From the Matter Unit:
Describe how matter is classified
Classify a substance as to type
Define Law of Definate Composition
Define Binary and Ternary
List names and formulas for polyatomic ions
Name ionic and covalent compounds (using the correct system-ex: roman numeral)
Write formulas for ionic and covalent compounds
Name and write formulas for acids
Part 1: Naming
Part 2: Formulas
1. Fe(NO3)2 iron(II) nitrate
11. cadmium(II) fluoride
CdF2
2. Mg(NO2)2 magnesium nitrite
12. barium phosphate
Ba3(PO4)2
3. KOH
potassium hydroxide
13. copper(I) chloride
CuCl
4. NH4NO3
ammonium nitrate
14. barium nitrate
Ba(NO3)2
5. CaSO4
calcium sulfate
15. iron(III) carbonate
Fe2(CO3)3
6. PbF2
lead(II) fluoride
16. calcium sulfite
CaSO3
7. PbO2
lead(IV) oxide
17. copper(II) hydroxide
Cu(OH)2
Part 3: What’s Wrong With These Formulas?
21. (NH3)2CO3
NH3 is not an ion. NH4 is ammonium.
22. Na2PO4
Because Na has a charge of +1 and phosphate is -3, the charges are
not equal. Should be Na3 PO4.
23. K(OH)2
Because K has a charge of +1 and hydroxide is -1, the charges are
not equal. Should be KOH.
From the Moles Unit:
Definition of a mole
Determine the molar mass of a substance
Make conversions from:
grams to moles
particles to moles (atoms/molecules)
liters of gas to moles
Calculate molarity of a compound
Calculate the % composition of a compund
Calculate an empirical formula of a compound
Determine the molecular formula from the empirical formula
3.01 x 1023 atoms of neon would be:
0.500 moles
10.1
grams
11.2
liters
Why atoms? Neon is an element..
6.02 x 1023 atoms of oxygen in a sample of oxygen gas would be:
0.500 moles of oxygen gas
16.0
grams of oxygen gas
11.2
liters of oxygen gas
Explain: Oxygen gas is diatomic, so 1 mole of oxygen gas is really 2
moles of oxygen atoms. Therefore, 1 mole of atoms is only 0.500 moles
of O2.
Mole Problems:
1.
What is the mass of one mole of the following?
Sodium sulfate
Na2SO4 = 142 grams
2.
What is the mass of 0.30 moles of aluminum?
0.30 moles Al x 27.0 g = 8.1 g
1 mole
3.
How many atoms of iron are in 0.250 grams of iron?
0.250 g Fe x 1 mole Fe x
6.03 x 1023 atoms Fe
55.85 g Fe
1 mole
4.
How many moles is 6.04 grams of manganese(II) sulfate?
6.04 g MnSO4 x
1 mole MnSO4
= 0.0400 moles MnSO4
151.0 g MnSO4
5.
What is the volume of a 25.0 gram sample of sulfur dioxide at STP?
= 2.69 x 1021 atoms
25.0 g SO2 x
1 mole SO2
64.06 g SO2
x
22.4 L SO2
1 mole SO2
= 8.74 L SO2
Molarity:
1.
What is the molarity if 82.0g of calcium nitrate is dissolved in 500.0 mL of
solution
82.0 g Ca(NO3)2 x 1 mole Ca(NO3)2 = 0.500 moles
164.1 g Ca(NO3)2
M =
0.500 moles
= 1.00 M
0.500 L solution
2.
Calculate the mass of solute in the following solution:
250.0mL of sodium sulfate that is 2.00M
2.00M = x moles
x = 0.500 moles x 142.04 g = 71.0 g Na2SO4
0.2500 L
1 mole
3.
How many milliliters of solution can be made from the following?
0.100M solution using 117g of sodium chloride
117 g NaCl x 1 mole = 2.00 moles
0.100M = 2.00 moles
58.44 g
x liters
x = 20.0 L
% Composition Problems:
1.
aluminum sulfate
Al2(SO4)3
54g Al x 100 = 16%
342g
2.
96g S x 100 = 28%
342g
192g O x 100 = 56%
342g
A compound is found to contain 74.0 grams of mercury and 6.0 grams of oxygen.
74.0g Hg x 100 = 92.5%
80.0g
6.0g O x 100 = 7.5%
80.0g
How many grams of oxygen are in 24.5 grams of this compound?
