Download Chemistry 115 Lecture Number Seventeen Test Two Review April 2

Chemistry 115
Lecture Number Seventeen
Test Two Review
April 2, 2009 [sk]
Note from Dr. Sevian: These notes were based on the material covered in Lecture 17, but
also augmented from the student’s own notes, so there is much more detail than what was
covered in lecture. Thank you to the student [sk] who put together these notes! They are an
excellent review to help with preparing for the test. The solution to the Hess’s law part (a) of
the Challenge Problem is also included at the end of these notes.
Categorizing Chemical Reactions-2 main ways to classify reactions:
-based on the pattern that they follow- exchange reactions, synthesis, decomposition
-based on underlying theoretical model- redox
-there are reactions that do not fall under these general rules-some decomposition
reactions, acid anhydride reactions with water
Exchange Reactions- No Oxidation Number Change-Precipitation reactions-The reagents anions and cations switch partners, then following the solubility
guidelines you determine which of the products is not soluble, this is your precipitate, then
you can determine the spectator ions who do not impact the reactions, this allows a net ionic
equation to be written with just the ions that impact the reaction.
-Acid-base reactions- Acid and base react to form a salt and water- you need to determine the strength or
weakness of the acid and the base (first semester chemistry will use only strong bases), the
strength or weakness of the acid will determine the extent to which the acid on the reactant
side will dissociate, this allows you to what the products will be, then determine spectator
ions to write net ionic equation.
-Gas forming reactionsAcid-Base Reactions-recognize acids by the H+ cation, recognize bases by the OH- anion
-pH-info about pH will be introductory- more complex discussed in second semester of
chemistry
-need to be able to go back and forth between the pH and the concentration of [H+];
calculate one if given the other
-concentration of [H+] is given in molarity- M=mol/L
-a pH of 7 means that the concentration of [H+] ions is 1.0 x 10-7- this allows us to calculate
the pH from the molarity, and the molarity from the pH
-to get from molarity to pH: take the –log of the molarity value
-to get from pH to molarity: take the antilog of the pH
-know the general scale of pH:
0------------------7-----------------14
Acidic
Neutral
Basic
-the weaker the acid the closer to 7- stronger acid the closer to 0
-the weaker the base the closer to 7- the stronger the base the closer to 14
-need to be able to predict the product of reactions between acids and bases
-***memorize the list of strong acids***
-HCl- hydrochloric acid
-HBr- hydrobromic acid
-HI- hydroiodic acid
-HClO3- chloric acid
-HClO4 – perchloric acid
-HNO3- nitric acid
-H2SO4 – sulfuric acid
-Predicting Products of Acid/Base Reactions- Exchange Reactions-need to know that in exchange reactions the anions and cations switch partners
-neutralization reactions the H+ and OH- combine to create H20
-need to know the list of strong acids to know if the acid will dissociate completely or only a
little- *memorize these*
-need to be able to give formula for the reagents and products listed -*memorize rules*
1)
HNO3 (aq) + NaOH (aq) > NaNO3 (aq) + H2O (l)
Strong Acid
Strong Base
Very Soluble
Molecule
Dis.Complety
Dis.Complety
Dis. Complety Remains Whole
H+ / NO3Na+ / OH- > Na+ / NO3H2O
-cross out spectator ions- (Na + and NO3-)
- Net ionic equation is H+ (aq) + OH- (aq) > H2O (l)
- this will always be the net ionic equation when a strong acid and a strong base react
because both dissociate completely – the anion from the acid and the cation from
the base will always be the spectator ions
2)
HCH3COO (aq) + KOH (aq)
> KCH3COO (aq) + H2O (l)
Weak Acid
Strong Base
Soluble
Molecule
Dis.Partially
Dis.Completly
Dis.Completly
Remains Whole
K+ / OHK+ / CH3COOH2O
Some [H+] ions
Some CH3COO ions
*Mostly HCH3COO*
-because the acid only dissociates partially, you use the most prevalent aspect of the
acid as it appears- meaning that because you mostly have HCH3COO present this is
what is used as the reagent, not its parts as you would if it were a strong acid
-cross out spectator ions (only K+)
-Net ionic equation is HCH3COO (aq) + OH- (aq) > CH3COO- (aq) + H2O (l)
3)
H2C2O4 (aq) + NaOH (aq) > NaHC2O4 (aq) + H2O (l)
-Net ionic equation is H2C2O4 (aq) + OH- (aq) > HC2O4- (aq) + H2O (l)
