How do I determine
the relative vapour pressure of different systems under
different temperatures and volumes?
Answer
In which of these
closed containers is the vapour pressure the greatest?
Explain your reasoning.
A) 1L container with 5 mL of water held at 20°C
B) 2L container with 15 mL of water held at 20°C
C) 3L container with 500 mL of water held at 15°C
D) 2L container with 15 mL of water held at 25°C.
The 2L container with 15 mL of water held at 25°C has the
highest vapour pressure.
The 1L container with 5 mL held at
20°oC and the 2L container with 15mL of water held at 20°C
are next with equal vapour pressure.
The 3L container with 500 mL held at
15°C has the lowest vapour pressure.
Here are the reasons:
 All containers contain the same liquid,
so they all have the same intermolecular force of attraction
in the liquid.
 The volume of
liquid in the container does not affect vapour pressure.
 The volume of the
container does not affect vapour pressure.

Only the
temperature affects vapour pressure. As temperature
increases, vapour pressure also increases.
In the given problem the only real
variable is the temperature of the liquid. The highest
temperature will yield the highest vapour pressure. The
medium temperature will yield a slightly lesser vapour
pressure. The low temperature will yield a third  and
lowest  vapour pressure.
Problem 2
How do I determine equilibrium
concentrations in a reaction when I am given initial
concentrations of the reactants?
Answer
The dissociation of ammonia at 300K has
a value of 2.63 x 10^{9}.
2NH_{3}(g) > N_{2}(g)
+ 3H_{2}(g)
If the initial concentration of ammonia was 0.100M,
calculate the equilibrium concentrations of each of the
substances.
Thought process
1. The first thing we need to ask is what am I supposed
to find?
We are asked to find the concentration of the
substances at equilibrium.
2. We next need to ask what information we have.
We have the initial concentration of the ammonia,
0.100M.
We have the equilibrium constant
for the dissociation of ammonia, K_{c} = 2.63 x
10^{9}.
We also have a balanced equation with the reactants
and the products. (If we did not have the equation we
would have to write the equation first.)
3. We now need to decide how we can use the information
that we have.
We can write the equation for the equilibrium
constant of the reaction.
This equation links the equilibrium constant with the
concentration of the reactants and products at equilibrium.
We have the equilibrium constant. We do not have any more
information to put into this equation; we have 3 unknowns
that we need to find.
Usually when we write an equation we try to find only one
unknown. So we need to find a way to write the equation with
one unknown to find.
We do this by preparing an Initial Change Equilibrium
(ICE) Table.
How do we know what to put in the table?
We need:
1. An initial concentration of
reactants and products.
2. An expression for the change in concentration of
reactants and products.
3. An expression for the concentration of the reactants
and products at equilibrium.
Initial Concentration
We know that the initial concentration of ammonia is
0.100M. When we put the ammonia in the container, we did not
put any hydrogen or nitrogen in the container. The initial
concentration of hydrogen and nitrogen would both be 0.
Change in Concentration
From the equation:
2NH_{3}(g) > N_{2}(g)
+ 3H_{2}(g)
Let x
be our single unknown concentration to be used in the
equilibrium constant equation. So use the mole ratios of the
reactants and products (2:1:3) from the equation of the
reaction.
The amount of ammonia that reacted
will be 2x. So the change in the concentration of
ammonia will be –2x, because 2x amount of
ammonia is no longer in the container.
Note: We usually try to keep the numbers in front of the
substances from the equation as a part of our answer.
When 2x ammonia reacts, we
have x amount of nitrogen and 3x amount of
hydrogen being produced.
Equilibrium Concentration
At equilibrium, x amount of
nitrogen and 3x amount of hydrogen is present.
Since 2x amount of ammonia
has reacted then 0.100 – 2x amount of ammonia is
present at equilibrium.
The concentrations of ammonia,
nitrogen and hydrogen at equilibrium can now be written with
one unknown x. We can
substitute the concentration of ammonia, nitrogen and
hydrogen at equilibrium from the ICE table.
We can now place these into the equation, along with the
equilibrium constant.
Find the square root of both sides of the equation:
We can use the quadratic formula to solve for x:
WE CAN NOW FIND THE
CONCENTRATION OF EACH OF THE SUBSTANCES AT EQUILIBRIUM.
