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Reactions in Aqueous Solution
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2
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)
Air(g)
Soft Solder(s)
H2O
N2
Pb
Sugar, CO2
O2, Ar, CH4
Snaqueous solutions
of KMnO4
3
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
4
Strong Electrolyte – 100% dissociation
NaCl(s) Na+(aq) + Cl-(aq)H2O
Weak Electrolyte – not completely dissociated
CH3COOH CH3COO-(aq) + H+(aq)
Conduct electricity in solution?
Cations (+) and Anions (-)
5
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.
6
Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner.
H2O
7
Nonelectrolyte does not conduct electricity?
No cations (+) and anions (-) in solution
C6H12O6(s) C6H12O6(aq)H2O
8
Precipitation Reactions
Precipitate – insoluble solid that separates from solution
molecular equation
ionic equation
net ionic equation
Pb2+ + 2NO3- + 2Na+ + 2I- PbI2(s) + 2Na+ + 2NO3
-
Na+ and NO3- are spectator ions
PbI2
Pb(NO3)2(aq) + 2NaI(aq) PbI2(s) + 2NaNO3(aq)
precipitate
Pb2+ + 2I- PbI2(s)
9
Solubility is the maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature.
10
Writing Net Ionic Equations1. 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
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
Ag+ + NO3- + Na+ + Cl- AgCl(s) + Na+ + NO3
-
Ag+ + Cl- AgCl(s)
Write the net ionic equation for the reaction of silver nitrate with sodium chloride.
11
Properties of Acids
Have a sour taste. Vinegar owes its taste to acetic acid. Citrusfruits contain citric acid.
React with certain metals to produce hydrogen gas.
React with carbonates and bicarbonatesto produce carbon dioxide gas
Cause color changes in plant dyes.
2HCl(aq) + Mg(s) MgCl2(aq) + H2(g)
2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)
Aqueous acid solutions conduct electricity.
12
Have a bitter taste.
Feel slippery. Many soaps contain bases.
Properties of Bases
Cause color changes in plant dyes.
Aqueous base solutions conduct electricity.
Examples:
13
Arrhenius acid is a substance that produces H+ (H3O+) in water
Arrhenius base is a substance that produces OH- in water
14
A Brønsted acid is a proton donorA Brønsted base is a proton acceptor
acidbase acid base
A Brønsted acid must contain at least one ionizable proton!
15
Monoprotic acids
HCl H+ + Cl-
HNO3 H+ + NO3-
CH3COOH H+ + CH3COO-
Strong electrolyte, strong acid
Strong electrolyte, strong acid
Weak electrolyte, weak acid
Diprotic acidsH2SO4 H+ + HSO4
-
HSO4- H+ + SO4
2-
Strong electrolyte, strong acid
Weak electrolyte, weak acid
Triprotic acidsH3PO4 H+ + H2PO4
-
H2PO4- H+ + HPO4
2-
HPO42- H+ + PO4
3-
Weak electrolyte, weak acid
Weak electrolyte, weak acid
Weak electrolyte, weak acid
16
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) Brønsted acid
CH3COO- (aq) + H+ (aq) CH3COOH (aq) Brønsted base
H2PO4- (aq) H+ (aq) + HPO4
2- (aq)
H2PO4- (aq) + H+ (aq) H3PO4 (aq)
Brønsted acid
Brønsted base
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Neutralization Reaction
acid + base salt + water
HCl(aq) + NaOH(aq) NaCl(aq) + H2O
H+ + Cl- + Na+ + OH- Na+ + Cl- + H2O
H+ + OH- H2O
18
Neutralization Reaction Involving a Weak Electrolyte
weak acid + base salt + water
HCN(aq) + NaOH(aq) NaCN(aq) + H2O
HCN + Na+ + OH- Na+ + CN- + H2O
HCN + OH- CN- + H2O
19
Neutralization Reaction Producing a Gas
acid + base salt + water + CO2
2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + H2O +CO2
2H+ + 2Cl- + 2Na+ + CO32- 2Na+ + 2Cl- + H2O + CO2
2H+ + CO32- H2O + CO2
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Oxidation-Reduction Reactions(electron transfer reactions)
2Mg 2Mg2+ + 4e-
O2 + 4e- 2O2-
Oxidation half-reaction (lose e-)
Reduction half-reaction (gain e-)
2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e-
2Mg + O2 2MgO
21
Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)
Zn is oxidizedZn Zn2+ + 2e-
Cu2+ is reducedCu2+ + 2e- Cu
Zn is the reducing agent
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(NO3)2(aq) + 2Ag(s)
Cu Cu2+ + 2e-
Ag+ + 1e- Ag Ag+ is reduced Ag+ is the oxidizing agent
22
Oxidation number
The charge the atom would have in a molecule (or anionic 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.
