Transcript Reactions in Aqueous Solution

Reactions in Aqueous Solution
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
A solution is a homogenous mixture of 2 or more
substances
The solute is(are) the substance(s) present in the
smaller amount(s)
The solvent is the substance present in the larger
amount
Solution
Solvent
Solute
Soft drink(l)
H2O
Sugar, CO2
Air(g)
N2
O2, Ar, CH4
Soft Solder(s)
Pb
Sn
aqueous solutions
of KMnO4
2
An electrolyte is a substance that, when dissolved in
water, results in a solution that can conduct electricity.
A nonelectrolyte is a substance that, when dissolved,
results in a solution that does not conduct electricity.
nonelectrolyte
weak electrolyte
strong electrolyte
3
Conduct electricity in solution?
Cations (+) and Anions (-)
Strong Electrolyte – 100% dissociation
NaCl(s)
H 2O
Na+(aq) + Cl-(aq)
Weak Electrolyte – not completely dissociated
CH3COOH
CH3COO-(aq) + H+(aq)
4
Ionization of acetic acid
CH3COOH
CH3COO-(aq) + H+(aq)
A reversible reaction. The reaction can
occur in both directions.
Acetic acid is a weak electrolyte because its
ionization in water is incomplete.
5
Hydration is the process in which an ion is surrounded
by water molecules arranged in a specific manner.
d-
d+
H2O
6
Nonelectrolyte does not conduct electricity?
No cations (+) and anions (-) in solution
C6H12O6(s)
H 2O
C6H12O6(aq)
7
Precipitation Reactions
Precipitate – insoluble solid that separates from solution
precipitate
Pb(NO3)2(aq) + 2NaI(aq)
PbI2(s) + 2NaNO3(aq)
molecular equation
Pb2+ + 2NO3- + 2Na+ + 2I-
PbI2(s) + 2Na+ + 2NO3-
ionic equation
Pb2+ + 2IPbI2
PbI2(s)
net ionic equation
Na+ and NO3- are spectator ions
8
Solubility is the maximum amount of solute that will dissolve
in a given quantity of solvent at a specific temperature.
9
Writing Net Ionic Equations
1. Write the balanced molecular equation.
2. Write the ionic equation showing the strong electrolytes
completely dissociated into cations and anions.
3. Cancel the spectator ions on both sides of the ionic equation
4. Check that charges and number of atoms are balanced in the
net ionic equation
Write the net ionic equation for the reaction of silver nitrate
with sodium chloride.
AgNO3(aq) + NaCl(aq)
AgCl(s) + NaNO3(aq)
Ag+ + NO3- + Na+ + Cl-
AgCl(s) + Na+ + NO3-
Ag+ + Cl-
AgCl(s)
10
Properties of Acids
Have a sour taste. Vinegar owes its taste to acetic acid. Citrus
fruits contain citric acid.
Cause color changes in plant dyes.
React with certain metals to produce hydrogen gas.
2HCl(aq) + Mg(s)
MgCl2(aq) + H2(g)
React with carbonates and bicarbonates
to produce carbon dioxide gas
2HCl(aq) + CaCO3(s)
CaCl2(aq) + CO2(g) + H2O(l)
Aqueous acid solutions conduct electricity.
11
Properties of Bases
Have a bitter taste.
Feel slippery. Many soaps contain bases.
Cause color changes in plant dyes.
Aqueous base solutions conduct electricity.
Examples:
12
Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water
13
A Brønsted acid is a proton donor
A Brønsted base is a proton acceptor
base
acid
acid
base
A Brønsted acid must contain at least one ionizable
proton!