The concentration of nitrogen at
equilibrium is x.
So the concentration of nitrogen at equilibrium is
0.000983M.
The concentration of hydrogen at
equilibrium is 3x.
So the concentration of hydrogen at equilibrium is 3 x
0.000983M = 0.00295M.
The concentration of ammonia at
equilibrium is 0.1 – 2x.
So the concentration of ammonia at equilibrium is 0.1 –
2(0.000983M) = 0.098M.
Problem 3
How do I determine
how Le Chatelier's Principle will affect a reaction when
different aspects of the reaction are changed?
Answer
The reaction
has an
enthalpy of +155.7kJ.
How will the amount of product present at equilibrium be
affected by the following?
Remember: figure out your equilibrium shift and then
state how the product will be affected.
a. Adding
(Increasing the concentration of the reactants)
According to Le Chatelier’s
Principle if you increase the concentration of the reactant
the reaction in equilibrium will try to remove the excess
reactants. So the reaction will produce more products. The
equilibrium will shift to the right to produce more
products.
This increases the production of NO.
b. Removing
(Decreasing the concentration of the reactants)
When the concentration of the reactants is decreased, the
reaction in equilibrium will try to increase the amount of
reactants. The product will break down to try to replace the
reactant that was removed. The equilibrium will shift to the
left to produce more reactants. This decreases the amount of
product formed.
c. Adding a Catalyst
Adding a catalyst increases the rate
of the reaction. This increases the rate of both the forward
and the backward reaction equally.
Remember that it will make the time taken to achieve
equilibrium faster, but it will not affect a system in
equilibrium.
d. Increasing the Temperature
When the reaction is written as
This means that the forward reaction is endothermic and the
backward reaction is exothermic. The forward reaction
absorbs heat from the surroundings and the exothermic
reaction adds heat to the surroundings.
When heat is increased, the system in equilibrium will try
to remove the excess heat. It does this by increasing the
endothermic reaction.
Since the forward reaction is endothermic, more products
will be produced.
e. Increasing the pressure.
When pressure is increased, the
system in equilibrium will try to remove the excess
pressure. There are 2 moles of molecules on the left hand
side of the reaction and 3 moles of molecules on the right
hand side of the reaction.
More molecules present will increase the pressure.
Less molecules present will decrease the pressure.
When pressure is increased on a system in equilibrium, the
system will try to decrease the pressure. It does this by
trying to decrease the number of molecules.
The system will try to remove the excess pressure by
decreasing the amount of products and shifting the
equilibrium to the left.
f. Decreasing the Volume
From Boyle’s Law, volume is
inversely proportional to pressure.
This means that when volume is increased, pressure is
decreased and when volume is decreased, pressure is
increased.
Therefore, the changes in volume will change the pressure.
When volume is decreased, pressure is increased and the
system will remove the excess pressure by decreasing the
number of molecules.
The reaction will decrease the pressure by decreasing the
amount of products. This is obtained by shifting the
equilibrium to the left.
Problem 4
How do I determine if a precipitate will
form when two aqueous solutions are mixed?
Answer
Will a precipitate
form when 100mL of 4.0 x 10^{3}M Barium chloride is
mixed with 300mL of 6.0 x 10^{4}M Sodium Sulfate?
The solubility product for Barium Sulfate is given by:
Calculate the ion
product. The equation for the ion
product is:
Q_{sp}
= [Ba^{2+}][SO_{4}^{2}]
Note that because BaSO_{4}
is a solid, you do not divide by the concentration of
BaSO_{4} in the ion product equation. To find
the ion product, Q_{sp} you
need to find the concentration of Barium and Sulfate
ions in solution.
Finding the concentration of Barium Ions
Find moles of Ba^{2+} in
solution, since Ba^{2+}: BaSO_{4} is
1:1, moles of BaSO_{4} = Ba^{2+}. Use
molarity equation to solve for moles of initial
solution. The concentration of Barium ions is the number
of moles of Barium ions in 1000mL of solution. 4.0 x 10^{3}M
Barium Chloride means that 1000 mL solution contains 4.0
x 10^{3} moles of Barium ions.