-Whenever reaction is with a weak acid- the weak acid is written as a whole molecule
Redox Reactions-utilize oxidation numbers
-oxidation numbers are a value assigned either to every element in an ion or molecule
involved in reaction, number can either have a negative or positive value, or could be zero
(can be fractions as well as whole numbers)
-reactions occur according to the theoretical model of oxidation numbers
-in ions these charges are real, in molecules the charges are fake because they are neutral
-allows to account for the movement of electrons in reactions that do not follow the
exchange pattern of reaction
-as something becomes more positive= oxidized
-as something becomes more negative= reduced
Assigning Oxidation Numbers-add to zero in neutral compounds, molecules are assigned fake charges
-add to total charge in ionic compounds
-atoms in free, neutral elements have oxidation number of zero (Zn, Ar, O2, S8, Ag, N2)
-ions with charges are assigned their charges as their oxidation numbers
FeCl3 :every Fe has a charge of 3+, where every Cl has a charge of 1so Fe has oxidation number of +3 and Cl has oxidation number of -1
FeCl2 :every Fe has a charge of 2+, where every Cl has a charge of 1so Fe has oxidation number of +2 and Cl has oxidation number of -1
Na+ : Na has charge of 1+
so Na has oxidation number of +1
-oxygen when in compounds almost always has oxidation number of 2Al2O3 :every O has oxidation number of 2-, so every Al must have oxidation
number of 3+ so that the compound remains neutral
MgO :every O has oxidation number of 2-, so Mg has oxidation number of 2+
-exception in peroxides when charge is 1- (H2O2-hydrogen peroxide, Na2O2-sodium
peroxide)
-remember if oxygen is in neutral state- O2 alone- it has an oxidation number of zero
-hydrogen when in compounds or as an ion has oxidation number of 1+
HCl :every H has oxidation number of 1+, so every Cl has oxidation number of
1- so that the compound remains neutral
-exceptions are when hydrogen becomes a anion- very rarely (CaH2, NaH)
-remember if hydrogen is in neutral state- H2 alone- it has an oxidation number of
zero
Practice Assigning Oxidation Numbers-CH3COOH
-every O has ox# of 2-every H has ox# of 1+
-compound is neutral, so ox#’s must equal zero
-use algebra equation to find ox# of C
2x + 4(1+) + 2(2-) = 0
2x + 4 + 4- =0
2x=0
x=0
-so C has oxidation number of 0
-C3H6
-ClO3-
-we know that every H has ox# of 1+
-so, algebraically
3x + 6(1+) = 0
3x +6= 0
3x=-6
x=2-so C has oxidation number of 2-we know that the overall charge has to be 1-we know that every O has ox# of 2-so, algebraically
x + 3(2-) = 1x -6 = 1x= 5
-so Cl has oxidation number of 5
Redox Reactions-redox reactions we cover in this semester of chemistry occur between a metal and an acid or
a metal and a metal salt
-follow this patternA + BX > AX + B
Zn(s) + 2 HBr(aq) > ZnBr2(aq) + H2 (g)
Mn(s) + Pb(NO3)2(aq) > Mn(NO3)2(aq) + Pb(s)
-element whose oxidation number becomes more positive is oxidized
-element whose oxidation number becomes more negative is reduced
-Example3Fe(NO3)2 (aq) + 2Al (s) > 3Fe (s) + 2 Al(NO3)3 (aq)
-first step is to assign oxidation numbers to the parts
-assign the charges on the parts that you know in order to assign charges on the
remainder-polyatomic ions- assign charge that the ions carry
-atoms alone have ox#- 0
-charged ions have ox# equal to their charge
-oxygen (usually) ox# -2
-hydrogen (usually) ox# +1
Reactant Side1) 3Fe(NO3)2 (aq) is made of Fe and NO3
-NO3 we know carries a charge of -1
a.We have 6 of them so that means that part is -6 ***the ox# of
NO3 is -1 though, not -6, we use -6 to determine the ox# of Fe****
b. So we know that the total charge must be 0, we can set our
unknown to x
c. 3X + -6 = 0
d. so X must be +2
2) 2Al (s) has oxidation number of 0 because it is a free element
Product Side3) 3Fe (s) has oxidation number of 0 because it is a free element
4) 2Al(NO3)3 (aq) is made of Al and NO3
- NO3 we know carries a charge of -1
a. We have 6 of them so that means that part is -6 ***the ox# of
NO3 is -1 though, not -6, we use -6 to determine the ox# of Fe****
b. So we know that the total charge must be 0, we can set our
unknown to x
c. 2X + -6 = 0
d. so X must be +3
5) Re-write the chemical formula with the oxidation numbers below so that you can
assess which is oxidized and which is reduced
-3Fe(NO3)2 (aq) + 2Al (s) > 3Fe (s) + 2Al(NO3)3 (aq)
+2 -1
0
0
+3 -1
-in summary- Fe +2 > Fe 0 so is reduced
- Al 0 > Al +3 so is oxidized
