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 Soft drink(
l
) Air(
g
) Soft Solder(
s
) Solvent H 2 O N 2 Pb Solute Sugar, CO 2 O 2 , Ar, CH 4 Sn aqueous solutions of KMnO 4 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 2 O Na + (
aq
) + Cl (
aq
) Weak Electrolyte – not completely dissociated CH 3 COOH CH 3 COO (
aq
) + H + (
aq
) 4
CH 3
Ionization
of acetic acid COOH CH 3 COO (
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 + H 2 O 6
Nonelectrolyte does not conduct electricity?
No cations (+) and anions (-) in solution H 2 O C 6 H 12 O 6 (
s
) C 6 H 12 O 6 (
aq
) 7
Precipitation Reactions
Precipitate – insoluble solid that separates from solution precipitate PbI 2 Pb(NO 3 ) 2 (
aq
) + 2NaI (
aq
) PbI 2 (
s
molecular equation
) + 2NaNO 3 (
aq
) Pb 2+ + 2NO 3 + 2Na + + 2I PbI 2 (
s
) + 2Na + + 2NO 3 -
ionic equation
Pb 2+ + 2I PbI 2 (
s
)
net ionic equation
Na + and NO 3 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.
AgNO 3 (
aq
) + NaCl(
aq
) AgCl(
s
) + NaNO 3 (
aq
) Ag + + NO 3 + Na + + Cl Ag + + Cl AgCl(
s
) + Na + + NO 3 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
) MgCl 2 (
aq
) + H 2 (
g
)
React with carbonates and bicarbonates to produce carbon dioxide gas
2HCl(
aq
) + CaCO 3 (
s
) CaCl 2 (
aq
) + CO 2 (
g
) + H 2 O(
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 + (H 3 O + ) in water Arrhenius base is a substance that produces OH in water 13
A
Br ønsted acid
is a proton donor A
Br ønsted base
is a proton acceptor base acid acid base A Br ønsted acid must contain at least one ionizable proton!
14
Monoprotic
acids HCl H + + Cl HNO 3 H + + NO 3 CH 3 COOH H + + CH 3 COO Strong electrolyte, strong acid Strong electrolyte, strong acid Weak electrolyte, weak acid
Diprotic
acids H 2 SO 4 H + + HSO 4 HSO 4 H + + SO 4 2-
Triprotic
acids H 3 PO 4 H 2 PO 4 HPO 4 2 H + + H 2 PO 4 H + H + + HPO 4 2 + PO 4 3 Strong electrolyte, strong acid Weak electrolyte, weak acid Weak electrolyte, weak acid Weak electrolyte, weak acid Weak electrolyte, weak acid 15
Identify each of the following species as a Br ønsted acid, base, or both. (a) HI, (b) CH 3 COO , (c) H 2 PO 4 HI (
aq
) H + (
aq
) + I (
aq
) Br ønsted acid CH 3 COO (
aq
) + H + (
aq
) CH 3 COOH (
aq
) Br ønsted base H 2 PO 4 (
aq
) H + (
aq
) + HPO 4 2 (
aq
) H 2 PO 4 (
aq
) + H + (
aq
) H 3 PO 4 (
aq
) Br ønsted acid Br ønsted base 16
Neutralization Reaction
acid + base salt + water HCl (
aq
) + NaOH (
aq
) NaCl(
aq
) + H 2 O H + + Cl + Na + + OH Na + + Cl + H 2 O H + + OH H 2 O 17
Neutralization Reaction Involving a Weak Electrolyte
weak acid + base salt + water HCN (
aq
) + NaOH (
aq
) NaCN(
aq
) + H 2 O HCN + Na + + OH Na + + CN + H 2 O HCN + OH CN + H 2 O 18
Neutralization Reaction Producing a Gas
acid + base salt + water + CO 2 2HCl (
aq
) + Na 2 CO 3 (
aq
) 2NaCl(
aq
) + H 2 O +CO 2 2H + + 2Cl + 2Na + + CO 3 2 2H + + CO 3 2 2Na + + 2Cl + H 2 O + CO 2 H 2 O + CO 2 19
Oxidation-Reduction Reactions
(electron transfer reactions) 2Mg 2Mg 2+ + 4e -
Oxidation
half-reaction (lose e ) O 2 + 4e 2O 2 2Mg + O 2 + 4e -
Reduction
2Mg + O 2 2Mg 2+ half-reaction (gain e ) + 2O 2MgO 2 + 4e 20
Zn(
s
) + CuSO 4 (
aq
) ZnSO 4 (
aq
) + Cu(
s
) Zn Zn 2+ + 2e Zn is oxidized Zn is the
reducing agent
Cu 2+ + 2e Cu Cu 2+ is reduced Cu 2+ is the
oxidizing agent
Copper wire reacts with silver nitrate to form silver metal.