24.5g x
3.
7.5g = 1.8g Oxygen
100.0g
A 39.2 gram sample of a compound is found to contain 29.4 grams of carbon.
The only other element found to be present is hydrogen. Is this compound CH4?
Explain.
29.4g C x 100 = 75% C
CH4 would be: 12g C x 100 = 75%
39.2g
16g
Yes. The compound is CH4 It has the same % composition.
Empirical Formulas:
1.
A compound contains 67.6% mercury, 10.8 % sulfur, and 21.6 % oxygen. What
is the empirical formula?
67.6g Hg x 1 mole = 0.337 mole / 0.337 = 1
200.6g
10.8g S x 1 mole = 0.337 mole / 0.337 = 1
HgSO4
32.07g
21.6g O x 1 mole = 1.35 mole / 0.337 = 4
mercury(II) sulfate
16.00g
2.
100 grams of a compound containing aluminum and oxygen is found to have
52.94 grams of aluminum.
52.94g Al x 1 mole = 1.962 mole / 1.962 = 1 x 2 = 2
26.981g
Al2O3
47.06g O x 1 mole = 2.941 mole / 1.962 = 1.5 x 2 = 3
15.999g
aluminum oxide
3.
50.0 grams of a compound containing only oxygen and hydrogen is found to have
47.05 grams of oxygen. What is the empirical formula?
47.05g O x 1 mole = 2.941 mole / 2.93 = 1
15.999g
HO
2.95g H x 1 mole = 2.93 mole / 2.93 = 1
1.008g
If the molar mass of the compound is 34g, what is the molecular formula?
From the Reactions Unit:
List and describe the types of reactions
Identify a reaction as to type
Write and balance equations for reactions
Predict the products of a single replacement reaction using a reactivity series
Predict the products of a double replacement reaction
using a solubility chart
Write net ionic equation for a double replacement reaction
***The 1’s don’t have to be written, they are implied.
1.
2 Al (s) + 3 CuSO4 (aq)  1 Al2(SO4)3 (aq) + 3 Cu (s)
2.
3 H2SO4 (aq) + 2 Al (s)  1 Al2(SO4)3 (aq) + 3 H2 (g)
Determine Reaction Type, Predict Products, and Balance:
1.
strontium (s) + nitrogen (g)  synthesis
3 Sr (s) + N2 (g)  Sr3N2 (s)
2.
benzene (C6H6) (l) + oxygen (g)  combustion
2 C6H6 (l) + 15 O2 (g)  12 CO2 (g) + 6 H2O (g)
3.
cobalt(II) bromide (s)  decomposition
CoBr2 (s)  Co (s) + Br2 (l)
4.
phosphoric acid (aq) + aluminum hydroxide (aq)  double replacement
H3PO4 (aq) + Al(OH)3 (aq)  AlPO4 (s) + 3 HOH (l)
5.
zinc (s) + copper(II) chloride (aq)  single replacement
Zn (s) + CuCl2 (aq)  ZnCl2 (aq) + Cu (s)
Single Replacement:
Complete and balance the following metal exchange reactions. No reaction is a possible
answer.
Cr(NO3)3 (aq) +
Cu (s)  No Reaction (Cr more active than Cu)
Ca (s) +
H2SO4 (aq)  CaSO4 (s) + H2 (g)
Pb(C2H3O2)2 (aq) +
Ag (s)  No Reaction (Pb more active than Ag)
Aluminum nitrate solution and copper  No Reaction (Al more active than Cu)
Magnesium solid and nitric acid solution 
Mg (s) + 2 HNO3 (aq)  Mg(NO3)2 (aq) + H2 (g)
Double Replacement:
1.
Lithium fluoride solution is mixed with ammonium hydroxide solution.
LiF (aq) + NH4OH (aq)  No Reaction (both products are soluble)
2.
Potassium chloride solution is mixed with copper(I) nitrate solution.
KCl (aq) + CuNO3 (aq)  CuCl (s) + KNO3 (aq)
3.
(CuCl is ppt)
Hydrochloric acid solution is mixed with a solution of lithium hydroxide.
HCl (aq) + LiOH (aq)  LiCl (aq) + HOH (l)
(water is formed)
What would be observed in each reaction above? Explain why these observations
could be made. (What is happening at an atomic level?) Write the net ionic:
1. You would see nothing. There is no reaction.
2. You would see CuCl solid form (a precipitate).
3. You would see nothing, but there is a reaction because the H+ and OH- were
removed from the mixture.