4.4
23
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.
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.
5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1.
HCO3-
O = –2 H = +1
3x(–2) + 1 + ? = –1
C = +4
What are the oxidation numbers of all the elements in HCO3
- ?
7. Oxidation numbers do not have to be integers. Oxidation number of oxygen in the superoxide ion, O2
-, is –½.
24
The Oxidation Numbers of Elements in their Compounds
25
NaIO3
Na = +1 O = -2
3x(-2) + 1 + ? = 0
I = +5
IF7
F = -1
7x(-1) + ? = 0
I = +7
K2Cr2O7
O = -2 K = +1
7x(-2) + 2x(+1) + 2x(?) = 0
Cr = +6
What are the oxidation numbers of all the elements in each of these compounds? NaIO3 IF7 K2Cr2O7
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Types of Oxidation-Reduction Reactions
Combination Reaction
A + B C
3Mg + N2 MgN2
Decomposition Reaction
2KClO3 2KCl + 3O2
C A + B
0 0 +2 -3
+1 +5 -2 +1 -1 0
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Types of Oxidation-Reduction Reactions
Combustion Reaction
A + O2 B
S + O2 SO2
0 0 +4 -2
2Mg + O2 2MgO0 0 +2 -2
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Displacement Reaction
A + BC AC + B
Sr + 2H2O Sr(OH)2 + H2
TiCl4 + 2Mg Ti + 2MgCl2
Cl2 + 2KBr 2KCl + Br2
Hydrogen Displacement
Metal Displacement
Halogen Displacement
Types of Oxidation-Reduction Reactions
0 +1 +2 0
0+4 0 +2
0 -1 -1 0
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The Activity Series for Metals
M + BC MC + B
Hydrogen Displacement Reaction
M is metalBC 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
Halogen Displacement Reaction
Cl2 + 2KBr 2KCl + Br2
0 -1 -1 0
F2 > Cl2 > Br2 > I2
I2 + 2KBr 2KI + Br2
31
Ca2+ + CO32- CaCO3
NH3 + H+ NH4+
Zn + 2HCl ZnCl2 + H2
Ca + F2 CaF2
Precipitation
Acid-Base
Redox (H2 Displacement)
Redox (Combination)
Classify each of the following reactions.
32
What type of reaction is shown below?2Mg + O2 2MgO
a. combinationb.single replacementc. neutralizationd.decomposition
33
What type of reaction is shown below?
2Na + Cl2 2NaCl
a. combinationb.single replacementc. neutralizationd.decomposition
34
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?
volume of KI solution moles KI grams KIM KI M KI
500. mL = 232 g KI166 g KI
1 mol KIx
2.80 mol KI
1 L solnx
1 L
1000 mLx
35
Preparing a Solution of Known Concentration
36
Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.
DilutionAdd Solvent
Moles of solutebefore dilution (i)
Moles of soluteafter dilution (f)=
MiVi MfVf=
37
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 = ? L
Vi =MfVf
Mi
= 0.200 M x 0.0600 L4.00 M
= 0.00300 L = 3.00 mL
Dilute 3.00 mL of acid with water to a total volume of 60.0 mL.
38
TitrationsIn 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 baseto unknown acid
UNTIL
the indicatorchanges color
39
Titrations can be used in the analysis of
Acid-base reactions
Redox reactions
H2SO4 + 2NaOH 2H2O + Na2SO4
5Fe2+ + MnO4- + 8H+ Mn2+ + 5Fe3+ + 4H2O
40
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!
volume acid moles red moles base volume base
H2SO4 + 2NaOH 2H2O + Na2SO4
4.50 mol H2SO4
1000 mL solnx
2 mol NaOH
1 mol H2SO4
x1000 ml soln
1.420 mol NaOHx25.00 mL = 158 mL
M
acid
rxn
coef.
M
base
41
WRITE THE CHEMICAL EQUATION!
volume red moles red moles oxid M oxid
0.1327 mol KMnO4
1 Lx
5 mol Fe2+
1 mol KMnO4
x1
0.02500 L Fe2+x0.01642 L = 0.4358 M
M
red
rxn
coef.