What is the oxidizing agent in the reaction?
Cu(
s
) + 2AgNO 3 (
aq
) Cu(NO 3 ) 2 (
aq
) + 2Ag(
s
) Cu Cu 2+ + 2e Ag + + 1e Ag 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, H 2 , O 2 , P 4 = 0 2. In monatomic ions, the oxidation number is equal to the charge on the ion.
Li + , Li = +1 ; Fe 3+ , Fe = +3 ; O 2 , O = -2 3. The oxidation number of oxygen is and O 2 2 it is –1 .
usually
–2 . In H 2 O 2 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 always –1 .
and fluorine is 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, O 2 , is –½ .
HCO 3 What are the oxidation numbers of all the elements in HCO 3 ?
O = – 2 3x( – 2) H = + 1 + ?
+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?
NaIO 3 IF 7 K 2 Cr 2 O 7 NaIO 3 Na = +1 O = -2 3x( -2 ) + 1 + ?
= 0 I = +5 IF F = I = 7 -1 7x( -1 ) + ?
+7 = 0 K 2 Cr 2 O 7 O = -2 K = +1 7x( -2 ) + 2x( +1 ) + 2x( ?) = 0 Cr = +6 25
Types of Oxidation-Reduction Reactions
Combination Reaction A + B C 0 0 3Mg + N 2 +2 -3 MgN 2 Decomposition Reaction C A + B +1 +5 -2 2KClO 3 +1 -1 0 2KCl + 3O 2 26
Types of Oxidation-Reduction Reactions
Combustion Reaction A + O 2 0 0 S + O 2 B +4 -2 SO 2 0 0 2Mg + O 2 +2 -2 2MgO 27
Types of Oxidation-Reduction Reactions
Displacement Reaction A + BC AC + B 0 +1 +2 Sr + 2H 2 O Sr(OH) 2 0 + H 2 +4 TiCl 4 0 0 +2 + 2Mg Ti + 2MgCl 2 Hydrogen Displacement Metal Displacement 0 Cl 2 -1 -1 0 + 2KBr 2KCl + Br 2 Halogen Displacement 28
The Activity Series for Metals
Hydrogen Displacement Reaction M + BC MC + B M is metal BC is acid or H 2 O B is H 2 Ca + 2H 2 O Ca(OH) 2 + H 2 Pb + 2H 2 O Pb(OH) 2 + H 2 29
The Activity Series for Halogens F
2
> Cl
2
> Br
2
> I
2 Halogen Displacement Reaction 0 Cl 2 -1 -1 0 + 2KBr 2KCl + Br 2 I 2 + 2KBr 2KI + Br 2 30
Classify each of the following reactions.
Ca 2+ + CO 3 2 NH 3 + H + CaCO 3 NH 4 + Precipitation Acid-Base Zn + 2HCl ZnCl 2 + H 2 Redox (H 2 Displacement) Ca + F 2 CaF 2 Redox (Combination) 31
What type of reaction is shown below?
2Mg + O 2 2MgO a. combination b.single replacement c. neutralization d.decomposition
32
What type of reaction is shown below?
2Na + Cl 2 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
M
KI moles KI grams KI 500. mL 1 L x 1000 mL x 2.80 mol KI x 1 L soln 166 g KI 1 mol 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)
M
i V i
= =
Moles of solute after dilution (f)
M
f V f 36
How would you prepare 60.0 mL of 0.200
M
HNO 3 from a stock solution of 4.00
M
HNO 3 ?
M
i V i =
M
f V f
M
i = 4.00
M M
f = 0.200
M
V f = 0.0600 L V i = ? L V i =
M
f V f
M
i = 0.200
M
x 0.0600 L 4.00
M
= 0.00300 L = 3.00 mL 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
H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4
Redox reactions
5Fe 2+ + MnO 4 + 8H + Mn 2+ + 5Fe 3+ + 4H 2 O 39
What volume of a 1.420
M
NaOH solution is required to titrate 25.00 mL of a 4.50
M
H 2 SO 4 solution?
WRITE THE CHEMICAL EQUATION!
H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4
M
volume acid acid moles red rxn coef.
M
moles base base volume base 25.00 mL x 4.50 mol H 2 SO 4 1000 mL soln x 2 mol NaOH 1 mol H 2 SO 4 1000 ml soln x 1.420 mol NaOH = 158 mL 40
16.42 mL of 0.1327
M
KMnO 4 solution is needed to oxidize 25.00 mL of an acidic FeSO 4 the molarity of the iron solution?
solution. What is 5Fe WRITE THE CHEMICAL EQUATION!
2+ + MnO 4 + 8H + Mn 2+ + 5Fe 3+ + 4H 2 O
M
volume red red rxn moles red coef.