14
Monoprotic acids
HCl
H+ + Cl-
HNO3
H+ + NO3H+ + CH3COO-
CH3COOH
Strong electrolyte, strong acid
Strong electrolyte, strong acid
Weak electrolyte, weak acid
Diprotic acids
H2SO4
H+ + HSO4-
Strong electrolyte, strong acid
HSO4-
H+ + SO42-
Weak electrolyte, weak acid
Triprotic acids
H3PO4
H2PO4HPO42-
H+ + H2PO4H+ + HPO42H+ + PO43-
Weak electrolyte, weak acid
Weak electrolyte, weak acid
Weak electrolyte, weak acid
15
Identify each of the following species as a Brønsted acid, base,
or both. (a) HI, (b) CH3COO-, (c) H2PO4-
HI (aq)
H+ (aq) + I- (aq)
CH3COO- (aq) + H+ (aq)
H2PO4- (aq)
Brønsted acid
CH3COOH (aq)
H+ (aq) + HPO42- (aq)
H2PO4- (aq) + H+ (aq)
H3PO4 (aq)
Brønsted base
Brønsted acid
Brønsted base
16
Neutralization Reaction
acid + base
salt + water
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O
H+ + Cl- + Na+ + OH-
Na+ + Cl- + H2O
H+ + OH-
H2O
17
Neutralization Reaction Involving a Weak
Electrolyte
weak acid + base
HCN(aq) + NaOH(aq)
HCN + Na+ + OH-
HCN + OH-
salt + water
NaCN(aq) + H2O
Na+ + CN- + H2O
CN- + H2O
18
Neutralization Reaction Producing a Gas
acid + base
2HCl(aq) + Na2CO3(aq)
2H+ + 2Cl- + 2Na+ + CO32-
2H+ + CO32-
salt + water + CO2
2NaCl(aq) + H2O +CO2
2Na+ + 2Cl- + H2O + CO2
H2O + CO2
19
Oxidation-Reduction Reactions
(electron transfer reactions)
2Mg
O2 + 4e-
2Mg2+ + 4e- Oxidation half-reaction (lose e-)
2O2Reduction half-reaction (gain e-)
2Mg + O2 + 4e2Mg2+ + 2O2- + 4e20
2Mg + O2
2MgO
Zn(s) + CuSO4(aq)
ZnSO4(aq) + Cu(s)
Zn2+ + 2e- Zn is oxidized
Zn
Cu2+ + 2e-
Zn is the reducing agent
Cu Cu2+ is reduced Cu2+ is the oxidizing agent
Copper wire reacts with silver nitrate to form silver metal.
What is the oxidizing agent in the reaction?
Cu(s) + 2AgNO3(aq)
Cu
Ag+ + 1e-
Cu(NO3)2(aq) + 2Ag(s)
Cu2+ + 2eAg Ag+ is reduced
Ag+ is the oxidizing agent
21
Oxidation number
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
1. Free elements (uncombined state) have an oxidation
number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2. In monatomic ions, the oxidation number is equal to
the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3. The oxidation number of oxygen is usually –2. In H2O2
and O22- it is –1.
22
4.4
4. The oxidation number of hydrogen is +1 except when
it is bonded to metals in binary compounds. In these
cases, its oxidation number is –1.
5. Group IA metals are +1, IIA metals are +2 and fluorine is
always –1.
6. The sum of the oxidation numbers of all the atoms in a
molecule or ion is equal to the charge on the
molecule or ion.
7. Oxidation numbers do not have to be integers.
Oxidation number of oxygen in the superoxide ion,
O2-, is –½.
-
HCO3
What are the oxidation numbers
of all the elements in HCO3- ?
O = –2
H = +1
3x(–2) + 1 + ? = –1
C = +4
23
The Oxidation Numbers of Elements in their Compounds
24
What are the oxidation numbers of
all the elements in each of these
compounds?
NaIO3
IF7
K2Cr2O7
NaIO3
IF7
F = -1
7x(-1) + ? = 0
I = +7
Na = +1 O = -2
3x(-2) + 1 + ? = 0
I = +5
K2Cr2O7
O = -2
K = +1
7x(-2) + 2x(+1) + 2x(?) = 0
Cr = +6
25
Types of Oxidation-Reduction Reactions
Combination Reaction
A+B
0
0
3Mg + N2
C
+2 -3
MgN2
Decomposition Reaction
C
+1 +5 -2
2KClO3
A+B
+1 -1
0
2KCl + 3O2
26
Types of Oxidation-Reduction Reactions
Combustion Reaction
A + O2
B
0
0
S + O2
0
0
2Mg + O2
+4 -2
SO2
+2 -2
2MgO
27
Types of Oxidation-Reduction Reactions
Displacement Reaction
A + BC
0
+1
+2
Sr + 2H2O
+4
0
TiCl4 + 2Mg
0
AC + B
-1
Cl2 + 2KBr
0
Sr(OH)2 + H2 Hydrogen Displacement
0
+2
Ti + 2MgCl2
-1
Metal Displacement
0
2KCl + Br2
Halogen Displacement
28
The Activity Series for Metals
Hydrogen Displacement Reaction
M + BC
MC + B
M is metal
BC is acid or H2O
B is H2
Ca + 2H2O
Ca(OH)2 + H2
Pb + 2H2O
Pb(OH)2 + H2
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The Activity Series for Halogens
F2 > Cl2 > Br2 > I2
Halogen Displacement Reaction
0
-1
Cl2 + 2KBr
I2 + 2KBr
-1
0
2KCl + Br2
2KI + Br2
30
Classify each of the following reactions.