Predicting Redox Reactions-using the metal reactivity series, we can predict what reaction will occur
-metal reactivity series is a table where the most reactive(easily oxidized) metals are higher –
called the active metals, and the less reactive(not easily oxidized) metals are lower- called
noble metals
-this allows you to predict what reaction will occur between two metals or a metal and an
acid or particular salt
-the metal that appears higher on the list will be oxidized
- ExampleMg(s) reacting with Ni(s) will follow:
Mg(s) + Ni(s) > Ni(s) + Mg(s)
0
2+
0
2+
-because Mg is higher on the activity series it will drive reaction and become oxidized
-the metal which is less active will remain as ion on reactant side
-*don’t need to memorize the reactivity table*
Writing Net Ionic Equations for Redox Reactions-Anions in redox reactions are the spectator ions and do not appear in the net ionic equation
- Example- Chromium with hydrobromic acid-Balanced Equation- 2Cr (s) + 6HBr (aq) > 2CrBr3 (aq) + 3H2 (g)
-Net Ionic Equation- 2Cr (s) + 6H+ (aq) > 2Cr3+ (aq) + 3H2
Solution Concentrations-concentration means how many particles are present in s given amount of space
-molarity (M) is equivalent to the moles of solute/liter of solution
-when you need to find concentration of solution you use the concentration formula:
M =moles solute
Liter of solution
-**in order to use formula you must have the number of moles, if you are given the
information in grams you will first need to convert to moles using the formula weight (f.w)
which is the same as the molar mass**
-**also be sure that the volume of solutions are converted to liters in order to use formula**
-ExampleWhat is the concentration of a solution that was made by adding 1.95g of Na2SO4 to
250. ml of water?
Given info1.95g of Na2SO4
250 ml of water
f.w.= 142.05
1) Convert grams to moles using f.w
1.95 g Na2SO4 x 1mol Na2SO4 = 0.01373 mol of Na2SO4
142.05 g
2) Convert 250 ml of water to L water
250. ml x 1 Liter = 0.25 L
1000 mL
3) Use this converted info in concentration formulaM =moles solute
M= 0.01373 mol
M= 0.0549 mol/L
Liter of solution
0.25 L
Knowing this molarity, what is the concentration of Na+ ions in solution?
1)The answer depends on the molar ratio of ions present in Na2SO4
2) We have 0.0549 mol/L Na2SO4
0.0549 mol Na2SO4 x 2 mol Na+ = 0.110 moles Na+/ L
1 Liter
1 mol Na2SO4
*** Important to pay close attention to the molar ratio to find the concentration of
specific ions within a solution, easy to overlook and get wrong answer***
-when finding concentration of two solutions combined you need to add the moles present
from each solution and divide by the sum of the volumes of the solutions added together
- ExampleWhat is the concentration of Cl- when 250. ml of 0.100 M NaCl is mixed with 250.
ml of 0.300 M CaCl2 ?
Given info250. mL of 0.100 M NaCl
250. mL of 0.300 M CaCl2
1) Find moles of Cl- present in NaCl
.250L x 0.100 mol = 0.0250 mol of Cl1 Liter
2) Find moles of Cl- present in CaCl2
.250L x 0.300 mol =0.075 mol of Cl1 Liter
Because it is Cl2 you must multiply the moles by 2
2x0.075 mol of Cl- = 0.150 mol Cl3) Find the total volume of solution
.250 L of + .250L = .500 L
4) Molarity of Cl- :
M= total moles of ClM= 0.0250 mol +0.150 mol M=0.350
L of solution
.500 L
Solution Stoichiometry-follows same methodology of Stoichiometry
-need to first find the moles of A- using the volume and the molarity to find moles
-next using the molar ratio given by the stoichiometric coefficients from the balance
equation you can determine the number of moles of B that are produced
-then you can convert moles to grams using the molar mass of B
-Example2AgNO3 (aq) + Cu (s) > Cu(NO3)2 (aq) + 2Ag (s)
-If you used 34.2 mL of 0.100 M AgNO3 solution, how many grams of copper metal would
react?
1) Using the volume and molarity of AgNO3 we find the moles of AgNO3
34.2 mL to L - .0342 L AgNO3
0.0342 L x 0.100 moles AgNO3 = 0.00342 moles AgNO3
1L
2) Using the molar ratio of moles AgNO3 to Cu find the moles of Cu
0.00342 moles AgNO3 x 1 mole Cu = 0.00171 moles Cu
2 moles AgNO3
3) Using the molecular weight of Cu find the grams of Cu
0.00171 moles Cu x 63.55 g Cu = 0.109 g Cu
1 mole Cu
Titration-a method used by chemists to determine the concentration of a particular solution by
adding a solution of known concentration- a standard solution
- ExampleYou are given 0.05182 g of KHP (f.w- 204.23 g/mol) and 23.4 mL of NaOH exactly
neutralizes the KHP, what is the concentration (M) of NaOH?