V
oxid
5Fe2+ + MnO4- + 8H+ Mn2+ + 5Fe3+ + 4H2O
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?
16.42 mL = 0.01642 L 25.00 mL = 0.02500 L
42
43
Gases
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
44
Elements that exist as gases at 250C and 1 atmosphere
45
46
• 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.
Physical Characteristics of Gases
NO2 gas
47
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
Pressure = ForceArea
(force = mass x acceleration)
48
P x V = constant
P1 x V1 = P2 x V2
Boyle’s Law
Constant temperatureConstant amount of gas
P 1/V For a fixed amount of an ideal gas kept at a fixed temperature, P and V are inversely proportional
49
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?
P1 x V1 = P2 x V2
P1 = 726 mmHg
V1 = 946 mL
P2 = ?
V2 = 154 mL
P2 = P1 x V1
V2
726 mmHg x 946 mL154 mL
= = 4460 mmHg
P x V = constant
50As T increases V increases
Variation in Gas Volume with Temperature at Constant Pressure
V T
51
Variation of Gas Volume with Temperatureat Constant Pressure
V T
V/T = constant
V1/T1 = V2 /T2T (K) = t (°C) + 273.15
Charles’ & Gay-Lussac’s
Law
Temperature must bein Kelvin
52
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 = 3.20 L
T1 = 398.15 K
V2 = 1.54 L
T2 = ?
T2 = V2 x T1
V1
1.54 L x 398.15 K3.20 L
= = 192 K
V1 /T1 = V2 /T2
T1 = 125 (°C) + 273.15 (K) = 398.15 K
53
Avogadro’s Law
V number of moles (n)
V = constant x n
V1 / n1 = V2 / n2
Constant temperatureConstant pressure
54
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 4NO + 6H2O
1 mole NH3 1 mole NO
At constant T and P
1 volume NH3 1 volume NO
55
Summary of Gas Laws
Boyle’s Law
56
Charles Law
57
Avogadro’s Law
58
Ideal Gas Equation
Charles’ law: V T(at constant n and P)
Avogadro’s law: V n(at constant P and T)
Boyle’s law: P (at constant n and T)1V
V nT
P
V = constant x = RnT
P
nT
PR is the gas constant
PV = nRT
59
The conditions 0 °C and 1 atm are called standard temperature and pressure (STP).
PV = nRT
R = PVnT
=(1 atm)(22.414L)
(1 mol)(273.15 K)
R = 0.082057 L • atm / (mol • K)
Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.
60
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
PV = nRT
V = nRTP
T = 0 °C = 273.15 K
P = 1 atm
n = 49.8 g x 1 mol HCl36.45 g HCl
= 1.37 mol
V =1 atm
1.37 mol x 0.0821 x 273.15 KL•atmmol•K
V = 30.7 L
61
Density (d) Calculations
d = mV =
PMRT
m is the mass of the gas in g
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRTP
M = d is the density of the gas in g/L
62
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?
dRTP
M = d = mV
4.65 g2.10 L
= = 2.21 g
L
M =2.21
g
L
1 atm
x 0.0821 x 300.15 KL•atmmol•K
M = 54.5 g/mol
63
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 mol CO2 V CO2
5.60 g C6H12O6
1 mol C6H12O6
180 g C6H12O6
x6 mol CO2
1 mol C6H12O6
x = 0.187 mol CO2
V = nRT
P
0.187 mol x 0.0821 x 310.15 KL•atmmol•K
1.00 atm= = 4.76 L
64
Dalton’s Law of Partial Pressures
V and T are constant
P1P2 Ptotal = P1 + P2
65
Consider a case in which two gases, A and B, are in a container of volume V.
PA = nART
V
PB = nBRT
V
nA is the number of moles of A
nB is the number of moles of B
PT = PA + PB XA = nA
nA + nB
XB = nB
nA + nB
PA = XA PT PB = XB PT
Pi = Xi PT mole fraction (Xi ) = ni
nT
66
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
Xpropane = 0.116
8.24 + 0.421 + 0.116
PT = 1.37 atm
= 0.0132
Ppropane = 0.0132 x 1.37 atm = 0.0181 atm
67
68
Kinetic Molecular Theory of Gases1. 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
69
Kinetic theory of gases and …
• Compressibility of Gases
• Boyle’s Law
P collision rate with wall
Collision rate number densityNumber density 1/VP 1/V
• Charles’ LawP collision rate with wall
Collision rate average kinetic energy of gas molecules
Average kinetic energy T
P T
70
Kinetic theory of gases and …
• Avogadro’s Law
P collision rate with wall
Collision rate number densityNumber density nP 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 = Pi
71
The distribution of speedsfor nitrogen gas molecules
at three different temperatures
The distribution of speedsof three different gases
at the same temperature
72
Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.