Ca2+ + CO32NH3 + H+
Zn + 2HCl
Ca + F2
CaCO3
NH4+
ZnCl2 + H2
CaF2
Precipitation
Acid-Base
Redox (H2 Displacement)
Redox (Combination)
31
What type of reaction is shown below?
2Mg + O2 2MgO
a. combination
b.single replacement
c. neutralization
d.decomposition
32
What type of reaction is shown
below?
2Na + Cl2 2NaCl
a. combination
b.single replacement
c. neutralization
d.decomposition
33
Solution Stoichiometry
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
M = molarity =
moles of solute
liters of solution
What mass of KI is required to make 500. mL of a
2.80 M KI solution?
M KI
volume of KI solution
500. mL x
1L
1000 mL
moles KI
x
2.80 mol KI
1 L soln
x
M KI
166 g KI
1 mol KI
grams KI
= 232 g KI
34
Preparing a Solution of Known Concentration
35
Dilution is the procedure for preparing a less concentrated
solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solute
before dilution (i)
=
Moles of solute
after dilution (f)
MiVi
=
MfVf
36
How would you prepare 60.0 mL of 0.200 M HNO3
from a stock solution of 4.00 M HNO3?
MiVi = MfVf
Mi = 4.00 M Mf = 0.200 M Vf = 0.0600 L
Vi =
MfVf
Mi
Vi = ? L
= 0.200 M x 0.0600 L = 0.00300 L = 3.00 mL
4.00 M
Dilute 3.00 mL of acid with water to a total volume
of 60.0 mL.
37
Titrations
In a titration a solution of accurately known concentration is
added gradually added to another solution of unknown
concentration until the chemical reaction between the two
solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
the indicator
changes color
38
Titrations can be used in the analysis of
Acid-base reactions
H2SO4 + 2NaOH
2H2O + Na2SO4
Redox reactions
5Fe2+ + MnO4- + 8H+
Mn2+ + 5Fe3+ + 4H2O
39
What volume of a 1.420 M NaOH solution is required
to titrate 25.00 mL of a 4.50 M H2SO4 solution?
WRITE THE CHEMICAL EQUATION!
H2SO4 + 2NaOH
M
volume acid
25.00 mL x
acid
2H2O + Na2SO4
rxn
moles red
4.50 mol H2SO4
1000 mL soln
x
coef.
M
moles base
2 mol NaOH
1 mol H2SO4
x
base
volume base
1000 ml soln
1.420 mol NaOH
= 158 mL
40
16.42 mL of 0.1327 M KMnO4 solution is needed to
oxidize 25.00 mL of an acidic FeSO4 solution. What is
the molarity of the iron solution?
WRITE THE CHEMICAL EQUATION!
5Fe2+ + MnO4- + 8H+
Mn2+ + 5Fe3+ + 4H2O
M
volume red
red
rxn
moles red
16.42 mL = 0.01642 L
0.01642 L x
0.1327 mol KMnO4
1L
coef.
V
moles oxid
M oxid
oxid
25.00 mL = 0.02500 L
x
5 mol Fe2+
1 mol KMnO4
x
1
0.02500 L
Fe2+
= 0.4358 M
41
42
Gases
43
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Elements that exist as gases at 250C and 1 atmosphere
44
45
Physical Characteristics of Gases
•
Gases assume the volume and shape of their
containers.