Net Ionic:
1.
lead(II) nitrate (aq) + potassium iodide (aq) 
Pb+2 (aq) + 2 I- (aq)  PbI2 (s)
2.
(NO3-1 and K+1 are spectators)
sulfuric acid (aq) + barium hydroxide (aq) 
2 H+ (aq) + SO4-2 (aq) + Ba+2 (aq) + 2 OH-1 (aq)  2 HOH (l) + BaSO4 (s)
3.
copper(I) nitrate (aq) + sodium chloride (aq) 
Cu+1 (aq) + Cl-1 (aq)  CuCl (s)
4.
(NO3-1 and Na+1 are spectators)
calcium acetate (aq) + magnesium nitrate (aq) 
No Reactions because Ca(NO3)2 and Mg(C2H3O2)2 are soluble.
5.
silver nitrate (aq) + potassium chloride (aq) 
Ag+1 (aq) + Cl-1 (aq)  AgCl (s)
(NO3-1 and K+1 are spectators)
From the Stoichiometry Unit:
Predict the amount of product formed or reactant used with mole ratios of an
equation (calculate amounts in moles or grams)
Use molarity in stoichiometry calculations
Determine the limiting reactant (and reactant in excess)
Calculate a theoretical yield
Calculate a % yield
The mole to mole ratio in which substance react and are formed is called the
stoichiometric ratio.
1. Balance the following equation. Sketch the atoms or molecules involved:
2 H2 (g) + O2 (g) -----
2 H2O (g)
Why can the unit of moles be substituted for particles in a reaction?
Because a mole is a certain number of particles, the ratio
would be the same as the particle ratio. (1 dozen: 2 dozen is the same
ration as 1 particle; 2 particle)
If 2 moles of H2 reacts, how many moles of H2O are formed?
2 moles
If 4 moles of H2 reacts, how many moles of H2O are formed?
4 moles
If 1 mole of H2 reacts, how many moles of O2 react?
0.5 moles
2. This reaction is used to reduce iron from iron(III) oxide, and iron ore. Balance
and answer the following questions:
Fe2O3 (s) +
3 H2 (g) ------
Fe (s)
(160 g)
3(2 g)
2(56 g)
+
=
+
+
H2O (g)
3(18 g)
State the Law of Conservation of Mass. Show mathematically that the equation
above does not lose or gain in mass.
Mass is not created or destroyed. The total mass of the products equals
the mass of the reactants that was started with.
166 g = 166 g in the reaction above
How many moles of Fe will be produced when 4.5 moles of H2 reacts? 3 moles
How many moles of Fe will be produced when 1 mole of Fe2O3 reacts? 2 moles
3.
How many grams of carbon dioxide are produced when 100.0g of ethane (C2H6)
undergoes combustion?
2 C2H6 (g) + 7 O2 (g)  4 CO2 (g) + 6 H2O (g)
100.0g C2H6 x
1 mole C2H6 x 4 moles CO2 x 44.009g CO2 = 292.7g
CO2
30.07g C2H6
4.
1 mole CO2
What is the theoretical yield of lead(II) sulfide will be obtained by reacting 3.31g
of lead(II) nitrate with hydrosulfuric acid in a double replacement reaction?
Pb(NO3)2 (aq) + H2S (aq)  PbS (s) + 2 HNO3 (aq)
3.31g Pb(NO3)2 x
5.
2 moles C2H6
1 mole Pb(NO3)2 x
1 mole PbS
x 239.27g PbS
331.2g Pb(NO3)2
1 mole Pb(NO3)2
1 mole PbS
= 2.39g
How many liters of carbon dioxide will be formed when 144.0g of pentane
(C5H12) undergoes combustion?
C5H12 (l) + 8 O2 (g)  5 CO2 (g) + 6 H2O (g)
144.0g C5H12 x
1 mole C5H12 x 5 mole CO2 x 22.414L CO2 = 223.7L
CO2
72.149g C5H12
6.
1 mole C5H12
1 mole CO2
200.0mL of 2.0M solution of potassium iodide are reacted with excess lead(II)
nitrate. What is the theoretical yield of lead(II) iodide solid?