1) Using the f.w for KHP find the moles of KHP
0.05182 g KHP x 1 mol KHP = 0.00025373 mol KHP
204.23 g
2) Using the molar ratio determine the moles of NaOH
0.00025373 mol KHP x 1 mol NaOH = 0.00025373 mol NaOH
1 mol KHP
3) Using the concentration formula find the molarity of NaOH
23.4 mL NaOH= 0.0234 L NaOH
M = 0.00025373 mol NaOH = 0.0108 M NaOH
0.0234 L
Thermodynamics-Conservation of Energy- important, energy is neither created nor destroyed- it changes
forms, we can keep track of changes in energy by measuring enthalpy (heat energy) changes
-∆H is change in energy as measured by temperature change
-Exothermic reactions- give off energy: start at levels of higher energy and end at levels of
lower energy, therefore the overall ∆H for an exothermic reaction will be negative, because
the system is loosing energy
-Endothermic reactions- take in energy: start off at levels of lower energy and end at levels
of higher energy, therefore the overall ∆H is positive, because the system is gaining energy
-In this way we can measure the change in energy by the change in temperature
Calorimetry-the method by which we can determine the heat lost or gained by a system
-a system is surrounded by a substance with a known heat capacity (C)- typically water
because it is well understood
-amount of heat transferred (q) is dependent on three variables
-the quantity of material in the object- m
-the identity of material- the specific heat- C *specific heats will be provided*
-the size of temperature change- ∆T (final T- initial T)
-formula- q= m x C x ∆T
-once you determine the change in temp of the surrounding of the system- (the qw,) you can
find the change of the system; the determined value for qsys is the opposite of qw- because we
known that the calculated change of the surrounding is equal to and opposite of the change
within the system
-Example- Measuring heat transferred from system
Given info:
Mass of water: 100.0g
Temp. of water initial: 23.00° C
Temp of water final: 23.71° C
C water: 4.184 J/g x K
q=m x C x ∆T
q= 100.0 g x 4.184 J/g•K x ( 23.71° C -23.00° C)
qw= 10041.6 J
qw =10.0 kJ
**the K degrees cancels out the C degrees because we are measuring changes in temp- the
degrees are the same size**
Hess’s Law-allows us to find the ∆H for a reaction from looking at the ∆H of formation of the
reactants and the products
-look at the challenge question to illustrate Hess’s Law and the ways we use them to solve
problemsChallenge Problem- we were asked to determine the amount of energy attained relative to
the moles of CO2 produced when given three different fossil fuels-We approach this problem by using the ∆H°f for the parts of the equations
∆H°f CO2 (g) = -393.5 kj/mol
∆H°f H2O (l) = -285.83 kj/mol
∆H°f C (s) = 0 kj/mol
∆H°f C8H18 (l) = -250 kj/mol
∆H°f O2 (g) = 0 kj/mol
∆H°f CH4 (g) = -74.8 kj/mol
1) Combustion of CoalC (s) + O2 (g) > CO2
∆Hrxn = ∑∆H°f products - ∑∆H°f reactants
∆Hrxn= [ -393.5]- [0+0] = -393.5 kj
-because this reaction occurs in a 1:1 mole ratio this means that the combustion of coal
releases 393.5 kj of energy per mole of CO2
2) Combustion of MethaneCH4 (g) + 2 O2 (g) > CO2 (g) + 2 H2O (l)
∆Hrxn = ∑∆H°f products - ∑∆H°f reactants
∆Hrxn= [ 2(-285.83) + -393.5] – [-74.8 + 2(0)] = -890.6 kj
-because this reaction occurs in a 1:1 ratio this means that the combustion of methane
releases 890.6 kj of energy per mole of CO2
Note: if you used H2O (g) in the reaction, then you would have used the ∆H°f for forming
the gas phase of H2O and your answer would be a little bit different. But as long as you are
using the same phase of H2O in the octane reaction, you can compare them appropriately.
3) Combustion of Octane
C8H18 (l) + 12.5 O2 (g) > 8 CO2 (g) + 9 H2O (l)
∆Hrxn = ∑∆H°f products - ∑∆H°f reactants
∆Hrxn= [ 8(-393.5) + 9(-285.83)] – [-250 + 12.5(0)] = -5470.5 kj
-because this reaction occurs in a 1:8 ratio, we need to divide the total energy released by 8
to get the kj/mol of CO2 released by this combustion= 683.8 kj/mol
Notice that all of these ∆Hrxn values are negative. It is because they are all exothermic
reactions.
From comparing the kj/mol CO2, methane is the fuel that produces the most heat for the
least amount of CO2 made.