dt = tfinal – tinitial = 5oC – 94oC = -89oC
q = msDt = 869 g x 0.444 J/g • oC x –89oC = -34,000 J
3
Problem 6.33
• A 6.22 kg piece of copper metal is heated
from 20.5 ºC and 324.3 ºC. Calculate the
heat absorbed (in kJ) by the metal.
4
5
Heating Curve for Water
1.
2.
3.
4.
Ice
Ice + water
Water
Water + steam
ΔHvap = 40.6 kJ/mol, ΔHfus = 6.02 kJ/mol, s = 4.184 J/g˚C
Calculate the energy to heat 17.9g of ice from 0.00˚C to
liquid water at 100.˚C.
(ΔHvap = 40.6 kJ/mol, ΔHfus = 6.02 kJ/mol, s = 4.184 J/g˚C)
7
Constant-Volume Calorimetry
qsys = qcal+ qrxn
qsys = 0
qrxn = - qcal
qcal = m x s x Dt
qcal = Ccal x Dt
Reaction at Constant V
DH = qrxn
DH ~ qrxn
No heat enters or leaves!
8
A quantity of 1.435 g of naphthalene (C10H8), a pungent
smelling substance used in moth repellents, was burned in
a constant-volume bomb calorimeter. Consequently the
temperature of the water rose from 20.28ºC to 25.95ºC. If
the heat capacity of the bomb plus water was 10.17 kJ/ºC,
calculate the heat of combustion of naphthalene on a
molar basis; that is find the molar heat of combustion.
qcal = Ccal Dt
Ccal= 10.17 kJ/oC
Dt = tfinal – tinitial = 25.95oC – 20.28oC = 5.67oC
qcal = Ccal Dt = 10.17 kJ/oC x 5.67oC = 57.66 kJ
Dhcomb = qrxn
moles
128.2 g
-57.66 kJ
=
x
3
mole = -5.151 x 10 kJ/mol
1.435 g
9
Problem 6.37
A 0.1375 g sample of solid magnesium is burned in a
constant volume bomb calorimeter that has a heat capacity of
3024 J/ºC. The temperature increases by 1.126ºC. Calculate
the heat given off by the burning Mg in kJ/mol.
10
Constant-Pressure Calorimetry
qsys = qcal + qrxn
qsys = 0
qrxn = - (qcal)
qcal = m x s x Dt
qcal = Ccal x Dt
Reaction at Constant P
DH = qrxn
No heat enters or leaves!
11
A lead (Pb) pellet having a mass of 26.47 g at 89.98ºC was
placed in a constant pressure calorimeter of negligible heat
capacity containing 100.0 mL of water. The water
temperature rose from 22.50ºC to 23.17ºC. What is the
specific heat of the lead pellet?
qPb + qwater = 0
qPb = - qwater
qwater = ms∆t
Dt = tfinal – tinitial = 23.17oC – 22.50oC = 0.67oC
qwater = ms∆t = 100.0g x 4.184 J/goC x 0.67oC = 280.3 J
12
The heat lost be the pellet is equal to the heat gained by
the water, so
qPb = -280.3 J
Dt = tfinal – tinitial = 23.17oC – 89.98oC = -66.81oC
qPb = ms∆t
-280.3 J= 26.47g x s x -66.81oC
s = 0.158 J/goC
13
Problem 6.82
A 44.0 g sample of an unknown metal at 99.0ºC was placed in a
constant pressure calorimeter containing 80.0 g of water at 24.0ºC.
The final temperature of the system was found to be 28.4ºC.
Calculate the specific heat capacity of the metal. (the specific heat
of water is 4.184 J/gºC)
14
Chemistry in Action:
Fuel Values of Foods and Other Substances
6CO2 (g) + 6H2O (l) DH = -2801 kJ/mol
C6H12O6 (s) + 6O2 (g)
1 cal = 4.184 J
1 Cal = 1000 cal = 4184 J
Substance
DHcombustion (kJ/g)
Apple
-2
Beef
-8
Beer
-1.5
Gasoline
-34
15
Enthalpy of Formations
16
Because there is no way to measure the absolute value of
the enthalpy of a substance, must I measure the enthalpy
change for every reaction of interest?
Establish an arbitrary scale with the standard enthalpy of
formation (DH0f ) as a reference point for all enthalpy
expressions.
Standard enthalpy of formation (DH0f) is the heat change
that results when one mole of a compound is formed from
its elements at a pressure of 1 atm.
The standard enthalpy of formation of any element in its
most stable form is zero.