Pb(NO3)2 (aq) + 2 KI (aq)  PbI2 (s) + 2 KNO3 (aq)
2.0M = x moles KI = 0.40 moles KI
0.2000L soln
0.40 moles KI x
1 mole PbI2
2 mole KI
2.0 moles KI x
1L solution
x
461.01g PbI2
1 mole PbI2
1 moles PbI2
2 mole KI
x
= 92g PbI2
OR:
0.2000L x
461.01g PbI2
1 mole PbI2
= 92g PbI2
7.
How many grams of calcium chloride can be produced by reacting 150ml of .25M
hydrochloric acid solution with calcium metal?
Ca (s) + 2 HCl (aq)  CaCl2 (aq) + H2 (g)
0.25M = x moles HCl = 0.0375 moles HCl
0.15L soln
0.0375 moles HCl x
1 mole CaCl2 x
2 moles HCl
110.98g CaCl2
1 mole CaCl2
= 2.1g CaCl2
OR: 0.15L x
0.25 moles HCl
1L solution
x
1 mole CaCl2 x 110.98g CaCl2 = 2.1g CaCl2
Limiting Reactant:
1. The unbalanced equation for the synthesis of sodium amide (NaNH2) is as
follows:
2 Na (s)
+
2 NH3 (g) ------ 2 NaNH2 (s)
+
H2 (g)
How much NH3 is required to react with 50.0 grams of Na?
50.0g Na x
1 mol Na x
22.9898g Na
2 mol NH3 x
2 mol Na
If you have 50.0 grams of NH3, is that enough?
17.03g NH3
1 mol NH3
= 37.0g NH3
Yes
What is the limiting reactant? Why does it matter? The Na would run out and so
is the limiting reactant. I have to use the mass of Na to calculate a yield because
not all 50.0grams of the NH3 will be reacted (some will be left over at the end of
the reaction).
What is the theoretical yield of NaNH2?
50.0g Na x
2.
1 mol Na x 2 mol NaNH3
22.9898g Na
2 mol Na
x
39.01g NaNH2
1 mol NaNH2
= 84.8g NaNH2
When 125mL of 0.55M silver nitrate solution is reacted with 85.0mL of 0.25M
aluminum chloride solution, solid silver chloride is formed.
a. Write the balanced equation for the reaction.
3 AgNO3 (aq) + AlCl3 (aq)  3 AgCl (s) + Al(NO3)3 (aq)
b. Which reactant is the limiting reactant?
0.55M = x mol AgNO3
0.125L soln
0.0688 mol AgNO3 x
= 0.0688 mol AgNO3
1 mol AlCl3 = 0.0229 moles AlCl3 needed so…
3 mol AgNO3
0.25M = 0.0229 mol AlCl3
x L solution
= 0.0917L of solution needed (91.7mL)
You need 91.7mL of the AlCl3 solution to react all the AgNO3 solution.
You only have 85.0mL, so you don’t have enough. AlCl3 limits!
c. What mass of solid can be produced?
0.25M = x mol AlCl3
0.0850L soln
0.0213 mol AlCl3 x
= 0.0213 mol AlCl3
3 mol AgCl = 0.0638 mol AgCl
1 mol AlCl3
0.0638 mol AgCl x
143.32g AgCl
1 mol AgCl
= 9.14g AgCl
From the Thermodynamics Unit:
Define exothermic and endothermic reactions
Define enthalpy
Write a thermochemical equation
Perform stoichiometric calculation using heat as part of the equation
Read a reaction chart to determine the change in enthalpy
Calculate the change in enthalpy using a standard heat of formation chart
Describe Hess’ Law
Define entropy
Use Gibbs’ Free energy to predict whether a reaction is spontaneous
1.
2.
How much heat does 32.0 g of water absorb when it is heated from 25.0oC to
80.0oC?
q = 32.0g x 4.18 j x 55.0oC = 7360 j
g oC
When 435 J of heat is added to 3.4 g of olive oil at 21oC, the temperature
increases to 85.oC. What is the specific heat of olive oil?
435 j = 3.4g x C x 64oC = 2.00
j
g oC
Would olive oil or water have a larger temperature change given the same amount
of energy?
The olive oil would have a greater temperature change because it has
a smaller specific heat. (Takes less energy to raise the temp 1oC)
3.
1500 joules of heat are added to 50.0g of water. How much would the
temperature change? Would the temperature increase or decrease?
1500 j = 50.0g x 4.18 j x t = 7.2oC
g oC
Phase Changes:
1.