1 ml of the solution will
contain 4.0 x 10^{3} / 1000 moles of Barium
ions. 100mL of the solution will contain 100 x 4.0 x 10^{3}
/ 1000 moles of Barium ions, or 4 x 10^{4}
moles of Barium ions.
Determine
concentration of ions by dividing moles of Ba2+ by the
total volume of both solutions that have now been mixed.
Use molarity equation and divide moles by total volume
of 400mL or 0.4L
You added 100 mL and 300 mL
of solution together so you have a total volume of 400
mL.
One solution had 4 x 10^{4}
moles of Barium ions, while the other solution had no
Barium ions.
As a result, 4 x 10^{4}
moles of Barium ions is now in 400 mL of solution. Since
we need the concentration of Barium ions present in the
new solution, we need to find the number of moles of
Barium ions that would be present in 1000 mL.
400 mL solution has 4 x 10^{4}
moles of Barium ions. 1mL solution has 4 x 10^{4}/400
moles of Barium ions. 1000mL solution has 1000 x 4 x 10^{4}/400
moles of Barium ions.
The concentration of Barium ions is 0.001 M
Finding the concentration of Sulfate ions
Find moles of SO4^{2}
in solution, since SO4^{2} : Na_{2}SO_{4}
is 1:1, moles of Na_{2}SO_{4} = SO4^{2}.
Use molarity equation to solve for moles of initial
solution.
Stepbystep solution
The concentration of Sulfate
ions is the number of moles of Sulfate ions in 100
mL of solution. 1000 mL Sodium Sulfate has 6.0 x 10^{4}
moles Sulfate ions. 1mL solution of Sodium Sulfate
has 6 x 10^{4}/1000 moles of Sulfate ions.
300 mL of Sodium Sulfate solution has 300 x 6 x 10^{4}/1000
moles of Sulfate ions, or 1.8 x 10^{4}
moles of Sulfate ions.
Determine concentration of ions
by dividing moles of SO_{4}4^{2} by
the total volume of both solutions that have now
been mixed. Use molarity equation and divide moles
by total volume of 400 mL or 0.4 L.
Stepbystep solution
You added 100 mL and 300 mL of
solution together so you have a total volume of 400
mL.
One solution had 1.8 x 10^{4}
moles of Sulfate ions, while the other solution had
no Sulfate ions. As a result 1.8 x 10^{4}
moles of Sulfate ions is now in 400 mL of solution
You need the concentration of the Sulfate ions,
so you need to find out how many moles would be in
1000 mL of solution.
400 mL of the new solution has
1.8 x 10^{4} moles of Sulfate ions. 1 mL of
the new solution has 1.8 x 10^{4}/400 moles
of Sulfate ions. 1000mL of the new solution has 1000
x 1.8 x 10^{4}/400 moles of Sulfate ions.
The concentration of Sulfate
ions is 4.5 x 10^{4}
You now have the
concentration of the Barium ions and the Sulfate ions in
the solution.
Determine Q_{sp}
You can now place these values in the equation to
find the ion product:
Q_{sp} = [Ba^{2+}][SO_{4}^{2}]
Q_{sp} = (0.001)(4.5 x
10^{4})
Q_{sp} = 4.5 x 10^{7}
Compare the ion product Q^{sp}
to the solubility product K^{sp},
Q_{sp} = 4.5 x 10^{7}
Recall that you were given K_{sp}
= 1.1 x 10^{10}
Now Q_{sp} > K_{sp}
This means that a Barium Sulfate precipitate will be
formed.
Problem 5
How do I determine the pH of a weak acid
solution?
Answer
Determine the pH of a 0.500M solution
of acetic acid
Step 1
Write the chemical equation for dissolving acetic acid in
water.
Step 2
Set up a table comparing the concentration of the acetic
acid and the ions formed on the right hand side of the
equation.
Step 3
Write the equation based on the acid constant.
Step 4
Now x is the concentration of
both the
and the ions.
Replace the concentrations in the equation with the
concentrations at equilibrium and with the Ka value for
acetic acid.