•
Gases are the most compressible state of matter.
•
Gases will mix evenly and completely when confined to
the same container.
•
Gases have much lower densities than liquids and
solids.
NO2 gas
46
Force
Pressure = Area
(force = mass x acceleration)
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
47
Boyle’s Law
P  1/V
For a fixed amount of an ideal gas kept at a fixed
temperature, P and V are inversely proportional
P x V = constant
P1 x V1 = P2 x V2
Constant temperature
Constant amount of gas
48
A sample of chlorine gas occupies a volume of 946 mL at a
pressure of 726 mmHg. What is the pressure of the gas (in
mmHg) if the volume is reduced at constant temperature to 154
mL?
P x V = constant
P1 x V1 = P2 x V2
P2 =
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1
V2
726 mmHg x 946 mL
=
= 4460 mmHg
154 mL
49
Variation in Gas Volume with Temperature at Constant Pressure
VT
As T increases
V increases
50
Variation of Gas Volume with Temperature
at Constant Pressure
Charles’ &
Gay-Lussac’s
Law
VT
V/T = constant
V1/T1 = V2 /T2
Temperature must be
in Kelvin
T (K) = t (°C) + 273.15
51
A sample of carbon monoxide gas occupies 3.20 L at 125 °C.
At what temperature will the gas occupy a volume of 1.54 L if
the pressure remains constant?
V1 /T1 = V2 /T2
V1 = 3.20 L
V2 = 1.54 L
T1 = 398.15 K
T2 = ?
T1 = 125 (°C) + 273.15 (K) = 398.15 K
T2 =
V2 x T1
V1
=
1.54 L x 398.15 K
3.20 L
= 192 K
52
Avogadro’s Law
V  number of moles (n)
Constant temperature
Constant pressure
V = constant x n
V1 / n1 = V2 / n2
53
Ammonia burns in oxygen to form nitric oxide (NO) and water
vapor. How many volumes of NO are obtained from one volume
of ammonia at the same temperature and pressure?
4NH3 + 5O2
1 mole NH3
4NO + 6H2O
1 mole NO
At constant T and P
1 volume NH3
1 volume NO
54
Summary of Gas Laws
Boyle’s Law
55
Charles Law
56
Avogadro’s Law
57
Ideal Gas Equation
1
(at constant n and T)
V
Charles’ law: V  T (at constant n and P)
Boyle’s law: P 
Avogadro’s law: V  n (at constant P and
T)
nT
V
P
nT
nT
V = constant x
=R
R is the gas constant
P
P
PV = nRT
58
The conditions 0 °C and 1 atm are called standard
temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal
gas occupies 22.414 L.
PV = nRT
(1 atm)(22.414L)
PV
R=
=
nT
(1 mol)(273.15 K)
R = 0.082057 L • atm / (mol • K)
59
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 °C = 273.15 K
P = 1 atm
PV = nRT
nRT
V=
P
1 mol HCl
n = 49.8 g x
= 1.37 mol
36.45 g HCl
1.37 mol x 0.0821
V=
L•atm
mol•K
x 273.15 K
1 atm
V = 30.7 L
60
Density (d) Calculations
PM
m
d=
=
V
RT
m is the mass of the gas in g
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRT
M=
P
d is the density of the gas in g/L
61
A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0
°C. What is the molar mass of the gas?
dRT
M=
P
M=
g
2.21
L
4.65 g
m
=
= 2.21
d=
2.10
L
V
x 0.0821
L•atm
mol•K
g
L
x 300.15 K
1 atm
M = 54.5 g/mol
62
Gas Stoichiometry
What is the volume of CO2 produced at 37 0C and 1.00 atm
when 5.60 g of glucose are used up in the reaction:
C6H12O6(s) + 6O2(g)
6CO2(g) + 6H2O(l)
g C6H12O6
mol C6H12O6
5.60 g C6H12O6 x
6 mol CO2
1 mol C6H12O6
x
= 0.187 mol CO2
180 g C6H12O6
1 mol C6H12O6
V=
nRT
=
P
mol CO2
V CO2
L•atm
x 310.15 K
mol•K
1.00 atm
0.187 mol x 0.0821
= 4.76 L
63
Dalton’s Law of Partial Pressures
V and T are constant
P1
P2
Ptotal = P1 + P2
64
Consider a case in which two gases, A and B, are in a
container of volume V.