NH3
17 g/molHCl
36 g/mol
NH4Cl
r1
r2
M2
M1=
molecular path
73
Deviations from Ideal Behavior
1 mole of ideal gas
PV = nRT
n = PVRT
= 1.0
Repulsive Forces
Attractive Forces
74
Effect of intermolecular forces on the pressure exerted by a gas.
75
Van der Waals equationnonideal gas
P + (V – nb) = nRTan2
V2( )}
correctedpressure
}
correctedvolume
76
A 13 gram gaseous sample of an unknown hydrocarbon occupies a volume of 11.2 L at STP. What is the hydrocarbon
a. CHb. C2H4
c. C2H2
d. C3H3
77
A force is applied o a container of gas reducing its volume by half. The temperature of the gas:
a. decreasesb. increasesc. remains constantd. the temperature change depends upon the
amount of force used
78
Ammonia burns in air to form nitrogen dioxide and water.
4NH3+7O24NO2 + 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 atmb. 6 atmc. 11 atmd. 12 atm
79
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
Energy Relationships in Chemical Reactions
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
81
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
82
Heat is the transfer of thermal energy between two bodies that are at different temperatures.
Energy Changes in Chemical Reactions
Temperature is a measure of the thermal energy.
Temperature = Thermal Energy
83
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
mass & energyExchange:
closed
energy
isolated
nothing
84
Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.
Endothermic process is any process in which heat has to be supplied to the system from the surroundings.
2H2(g) + O2(g) 2H2O(l) + energy
H2O(g) H2O(l) + energy
energy + 2HgO(s) 2Hg(l) + O2(g)
energy + H2O(s) H2O(l)
85
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.
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.
energy, pressure, volume, temperature
U = Ufinal - Uinitial
P = Pfinal - Pinitial
V = Vfinal - Vinitial
T = Tfinal - Tinitial
86
First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed.
Usystem + Usurroundings = 0or
Usystem = -Usurroundings
C3H8 + 5O2 3CO2 + 4H2O
Exothermic chemical reaction!
Chemical energy lost by combustion = Energy gained by the surroundingssystem surroundings
U is the change in internal energy of a system
87
Enthalpy and the First Law of Thermodynamics
U = q + w
U = H - PV
H = U + PV
q = H and w = -PV
At constant pressure:
U is the change in internal energy of a system
88
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.
H = H (products) – H (reactants)
H = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
H < 0Hproducts > Hreactants
H > 0
89
Thermochemical Equations
H2O(s) H2O(l) H = 6.01 kJ/mol
Is H negative or positive?
System absorbs heat
Endothermic
H > 0
6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.
90
Thermochemical Equations
CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890.4 kJ/mol
Is H negative or positive?
System gives off heat
Exothermic
H < 0
890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.
91
H2O(s) H2O(l) H = 6.01 kJ/mol
• The stoichiometric coefficients always refer to the number of moles of a substance
Thermochemical Equations
• If you reverse a reaction, the sign of H changes
H2O(l) H2O(s) H = -6.01 kJ/mol
• If you multiply both sides of the equation by a factor n, then H must change by the same factor n.
2H2O(s) 2H2O(l) H = 2 x 6.01 = 12.0 kJ
92
H2O(s) H2O(l) H = 6.01 kJ/mol
• The physical states of all reactants and products must be specified in thermochemical equations.
Thermochemical Equations
H2O(l) H2O(g) H = 44.0 kJ/mol
How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4(s) + 5O2(g) P4O10(s) H = -3013 kJ/mol
266 g P4
1 mol P4
123.9 g P4
x3013 kJ1 mol P4
x = 6470 kJ
93
A Comparison of H and U
2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g) H = -367.5 kJ/mol
U = H - PV At 25 °C, 1 mole H2 = 24.5 L at 1 atm
PV = 1 atm x 24.5 L = 2.5 kJ
U = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol
94
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 t
q = C x t
t = tfinal - tinitial
95
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
t = tfinal – tinitial = 5°C – 94°C = -89°C
q = mst = 869 g x 0.444 J/g • °C x –89°C = -34,000 J
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