Step 5
This acid is weakly dissociated in
water, so x is very small, and 0.5 – x is
approximately equal to 0.5. Therefore you can rewrite this
equation as
And solve this equation for x
The concentration of the
ion is
0.003M.
Step 6
The pH is the negative logarithm to
the base 10 of the
ion
concentration.
Since you know the concentration of
the ion
you can put this value into the equation.
Problem 6
How do I use VSEPR
theory to explain the shape of a molecule?
Answer
How does VSEPR
explain the following?
CH_{4} is a
tetrahedral molecule
Part 1:
CH_{4} is
a tetrahedral molecule
The first thing to do with
this sort of question is to identify the central atom. The
central atom in this case is the carbon atom because all the
hydrogen atoms are attached to the carbon atom.
This is shown in the Lewis
dotcross diagram below:
The next stepis to determine
the number of electrons in the valence shell of the carbon
atom. There are 4 valence electrons in the carbon atom and
each of them is bonded to a hydrogen atom. There is a total
of 8 electrons surrounding the central atom of the
molecule:†† 4 pairs of electrons.
There are no lone pairs of
electrons on the carbon atom.
This also indicates that in the CH_{4}
molecule, only the bonding pairs of electrons will determine
the shape of the molecule.
Below is a Lewis Structure for
methane that indicates this.
When drawn in three dimensions,
this is the shape of methane:
The dashed lines represent a
hydrogen that protrudes backwards. The dark triangle
represents a hydrogen that protrudes forwards. (It is
sticking outwards towards you.) The plain lines
represent two hydrogens which are in the plane of the
computer screen.
The furthest these bonds can be
apart will be 109.5°.
This molecule is said to have a
tetrahedral shape because if you imagine drawing lines
joining only the atoms at the edge of the molecule (the
hydrogen atoms)  the resulting shape is a tetrahedron.
Note carefully
When describing the shapes of
molecules, lines joining only the atoms at the edges of the
molecules are considered. These are not the lines
representing the bonds. The lines representing the bonds are
only drawn from the central atom to the other atoms. The
lines representing the bonds are only used when the central
atom is one of the atoms at the edge of the molecule.
In the drawing below, the red lines
represent the lines showing the shape of the molecule, and
the lines in the black represent the bonds.
Problem 7
How can I predict
the relative boiling points in different molecules?
Answer
On the basis of
the relative strengths of forces, predict the order of
decreasing boiling points of the following three substances:
Methane, Methanol, and Hydrogen gas?
1. What type of intermolecular forces
of attraction exists between the molecules of the compounds?
The forces of attraction between
these molecules are as follows:
Methane has dispersion forces
because it is a nonpolar molecule.
Hydrogen gas has dispersion forces because it is a
covalently bonded molecule.
Methanol has hydrogen bonding because it is a polar
molecule, with an oxygenhydrogen bond.
2. How do these forces affect
boiling point?
The methanol has hydrogen
bonding between the molecules, so it has the highest
boiling point.
Both methane and hydrogen gas
has dispersion forces between the molecules.
3. How do I predict the difference
between the boiling points of substances with the same
forces of attraction between the molecules?
Methane (Molar Mass 16) is a
larger molecule than hydrogen gas (Molar Mass 2), so it
will have a larger boiling point.
Order of decreasing boiling points
Methanol – highest boiling
point
Methane
Hydrogen gas – lowest boiling point.
Problem 8
How can I
determine how to identify and name the different structural
formulas and functional groups in a molecule?
Answer
Example 1
Write the name for the following
structure.
Step 1: Identify
the longest carbon chain and identify the parent name
There are 5 C in the longest
chain, so the parent name is pentane (see table 1 below)
Step 2: Identify
the functional group
You have –COOH as a functional
group, so drop “e” from pentane and attach –oic acid
(see table 3). You get pentanoic acid.
Step 3: Assign the
position number
Step 4: Write the
position number of the functional group if necessary.
Since the functional group is
in the first C, you don’t have to write the position
number.
Step 5: Identify
the hydrocarbon branch.
There are three hydrocarbon
branches, two
(so need to write di) and one
.
The one of the
is
attached to the third carbon and other
to
the fourth carbon. The
is
attached to the second carbon. Use table 1.