How much heat is liberated when 18.0g of water vapor turns into rain?
18.0g H2O x 540 cal/g = 9720 cal = 9700cal to 2 sig fig
2.
How much heat is removed from your hand when 18g of snow melts on your
glove?
18g x 80.0 cal/g = 1440 cal = 1400 cal to 1 sig fig
3.
How much heat is liberated when 450.0g of steam at 100oC is changed to water at
20oC?
450.0g 540 cal/g = 243000cal
o
total = 280,000cal
to 2 sig fig
450.0g x 1 cal x 80.0 C = 36000cal
g oC
Thermochemical Equations:
1.
Define
a. endothermic a reaction in which energy is absorbed
b. exothermic a reaction in which energy is released
c. enthalpy heat (energy) changes in a reaction
2.
For the reaction:
C8H18 (l) + 25/2 O2 (g)  8 CO2 (g) + 9 H2O (g) + 1300 kJ
a. Is the reaction exothermic or endothermic? exothermic
b. How much energy is released when 456.0g of C8H18 burns?
456,0g C8H18 x 1 mole C8H18 x
1300kJ
= 5190kJ
114.22g C8H18
1 mole C8H18
c. H = - 1300kJ
d. How does the energy of the reactants compare to the energy of the products?
Draw a sketch of the graph of energy vs. time.
3.
Using a heat of formation chart, Calculate the H for each of the following
reactions:
2 C2H2 (g) + 5 O2 (g)  4 CO2 (g) + 2 H2O (g)
2 NH3 (g)  N2 (g) + 3 H2 (g)
4.
Methane (CH4) undergoes combustion.
a.
Use the heat of formation chart to calculate the H for the reaction.
CH4 (g)
+ 2 O2 (g)
1(-17.889kcal) + 2(0kcal)
= -17.889kcal
 CO2 (g)
+
2 H2O (g)
1(-94.0518kcal) + 2(-57.7979kcal)
= -209.6476kcal
-209.6476kcal – (-17.889kcal) = -192kcal
b.
Write the thermochemical equation.
CH4 (g)
+
2 O2 (g) 
CO2 (g)
+
2 H2O (g) + 192kcal
c. Is the reaction exothermic or endothermic? exothermic
d. Which has more energy, reactants or products? Draw a graph to show the
energy relationships on the reaction. Reactants have more energy!
e. How many grams of methane are needed to produce 4000.0 kcal of energy?
4000.0kcal x
1 mol CH4
192kcal
x
16.05g
= 334kcal
1 mol CH4
f. How much heat is released per gram of methane?
192kcal
= 12.0kcal/g
16.05g CH4
g. Would you expect this reaction to be spontaneous at room temperature? What
is the driving force?
The change in enthalpy is very favorable (a large negative number)
because the reaction is very exothermic. The change in entropy is
probably not significant as it goes from all gases to all gases. So, Gibbs
Free Energy is probably negative and the reaction is spontaneous.
5. Calculate the heat of reaction for the following:
2 C (graphite) + H2 (g)  C2H2 (g)
given:
x4: 4C (graphite) + 4O2 (g)  4CO2 (g)
2 C2H2 (g) + 5 O2 (g)  4 CO2 (g) + 2 H2O (l)
4 CO2 (g) + 2 H2O (l)  2 C2H2 (g) + 5 O2 (g)
H = -393.5 kJ x 4 =
H = -1574.0 kJ
H = -285.8 kJ x 2 =
H = -571.6 kJ
H = -2598.8 kJ
H = +2598.8 kJ
4 C (graphite) + 2 H2 (g)  2 C2H2 (g)
H = +453.2 kJ
C (graphite) + H2 (g)  C2H2 (g)
H = +226.6 kJ
x 2: 2H2 (g) + 1 O2 (g)  2H2O (l)
flip:
or:
Evaluate this reaction in terms of H, S, and G.
The H is positive, so not favorable.
The S is probably positive, but only marginally as it goes from one solid and one
gas to only one gas. This is favorable as it is an increase in entropy.
So G is negative at higher temperatures.
Do you think this reaction would be spontaneous?
The change in enthalpy is unfavorable (a positive number) because the reaction is
endothermic. The change in entropy is probably positive as it ends with only
gases. So, Gibbs Free Energy is negative at higher temperatures.