nART
PA =
V
nA is the number of moles of A
nBRT
PB =
V
nB is the number of moles of B
PT = PA + PB
PA = XA PT
nA
XA =
nA + nB
nB
XB =
nA + nB
PB = XB PT
Pi = Xi PT
mole fraction (Xi ) =
ni
nT
65
A sample of natural gas contains 8.24 moles of CH4, 0.421
moles of C2H6, and 0.116 moles of C3H8. If the total pressure
of the gases is 1.37 atm, what is the partial pressure of
propane (C3H8)?
Pi = Xi PT
PT = 1.37 atm
0.116
Xpropane =
8.24 + 0.421 + 0.116
= 0.0132
Ppropane = 0.0132 x 1.37 atm = 0.0181 atm
66
67
Kinetic Molecular Theory of Gases
1. A gas is composed of molecules that are separated from
each other by distances far greater than their own
dimensions. The molecules can be considered to be
points; that is, they possess mass but have negligible
volume.
2. Gas molecules are in constant motion in random
directions, and they frequently collide with one another.
Collisions among molecules are perfectly elastic.
3. Gas molecules exert neither attractive nor repulsive
forces on one another.
4. The average kinetic energy of the molecules is
proportional to the temperature of the gas in kelvins.
Any two gases at the same temperature will have the same
average kinetic energy
68
Kinetic theory of gases and …
• Compressibility of Gases
• Boyle’s Law
P  collision rate with wall
Collision rate  number density
Number density  1/V
P  1/V
• Charles’ Law
P  collision rate with wall
Collision rate  average kinetic energy of gas
molecules
Average kinetic energy  T
PT
69
Kinetic theory of gases and …
• Avogadro’s Law
P  collision rate with wall
Collision rate  number density
Number density  n
Pn
• Dalton’s Law of Partial Pressures
Molecules do not attract or repel one another
P exerted by one type of molecule is unaffected by the
presence of another gas
Ptotal = SPi
70
The distribution of speeds
of three different gases
at the same temperature
The distribution of speeds
for nitrogen gas molecules
at three different temperatures
71
Gas diffusion is the gradual mixing of molecules of one gas
with molecules of another by virtue of their kinetic properties.
r1
r2
=

M2
M1
molecular path
NH4Cl
NH3
17 g/mol
HCl
36 g/mol
72
Deviations from Ideal Behavior
1 mole of ideal gas
PV = nRT
PV = 1.0
n=
RT
Repulsive Forces
Attractive Forces
73
Effect of intermolecular forces on the pressure exerted by a gas.
74
Van der Waals equation
nonideal gas
2
an
( P + V2 ) (V – nb) = nRT
}
}
corrected
pressure
corrected
volume
75
A 13 gram gaseous sample of an
unknown hydrocarbon occupies a
volume of 11.2 L at STP. What is the
hydrocarbon
a. CH
b. C2H4
c. C2H2
d. C3H3
76
A force is applied o a container of gas reducing its
volume by half. The temperature of the gas:
a. decreases
b. increases
c. remains constant
d. the temperature change depends upon the
amount of force used
77
Ammonia burns in air to form nitrogen dioxide
and water.
4NH3+7O24NO2 + 6H2O
If 8 moles of NH3 are reacted with 14 moles of
O2 in a rigid container with an initial pressure of
11 atm, what is the partial pressure of NO2 in
the container when the reaction runs to
completion? ( Assume constant temperature)
a. 4 atm
b. 6 atm
c. 11 atm
d. 12 atm
78
At STP, one liter of which of the following gases
contains the most molecules?
a. H2
b. He2
c. N2
d. Each gas contains the same number of
molecules at STP
79
Energy Relationships
in Chemical Reactions
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Energy is the capacity to do work.