Therefore, the name of the
structure is
Example 2
Write the structure for the
following name.
3,4diethyl5methyl2heptanol
Step 1: Write the
number of carbons according to the parent name
Parent name is heptane, so
there are 7 C in the longest carbon chain.
C – C – C – C –
C – C – C
Step 2: Assign a
number to each carbon of the parent (left to right)
Step 3: Write the
functional group as indicated by the ending of the parent
name at the appropriate position.
The ending is –ol and there is
2 in front of the parent name, so you need to write –OH
at the 2nd carbon.
Step 4: Write any
hydrocarbon branches at the appropriate position number, if
any
You have ethyl at 3rd and 4th
carbon, and methyl at 5th carbon.
Step 5/step 6:
Complete the structure by adding
the appropriate numbers of hydrogen to each carbon. Carbon
must have four bonds. Delete the position number.
Problem 9
How do I determine
the hybrid orbitals that are formed when two or more
orbitals come together?
Answer
What type of
hybrid orbital is produced when:
One 2s orbital and one 2p
orbital are mixed?
One 2s orbital and two 2p
orbitals are mixed?
One 2s orbital and three 2p
orbitals are mixed?
One 2s orbital, three 3p
orbitals and one 3d orbital?
One 3s orbital, three 3p
orbitals and two 3d orbitals?
Part 1: One 2s orbital and
one 2p orbital are mixed.
When one 2s and one 2p orbital are
mixed, the sp hybrid orbitals are produced. The shapes of
the orbitals are demonstrated below.
There are two atomic orbitals
combined, so two hybrid orbitals are produced.
The diagram below represents a
hybrid orbital.
The following diagram represents
how the two sp hybrid orbitals are arranged on an atom. The
orbitals are represented in different colours just to
highlight the different hybrid orbitals.
Part 2: One 2s orbital and
two 2p orbitals are mixed.
When one 2s orbital and two 2p
orbitals are mixed, the hybrid orbitals are called sp^{2}
hybrid orbitals. The superscript 2 in sp^{2} shows
the number of p orbitals that were used in formation of the
hybrid orbitals. When there is only one orbital there is no
subscript.
For example, there is only one s
orbital in the hybrid orbital, so there is no 1 written as a
subscript. There are three original atomic orbitals (one s
and two p orbitals) and three hybrid orbitals.
One way to check the number of
hybrid orbitals formed is to write on your rough work paper
the s^{1}p^{2}.
Remember that although no 1 is
written as a subscript in the sp^{2} hybrid orbitals,
there is one s orbital. Then and add the superscripts 1 + 2
= 3.
There are three sp^{2}
hybrid orbitals.
The following diagram represents
how the three sp^{2} hybrid orbitals are arranged on
an atom. Each sp^{2} hybrid orbital is represented
in a different colour.
Three sp^{2} hybrid orbitals
Part 3: One 2s orbital and
three 2p orbitals are mixed.
When one 2s and three 2p orbitals
are mixed, the hybrid orbital is called sp^{3}
hybrid orbitals. There are four hybrid orbitals formed from
a total of four atomic orbitals (one s and three p).
In order to check the number of
hybrid orbitals formed, write on your rough work paper the s^{1}p^{3}.
Then add the superscripts 1 + 3 =
4.
There are four sp^{3}
hybrid orbitals.
The following diagram represents
how the four sp^{3} hybrid orbitals are arranged on
an atom. Each sp^{3} hybrid orbital is represented
in a different colour.
Four sp^{3} hybrid orbitals
Part 4: One 3s orbital,
three 3p orbitals and one 3d orbital.
When one 3s, three 3p and one 3d
orbital are mixed, the type of hybrid orbitals formed are
called dsp^{3} orbitals. There are five original
atomic orbitals (one s, three p and one d), so there are
five dsp^{3} hybrid orbitals produced.
One 3s + three 3p + one 3d "→" five
dsp^{3} hybrid orbitals.
The following is a representation
of the five dsp^{3} hybridized orbitals on one atom.
Five dsp^{3} hybrid orbitals
The way to check the number of
hybrid orbitals formed is to write on your rough work paper
the d^{1}s^{1}p^{3}. Then add the
superscripts 1 + 1 + 3 = 5.