From the Rate Unit:
Describe the factors affecting rate
Calculate rate of reaction in a specific time interval
Use stoichiometry to determine information about the rate of a reaction
Determine the rate law for a given reaction with data
Calculate a rate constant
Use a rate constant to predict rates
1. Above is the reaction graph for the following reaction:
A
+ 2B  AB2
Label the position of the reactants, products, and activated complex.
The reactants are at 50kJ on the left, the products are at 100kJ on the right and the
Activated complex it at the top of the hill at 250kJ of energy.
What is the activation energy (Ea) of the reaction?
250)
200kJ (from 50 to
What is the H for the reaction?
+ 50kJ (from 50 to 100)
Do the reactants or the products have a higher energy?
Products have higher
Is the reaction exothermic or endothermic?
What does a catalyst do to the graph?
List 4 things that would speed up the reaction.
Using a higher concentration of reactants
Heating the reaction
Using a catalyst (if there is one for that reaction)
Using small particles of the reactants
List 4 things that would slow down the reaction.
Using a lower concentration of reactants
endothermic
A catalyst lowers the
activation energy by making
the collisions more effective.
Coolong the reaction
Using an inhibitor (if there is one for that reaction)
Using large particles of the reactants
If .5 moles of A react in 2 minutes and 16 seconds, what is the rate in terms of
disappearance of A?
0.5 moles A = 4 x 10-3 mol A/sec
136 seconds
What is the rate in terms of disappearance of B? How do you know?
Because there are 2 mol of B reacted for every 1 of A, the rate for B is double the
rate for A (twice as many moles of B are used in the same time interval).
Rate = 2 (4 x 10-3 mol A/sec) = 8 x 10-3 mol B/sec
What is the rate in terms of appearance of the product? How do you know?
Because the mol ratio of A to AB2 is 1:1, the rates are equal (you produce the
same number of moles of AB2 as you use of A in the same time interval).
Rate = 4 x 10-3 mol AB2/sec
2. A series of data was collected for the following reaction. Use the coefficients of the
equation to fill in the blacks with your prediction of the molarities of the other species
in the reaction.
H2 (g) + I2 (g)  2 HI (g)
time (sec)
0
2
4
[H2]
4.00
2.00
1.00
[I2]
6.00
4.00
3.00
[HI]
0.000
4.00
6.00
What is the rate in terms of appearance of HI for the 1st 2 second interval?
Rate = 4.00 mole HI = 2.00 mol HI/sec
2 second
What would the rate for the 1st 2 second time interval be in terms of the
disappearance of H2?
The rate in terms of H2 is half of the rate in terms of HI because of the 1:2 mole
ratio.
Or:
Rate = 2.00 mole H2 = 1.00 mol H2/sec
2 second
What is the rate in terms of appearance of HI for the 2nd 2 second interval?
Rate = 4.00 mole HI = 2.00 mol HI/sec
2 second
Why did it change?
As the concentration of the reactants decreases, there are fewer collisions so the
reaction slows.
3. The following rate data describe the reaction below at a temperature of 250 K:
F2 (g) + 2 ClO2 (g)  2 FClO2 (g)
[F2]
0.10
0.10
0.20
[ClO2]
0.010
0.040
0.010
initial rate of disappearace of F2
1.2 x 10-3 mol. L.-1 sec-1
4.8 x 10-3 mol. L.-1 sec-1
2.4 x 10-3 mol. L.-1 sec-1
What is the rate law for the reaction?
Rate = k[F2]1 [ClO2]1
Calculate the rate law constant.
1.2 x 10-3 mol. L.-1 sec-1 = k [0.10]1 [0.010]1
k = 1.2
Calculate the rate when [F2] = 0.010 M and [ClO2] = 0.020 M.
Rate = 1.2 [0.010]1 [0.020]1
= 2.4 x 10-3 mol. L.-1 sec-1
What is the rate in the first trial in terms of FClO2 ?
Rate = 2(1.2 x 10-3 mol/L/sec) = 2.4 x 10-3 mol/L/sec
From the Equilibrium Unit:
Define reversible reaction
Describe chemical equilibrium (what is true at equilibrium)
Describe the factors affecting equilibrium
Write the equilibrium law expression for a reaction
Calculate the equilibrium constant given data (may need to use stoichiometry or
express the concentration as x)
1. Write the expression for the equilibrium constant for the production of HI gas from
hydrogen gas and iodine gas.