0 (C, graphite) = 0
DH
f
DH0f (O2) = 0
0 (C, diamond) = 1.90 kJ/mol
DH
0
f
DH (O ) = 142 kJ/mol
f
3
17
18
0 ) is the enthalpy of
The standard enthalpy of reaction (DHrxn
a reaction carried out at 1 atm.
aA + bB
cC + dD
DH0rxn = [ cDH0f (C) + dDH0f (D) ] - [ aDH0f (A) + bDH0f (B) ]
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
Hess’s Law: When reactants are converted to products, the
change in enthalpy is the same whether the reaction takes
place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get
there, only where you start and end.)
19
C (graphite) + 1/2O2 (g)
CO (g) + 1/2O2 (g)
C (graphite) + O2 (g)
CO (g)
CO2 (g)
CO2 (g)
20
Calculate the standard enthalpy of formation of CS2 (l) given
that:
C(graphite) + O2 (g)
CO2 (g) DH0rxn = -393.5 kJ/mol
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
SO2 (g)
DH0rxn = -296.1 kJ/mol
CO2 (g) + 2SO2 (g)
0 = -1072 kJ/mol
DHrxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic)
CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
C(graphite) + O2 (g)
2S(rhombic) + 2O2 (g)
+ CO2(g) + 2SO2 (g)
CO2 (g) DH0rxn = -393.5 kJ/mol
2SO2 (g) DH0rxn = -296.1 kJ/mol x 2
CS2 (l) + 3O2 (g)
0 = +1072 kJ/mol
DHrxn
C(graphite) + 2S(rhombic)
CS2 (l)
DH0rxn= -393.5 + (2x-296.1) + 1072 = 86.3 kJ/mol
21
Benzene (C6H6) burns in air to produce carbon dioxide and
liquid water. How much heat is released per mole of benzene
combusted? The standard enthalpy of formation of benzene is
49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
DH0rxn = [ 12DH0f (CO2) + 6DH0f (H2O)] - [ 2DH0f (C6H6)]
DH0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ
-5946 kJ
= - 2973 kJ/mol C6H6
2 mol
22
Chemistry in Action: Bombardier Beetle Defense
C6H4(OH)2 (aq) + H2O2 (aq)
C6H4O2 (aq) + H2 (g) DH0 = 177 kJ/mol
C6H4(OH)2 (aq)
H2O2 (aq)
C6H4O2 (aq) + 2H2O (l) DH0 = ?
H2O (l) + ½O2 (g) DH0 = -94.6 kJ/mol
H2 (g) + ½ O2 (g)
H2O (l) DH0 = -286 kJ/mol
DH0 = 177 - 94.6 – 286 = -204 kJ/mol
Exothermic!
23
Problem 6.62
From the following data
C(graphite) + O2(g)
CO2(g)
∆Hrxnº=-393.5 kJ/mol
H2(g) + ½O2(g)
H2O(l)
∆Hrxnº=-285.8 kJ/mol
2C2H6(g) + 7O2(g)
4CO2(g) + 6H2O(l)
∆Hrxnº=-3119.6 kJ/mol
calculate the enthalpy change for the reaction
2C(graphite) + 3H2(g)
C2H6(g)
24
The enthalpy of solution (DHsoln) is the heat generated or
absorbed when a certain amount of solute dissolves in a
certain amount of solvent.
DHsoln = Hsoln - Hcomponents
Which substance(s) could be
used for melting ice?
Which substance(s) could be
used for a cold pack?
25
The Solution Process for NaCl
DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol
26
The energy required to completely separate one mole of a solid
ionic compound into gaseous ions is called lattice energy (U).
energy NaCl(s) 
 Na  ( g )  Cl  ( g )
The energy change associated with the hydration process is
called the heat of hydration, ∆Hhyd
2O
Na  ( g )  Cl  ( g ) H

Na  (aq)  Cl  (aq)
∆Hsoln = U + ∆Hhyd
NaCl(s) 
 Na  ( g )  Cl  ( g )
U = 788 kJ/mol
2O
+ Na  ( g )  Cl  ( g ) H

Na  (aq)  Cl  (aq) ∆Hhydr = -784 kJ/mol
NaCl(s) 
 Na  (aq)  Cl  (aq)
DH0rxn= 788 + (-784) = 4 kJ/mol
∆Hsoln = 4 kJ/mol
27
Problem 6.133
From the following data, calculate the heat of
solution for KI:
NaCl
NaI
KCl
KI
Lattice energy
(kJ/mol)
788
686
699
632
Heat of Solution
(kJ/mol)
4.0
-5.1
17.2
?
28
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