There are five dsp^{3}
hybrid orbitals.
Part 5: One 3s orbital,
three 3p orbitals and two 3d orbitals.
When one 3s, three 3p and two 3d
orbital are mixed, the type of orbitals formed are called d^{2}sp^{3}
orbitals. There are six original atomic orbitals (one s,
three p and two d), so there are six d^{2}sp^{3}
hybrid orbitals produced.
One 3s + three 3p + two 3d "→" six
d^{2}sp^{3} hybrid orbitals.
The following is a representation
of the six d^{2}sp^{3} hybridized orbitals
on an atom.
Six d^{2}sp^{3} hybrid orbitals
You can check the number of hybrid
orbitals formed by writing on your rough work paper the d^{2}s^{1}p^{3}.
Then add the superscripts 2 + 1 + 3 = 6.
There are six d^{2}sp^{3}
hybrid orbitals.
Problem 10
How do I determine
the oxidation number of an unknown atom in a compound?
Answer
Example 1
You must use the rules of assigning
oxidation numbers to determine the oxidation number of Mn.
You already know all of the following things:
The oxidation state of O in
the compound is 2. There are 4 oxygen atoms so the
oxidation state of the oxygen has to be multiplied by 4
to get the total oxidation state of the 4 oxygen atoms.
The oxidation state of K in
the compound is +1 because it is a monovalent ion in
group 1.
When you add all the oxidation
numbers in a compound , the total will be 0.
From this information you can write an
equation to solve for the oxidation state of Mn. This
equation will represent the oxidation states of each atom in
the compound.
Let x be the oxidation number of
the Mn
Therefore:
1 + x + 4 (2) = 0 (Sum of
oxidation states in a compound is 0)
1 + x – 8 = 0
x – 7 = 0
x = +7
The oxidation state of Mn in KMnO_{4}
is +7.
Example 2
Find the oxidation state of S in the
ion SO_{3}^{2}.
You already know
The oxidation state of O in
the ion is 2. There are 3 oxygen atoms so you must
multiply the oxidation state by 3 to get the total
oxidation state of the 3 oxygen atoms.
The sum of the oxidation
states of the ions is equivalent to the oxidation state
of the ion. Therefore the oxidation state of the SO32
is 2.
From this information you can write an
equation to be solved.
Let x be the oxidation state of the
S.
x + 3 (2) = 2 (Sum of oxidation
states equivalent to charge of ion)
x – 6 = 2
x = 2 + 6
x = +4
The oxidation state of S in SO_{3}^{2}
is +4
Problem 11
When given a redox
chemical reaction, how do you know which substance is
reduced and which is oxidized?
Answer
Identify the
substance that is oxidized and the substance that is reduced
as well as the oxidizing and reducing agent.
O = 2, but there are
2 of them
Therefore,
(2) x 2 = 4
O = 2, but there are
4 of them
Therefore,
(2) x 2 = 8
Cl = 1
To find Mn; use the
following equation
Mn + (4) = 0 (because the net charge of the
substance is zero)
Therefore,
Mn = 4
H = +1, but there are
2 of them
Therefore,
(+1) x 2 = +2
To solve for S; use
the following equation
(+2) + S + (8) = 0
S  6 = 0
S = +6
Products:
K_{2}SO_{4}
MnSO_{4}
Cl_{2}
2H_{2}O
O = 2, but there are
4 of them
Therefore,
(2) x 2 = 8
O = 2, but there are
4 of them
Therefore,
(2) x 2 = 8
Since chlorine is in
its elemental state, its oxidation number is
zero
Cl = 0
O = 2
K = +1, but there are
2 of them
Therefore,
(+1) x 2 = +2
Since S is part of the
sulphate ion, its oxidation number stays the
same
S = +6
H = +1, and there are
two of them
To solve for S; use
the following equation
(+2) + S + (8) = 0
S  6 = 0
S = +6
To solve for Mn; use
the following equation
Mn + (+6) + (8) =0
Mn + (2) = 0
Mn = +2
** NOTE: The oxidation number
is the value before the number of atoms is multiplied to
it. Therefore, oxygen’s oxidation number is 2. We
multiply the oxidation number by the number of atoms to
determine other atom’s oxidation numbers.