H2 (g) + I2 (g)   2 HI (g)
K =
[HI]2
[H2] [I2]
=
3. For the equilibrium:
sulfur dioxide + oxygen <--> sulfur trioxide
at a certain temperature, [SO2] = 0.20 M, [O2] = 0.20 M, and [SO3] = 0.40 M
Calculate Keq.
2 SO2 (g) + O2 (g)   2 SO3 (g)
[0.20]
[0.20]
[0.40]
K =
4.
[SO3]2
=
[SO2]2[O2]
(0.40)2
= 20
(0.20)2 (0.20)
Consider the following reaction:
2 HCl (g)  H2 (g)
Cl2 (g)
The initial concentration of HCl is 2.0M and there is no H2 or Cl2 present. After
equilibrium conditions have been established, the concentration of Cl2 is 0.10M.
What the equilibrium concentrations for HCl and H2?
2 HCl (g)  H2 (g)
Cl2 (g)
[2.0]
0
0
-0.20
+0.10
+0.10
[1.80]
[0.10]
[0.10]
Calculate Keq .
Keq =
5.
[0.10] [0.10]
[1.80]2
= 0.0031
The reaction, A + B   C + D has a Keq of 6.57 x 10-3. Determine the
final equilibrium concentrations of all substances if 0.200M C and 0.200M D are
mixed. No A or B are initially present.
A
+
B  C
+
D
Initial
0
0
[0.200]
[0.200]
Change
+x
+x
-x
-x
Eq
x
x
0.2 - x
0.2 - x
[0.185]
[0.185]
[0.015] [0.015]
6.57 x 10-3 = [C] [D] =
[A] [B]
(0.200 -x)2
x2
x = 0.185
*simplify by taking the square root immediately. So:
6.
0.0811 = 0.2 – x
or 0.0811x = 0.20 – x
x
Use the following equation to answer the following questions:
N2 (g) +
3H2 (g)
2NH3 (g) + heat
Write the equilibrium constant expression
Keq = [NH3]2
[N2][H2]3
Calculate the value of Keq if at equilibrium nitrogen is 2.0M, hydrogen is 1.0M,
and ammonia is 0.50M.
Keq = [0.50]2
= 0.125
[2.0][1.0]3
Will there be more reactant or product?
Favors the reactant
Will the reactants or products be favored if:
a.
b.
c.
d.
e.
f.
the pressure is decreased. reactant
the reaction is cooled. product
a catalyst is added. neither
NH3 is added. reactant
H2 is added. product
some nitrogen is removed. reactant
How will the following change the value of Keq?
a.
b.
c.
d.
Ammonia is added. No change.
The reaction is heated. Value will decrease.
The pressure is increased. No change
A catalyst is added. No change.
7. Describe what is happening as a system goes toward equilibrium. What is true at
equilibrium and what happens at each shift shown below. (Use the terms: reversible
reaction, forward reaction, reverse reaction, rate, products, and reactants.)
A reversible reaction is a reaction in which the forward and reverse reactions occur at the
same time. The forward reaction begins quickly and slows as product is formed. The
reverse reaction begins slowly and speeds up. Eventually the rate of the forward reaction
is equal to the rate of the reverse reaction. When the rates are equal, the concentrations
of the reactants and products remain constant. (although not equal to each other) You
are making product as quickly as you are reacting the product away.
8. Describe La Chatelier’s Principle:
When a stress is applied to a system at equilibrium, the system shifts to minimize the
stress. (if you add a reactant, the system tries to use it up by making more product.)
Fill in the table below for the reversible exothermic reaction. A plus sign indicates that
the concentration of that substance was increased. A minus sign indicates that the
concentration of that substance was decreased.
3 O2(g) + 2 H2S (g)   2 H2O (g) + 2 SO2 (g) + heat
Stress
[O2]
[H2S]
Direction
Of Shift
[H2O]
[SO2]
Add
H2 S
-
+

+
+
Remove
H2 O
-
-

-
+
Increase
Temperature
+
+

-
-
Decrease
Pressure
+
+

-
-
9. In which trial(s) would the value of the equilibrium constant (Keq ) change? Would
the value increase or decrease?
The value of Keq changes with the increase in temperature.
When temperature increased, reactants favored, Keq decreases
10. What would you do to maximize the production of SO2? High temp, low pressure,
remove H2O and SO2, add O2 and H2S.