V
moles oxid oxid 16.42 mL = 0.01642 L 25.00 mL = 0.02500 L
M
oxid 0.01642 L x 0.1327 mol KMnO 4 1 L x 5 mol Fe 2+ 1 mol KMnO 4 x 1 0.02500 L Fe 2+ = 0.4358
M
41
42
Gases
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
43
Elements that exist as gases at 25 0 C 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.
NO 2 gas 46
Pressure = Force Area ( force = mass x acceleration) Units of Pressure 1 pascal (Pa) = 1 N/m 2
1 atm = 760 mmHg = 760 torr
47
Boyle’s Law
P
1/
V
P x V = constant
For a fixed amount of an ideal gas kept at a fixed temperature, P and V are inversely proportional
P
1 x
V
1 =
P
2 x
V
2 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 P
1 x
V
1 =
P
2 x
V
2
P
1 = 726 mmHg
P
2 = ?
V
1 = 946 mL
V
2 = 154 mL
P
2 =
P
1 x
V
1
V
2 = 726 mmHg x 946 mL 154 mL = 4460 mmHg 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
V
T V/T
= constant
V
1 /
T
1 =
V
2 /
T
2 Charles’ & Gay Lussac’s Law 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?
V
1 /
T
1 =
V
2 /
T
2
V
1 = 3.20 L
V
2 = 1.54 L
T
1 = 398.15 K
T
2 = ?
T
1 = 125 ( ° C) + 273.15 (K) = 398.15 K
T
2 =
V
2 x
T
1 =
V
1 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 V
1 /
n
1 =
V
2 /
n
2 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?
4NH 3 + 5O 2 4NO + 6H 2 O 1 mole NH 3 1 mole NO At constant
T
and
P
1 volume NH 3 1 volume NO 54
Boyle’s Law Summary of Gas Laws 55
Charles Law 56
Avogadro’s Law 57
Ideal Gas Equation
Boyle’s law: P Charles’ law:
V
T
1
V
(at constant
n
and (at constant
n
and
P
)
T
) Avogadro’s law: V
n T
)
V V
nT P nT
= constant x =
R P
(at constant
P
and
nT P R
is the
gas constant
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 R = PV nT
(1 atm)(22.414L) = (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?
PV = nRT V = nRT P T
= 0 ° C = 273.15 K
P = 1 atm n
= 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol
V
= 1.37 mol x 0.0821 x 273.15 K mol •K 1 atm
V
= 30.7 L 60
Density (
d
) Calculations
d = m V
=
P M 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
M
=
dRT 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?
M M
=
dRT P
=
d = m V
= 4.65 g 2.10 L = 2.21
g L 2.21
g L x 0.0821 x 300.15 K mol •K 1 atm
M
= 54.5 g/mol 62
Gas Stoichiometry
What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (
s
) + 6O 2 (
g
) 6CO 2 (
g
) + 6H 2 O(
l
) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2
V
CO 2 5.60 g C 6 H 12 O 6 x 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 = 0.187 mol CO 2
V
=
nRT P
= L •atm 0.187 mol x 0.0821 x 310.15 K mol •K 1.00 atm = 4.76 L 63
Dalton’s Law of Partial Pressures
V
and
T
are
constant
P
1
P
2
P
total
= P
1 +
P
2 64
Consider a case in which two gases, A and B , are in a container of volume V.
P
A =
n A
RT
V P
B =
n B
RT
V P
T =
P
A +
P
B
P
A =
X
A
P
T
n
A is the number of moles of A
n
B is the number of moles of B
X
A =
n
A
n
A +
n
B
X
B =
n
A
n
B +
n
B
P
B =
X
B
P
T
P i
=
X i P
T
mole fraction (X i ) =
n i n T
65
A sample of natural gas contains 8.24 moles of CH 4 , 0.421 moles of C 2 H 6 , and 0.116 moles of C 3 H 8 . If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )?
P i
=
X i P
T
P
T = 1.37 atm
X
propane = 0.116
8.24 + 0.421 + 0.116
= 0.0132
P
propane = 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
P
1/
V
1/
V
• Charles’ Law
P
collision rate with wall Collision rate molecules average kinetic energy of gas Average kinetic energy
P
T 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 P
exerted by one type of molecule is unaffected by the presence of another gas total = S
P
i 70
The distribution of speeds for nitrogen gas molecules at three different temperatures The distribution of speeds of three different gases at the same temperature 71
Gas diffusion
is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.
r 1 r 2 =
M 2 M 1
molecular path NH 4 Cl NH 3 17 g/mol HCl 36 g/mol 72
Deviations from Ideal Behavior 1 mole of ideal gas
PV
=
nRT n = PV RT
= 1.0
Repulsive Forces Attractive Forces 73
Effect of intermolecular forces on the pressure exerted by a gas.