Write the 2 halfreactions
Take the 1^{st}
atom that has an oxidation number that is changing
and write only it and the compound it is part of, in
an equation.
Chlorine has changed from
1 to zero
2KCl → Cl_{2}
(1) → (0)
Take the 2^{nd}
atom that has an oxidation number that is changing
and write only it and the compound it is part of, in
an equation.
Manganese has changed from
+4 to +2
MnO_{2} → MnSO_{4}
(+4) → (+2)
Add in the appropriate number
of electrons on either the reactant side if they have
gained electrons, or the product side if they lost
electrons.
2KCl → Cl_{2} + 2e^{}
MnO_{2} + 2e^{}
→ MnSO_{4}
Determine which substance is
oxidized or reduced:
Whichever substance gained electrons will be the
reduced substance, or the oxidizing agent. Whichever
substance lost electrons will be the oxidized substance,
or the reducing agent. 2KCl → Cl2 + 2e
 lost electrons; therefore, oxidized and is the
reducing agent MnO2 + 2e → MnSO4
 gained electrons; therefore, reduced and is the
oxidizing agent
Problem 12
How do I balance
Redox Equations using the ionelectron method? (General
terms using oxidation numbers in either an acidic or basic
medium.)
Answer
Thought process
Because this
equation is in a basic medium, we follow example c.
Ionelectron method in a basic medium.
You must follow
the 10 steps in order to ensure that you get the correct
answer to the question.
Now, go through
each step in the process.
i. Divide the
equation into halfreactions. Determine the
oxidization state/number for each atom in each
halfreaction
v. Balance the net charge by adding electrons (e). It is a
good idea to do a calculation to the side of your equation
to add up the charge on the left side (LS) and the right
side (RS).
Problem 13
How do I determine the balanced
chemical equation, complete ionic equation and net ionic
equation for a specific reaction?
Answer
Write a balanced equation, a
complete ionic equation and a net ionic equation for
each of the following reactions.
 A barium
nitrate solution reacts with a sodium sulphate solution.
Step 1
Write a balanced chemical equation indicating the state of
each species. You will need to use your solubility chart
for this.
Step 2
Rewrite the equation, separating the ions that are in
solution. At this stage you have written the complete
ionic equation.
Since all the
reactants are aqueous solutions, this means that the ions
are able to move freely. Therefore, you can separate the
compounds into their ions.
Note that (NO_{3})_{2}
from the Ba(NO_{3})_{2} is written as 2NO_{3}^{}
because the ions are now separated, and for each barium
nitrate in solution, two nitate ions and one barium ion are
produced.
The same applies to
all of the 2Na^{+} ions.
Note that
solids and pure liquids do not separate
into ions.
Step 3
Identify and cross out spectator ions. Spectator ions are ions that are exactly the same on
both sides of the reaction. They are ions that have the same
oxidation state and physical state on both sides of the
equation.
The ions that are
the same on both sides of the equation are 2Na+(aq) and
2NO3(aq) Therefore, these are the spectator ions.
ΔΗ^{o}_{reaction}
= {(1 mol Na_{2}CO_{3}(s) x ΔΗ^{o}_{f}
Na_{2}CO_{3}(s) ) + (1 mol H_{2}O(l)
x ΔΗ^{o}_{f} H_{2}O(l)) +
(1mol CO_{2}(g) x ΔΗ^{o}_{f}
CO_{2}(g))}  {(2 mol NaHCO_{3}
(s) x ΔΗ^{o}_{f} NaHCO_{3}(g))}
Step 3:
Substitute in the values, found
in the Chart of Thermodynamic Values (usually found in
the appendix of your text book) and the appropriate
coefficient.
ΔΗ^{o}_{reaction}
= {(1 mol Na_{2}CO_{3}(s) x 1131kJ/mol
) + (1 mol H_{2}O(l) x 285.9kJ/mol) +
(1 mol CO_{2}(g) x
393.5kJ/mol)}  {(2 mol NaHCO_{3} (s) x
947.7kJ/mol)}
Step 4:
Solve the equation
ΔΗ^{o}_{reaction}
= (1810 kJ) –(1895kJ)
ΔΗ^{o}_{reaction}
= +85kJ
Since the answer is positive,
the reaction is endothermic.