•
Radiant energy comes from the sun and is earth’s
primary energy source
•
Thermal energy is the energy associated with the
random motion of atoms and molecules
•
Chemical energy is the energy stored within the
bonds of chemical substances
•
Nuclear energy is the energy stored within the
collection of neutrons and protons in the atom
•
Potential energy is the energy available by virtue
of an object’s position
81
Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between
two bodies that are at different temperatures.
Temperature is a measure of the thermal
energy.
Temperature = Thermal Energy
82
Thermochemistry is the study of heat change in chemical
reactions.
The system is the specific part of the universe that is of
interest in the study.
open
Exchange: mass & energy
closed
isolated
energy
nothing
83
Exothermic process is any process that gives off heat –
transfers thermal energy from the system to the surroundings.
2H2(g) + O2(g)
H2O(g)
2H2O(l) + energy
H2O(l) + energy
Endothermic process is any process in which heat has to be
supplied to the system from the surroundings.
energy + 2HgO(s)
energy + H2O(s)
2Hg(l) + O2(g)
H2O(l)
84
Thermodynamics is the scientific study of the
interconversion of heat and other kinds of energy.
State functions are properties that are determined by the state
of the system, regardless of how that condition was achieved.
energy, pressure, volume, temperature
DU = Ufinal - Uinitial
DP = Pfinal - Pinitial
DV = Vfinal - Vinitial
DT = Tfinal - Tinitial
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
85
First law of thermodynamics – energy can be
converted from one form to another, but cannot be
created or destroyed.
DU is the change in internal energy of
a system
DUsystem + DUsurroundings = 0
or
DUsystem = -DUsurroundings
C3H8 + 5O2
3CO2 + 4H2O
Exothermic chemical reaction!
Chemical energy lost by combustion = Energy gained by the surroundings
86
system
surroundings
Enthalpy and the First Law of Thermodynamics
DU = q + w
At constant pressure:
DU is the change in
internal energy of a
system
q = DH and w = -PDV
DU = DH - PDV
DH = DU + PDV
87
Enthalpy (H) is used to quantify the heat flow into or out of a
system in a process that occurs at constant pressure.
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
DH < 0
Hproducts > Hreactants
DH > 0
88
Thermochemical Equations
Is DH negative or positive?
System absorbs heat
Endothermic
DH > 0
6.01 kJ are absorbed for every 1 mole of ice that
melts at 00C and 1 atm.
H2O(s)
H2O(l)
DH = 6.01 kJ/mol
89
Thermochemical Equations
Is DH negative or positive?
System gives off heat
Exothermic
DH < 0
890.4 kJ are released for every 1 mole of methane
that is combusted at 250C and 1 atm.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
DH = -890.4 kJ/mol
90
Thermochemical Equations
•
The stoichiometric coefficients always refer to the number
of moles of a substance
H2O(s)
•
DH = 6.01 kJ/mol
If you reverse a reaction, the sign of DH changes
H2O(l)
•
H2O(l)
H2O(s)
DH = -6.01 kJ/mol
If you multiply both sides of the equation by a factor n,
then DH must change by the same factor n.
2H2O(s)
2H2O(l)
DH = 2 x 6.01 = 12.0 kJ
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Thermochemical Equations
•
The physical states of all reactants and products must be
specified in thermochemical equations.
H2O(s)
H2O(l)
DH = 6.01 kJ/mol
H2O(l)
H2O(g)
DH = 44.0 kJ/mol
How much heat is evolved when 266 g of white phosphorus (P4)
burn in air?
P4(s) + 5O2(g)
266 g P4 x
P4O10(s)
1 mol P4
123.9 g P4
x
DH = -3013 kJ/mol
3013 kJ
= 6470 kJ
1 mol P4
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A Comparison of DH and DU
2Na(s) + 2H2O(l)
DU = DH - PDV
2NaOH(aq) + H2(g) DH = -367.5 kJ/mol
At 25 °C, 1 mole H2 = 24.5 L at 1 atm
PDV = 1 atm x 24.5 L = 2.5 kJ
DU = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol
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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
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How much heat is given off when an 869 g iron bar cools
from 94oC to 5oC?
s of Fe = 0.444 J/g • °C
Dt = tfinal – tinitial = 5°C – 94°C = -89°C
q = msDt = 869 g x 0.444 J/g • °C x –89°C = -34,000 J
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