74
Van der Waals equation
nonideal gas
(
P V
2 2
)
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. C 2 H 4 c. C 2 H 2 d. C 3 H 3 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.
4NH 3 +7O 2 4NO 2 + 6H 2 O If 8 moles of NH3 are reacted with 14 moles of O 2 in a rigid container with an initial pressure of 11 atm, what is the partial pressure of NO 2 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. H 2 b. He 2 c. N 2 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
energy.
is a measure of the thermal 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 energy isolated nothing 83
Exothermic process
is any process that gives off heat – transfers thermal energy from the system to the surroundings. 2H 2 (
g
) + O 2 (
g
) 2H 2 O (
l
) + energy H 2 O (
g
) H 2 O (
l
) + energy
Endothermic process
is any process in which heat has to be supplied to the system from the surroundings. energy + 2HgO (
s
) energy + H 2 O (
s
) 2Hg (
l
) H 2 O (
l
) + O 2 (
g
) 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 D
U
=
U final
-
U initial
D
P
=
P final
-
P initial
D
V
=
V final
-
V initial
D
T
=
T final
-
T initial
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.
D
U
is the change in internal energy of a system D
U system
D
U system
+ D
U surroundings
or =
-
D
U surroundings
= 0 C 3 H 8 + 5O 2 3CO 2 + 4H 2 O Exothermic chemical reaction!
Chemical energy lost by combustion = Energy gained by the surroundings 86 system surroundings
Enthalpy and the First Law of Thermodynamics D
U
=
q
+
w
At constant pressure:
q
= D
H
and
w
= -
P
D
V
D
U
= D
H
-
P
D
V
D
H
= D
U
+
P
D
V
D
U
is the change in internal energy of a system 87
Enthalpy (H)
is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.
D
H
=
H
(products) –
H
(reactants) D
H
= heat given off or absorbed during a reaction at constant pressure
H
products <
H
reactants D
H
< 0
H
products >
H
reactants D
H
> 0 88
Thermochemical Equations Is D
H
negative or positive?
System absorbs heat Endothermic D
H
> 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 0 0 C and 1 atm.
H 2 O (
s
) H 2 O (
l
) D
H
= 6.01 kJ/mol 89
Thermochemical Equations Is D
H
negative or positive?
System gives off heat Exothermic D
H
< 0 890.4 kJ are released for every 1 mole of methane that is combusted at 25 0 C and 1 atm.
CH 4 (
g
) + 2O 2 (
g)
CO 2 (
g) +
2H 2 O (
l
) D
H
= -890.4 kJ/mol 90
Thermochemical Equations • • • The stoichiometric coefficients always refer to the number of moles of a substance H 2 O (
s
) H 2 O (
l
) D
H
= 6.01 kJ/mol If you reverse a reaction, the sign of D
H
changes H 2 O (
l
) H 2 O (
s
) D
H
= 6.01
kJ/mol If you multiply both sides of the equation by a factor
n
, then D
H
must change by the same factor
n
.
2H 2 O (
s
) 2H 2 O (
l
) D
H
= 2 x 6.01 = 12.0 kJ 91
Thermochemical Equations • The physical states of all reactants and products must be specified in thermochemical equations.
H 2 O (
s
) H 2 O (
l
) D
H
= 6.01 kJ/mol H 2 O (
l
) H 2 O (
g
) D
H
= 44.0 kJ/mol How much heat is evolved when 266 g of white phosphorus (P 4 ) burn in air?
P 4 (
s
) + 5O 2 (
g
) P 4 O 10 (
s
) D
H
= -3013 kJ/mol 266 g P 4 x 1 mol P 4 123.9 g P 4 x 3013 kJ 1 mol P 4 = 6470 kJ 92
A Comparison of D
H
and D
U
2Na(
s
) + 2H 2 O(
l
) 2NaOH(
aq
) + H 2 (
g
) D H = -367.5 kJ/mol D
U
= D
H
-
P
D
V
At 25 ° C, 1 mole H 2 = 24.5 L at 1 atm
P
D
V
= 1 atm x 24.5 L = 2.5 kJ D
U
= -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol 93
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
=
m x s
Heat (
q
) absorbed or released:
q
=
m x s x
D
t q
=
C x
D
t
D
t
=
t
final -
t
initial 94
How much heat is given off when an 869 g iron bar cools from 94 o C to 5 o C?
s
of Fe = 0.444 J/g • ° C D
t
=
t
final –
t
initial = 5 ° C – 94 ° C = -89 ° C
q
=
ms
D
t
= 869 g x 0.444 J/g • ° C x –89 ° C = -34,000 J 95