Remember in doing these problems
Your final units should
be in kilojoules (kJ).
If your answer is
positive, the reaction is endothermic.
If your answer is
negative, the reaction is exothermic.
Different compounds,
and different states will be in your reaction.
Take note of the
coefficients in your balanced chemical equation.
These will change depending on your reaction and
affect the number of moles in the enthalpy
multiplication.
Problem 15
How do I determine
the standard entropy of a reaction?
Answer
Find the ΔS^{o}
for the decomposition of baking soda: NaHCO_{3}
Thought process
This question is asking you to find
the change in entropy for a decomposition reaction. You
should use the standard entropy change equation to solve
this problem.
The following steps will guide
you through to the answer
Step 1:
Write the balanced chemical
equation for the reaction
ΔS^{o}_{reaction}
= {(1 mol Na_{2}CO_{3}(s) x ΔS^{o}
Na_{2}CO_{3}(s)) + (1 mol H_{2}O(l)
x ΔS^{o} H_{2}O(l)) + (1 mol CO_{2}(g)
x ΔS^{o} CO_{2}(g))}  {(2 mol
NaHCO_{3} (s) x ΔS^{o} NaHCO_{3}(s))}
Step 3:
Substitute in the values, found
in the Chart of Thermodynamic Values. (Look in the
appendix of your text book.)
ΔS^{o}_{reaction}
= {(1 mol Na_{2}CO_{3}(s) x +136 J/molK
) + (1 mol H_{2}O(l) x +69.96 J/molK) +
(1 mol CO_{2}(g) x
+213.6J/molK)} {(2 mol NaHCO_{3} (s) x
+102 J/molK)}
Notice that the moles cancel
out, leaving your units as J/K
Step 4:
Solve the equation
ΔS^{o}_{reaction}
= +419.56J/K  (+204J/K)
ΔS^{o}_{reaction}
= +215.56 J/K
Since the answer is positive,
the reaction favours spontaneity.
Remember that your final units
should be in Joules/Kelvin.
Also, if your answer is
positive, the reaction favours spontaneity.
Problem 16
How do I determine the heat capacity
of a substance when I am given information about water (no
phase change involved)?
Answer
A sample of copper was heated to
and
then thrust into 200 g of water at
The
temperature of the mixture became
a) How much heat in joules was
absorbed by the water?
b) How much heat in joules was lost by the copper?
c) What was the mass in grams of the copper sample?
Draw a picture:
Since the temperature of the copper
is greater than the water, copper will release energy while
water absorbs.
a. You were asked to find the heat
energy, so you would use
b. You know that q (absorbed) = 
q(released), so
q(copper) =  1379.4 J
c. You found q(copper) =  1379.4 J
from part (b), and you also know that
, so
Problem 17
How do I sketch the heating/cooling
curve of a reaction? For example, what mass of water at 50°C
was cooled through freezing point to ice at 20°C if 84.7 KJ
of heat was transferred? Draw a cooling curve.
Answer
What mass of water
at 50°C was cooled through freezing point to ice at 20°C if
84.7 KJ of heat was transferred? Draw a cooling curve.
Thought process
In this example, the question
is giving you the starting and finishing temperature of
the pure substance, water. It has also told you that you
are starting with a liquid and are finishing with a
solid. Therefore, you know the main parts of the graph.
The question is also asking
you to solve for the mass. Therefore, you need to use
the Heat Capacity Equation to determine this value.
Remember that the equation is:
The second part of the
question asks you to sketch the cooling curve for the
reaction.
Cooling curve:
The water is being cooled,
thus the heat energy is being taken away and the
temperature begins to drop. Therefore, the kinetic
energy begins to decrease and the molecules begin to
lose speed, and the amount of space between
molecules decreases. As the space between molecules
decreases, the intermolecular force of attraction
begins to increase. This decrease in temperature,
speed and distance continues until the force of
attraction between the molecules is strong enough to
change the intermolecular bonds and create a solid.
At this point, the molecules are "solidifying" and
changing state from a liquid to a solid.
Now the temperature of the gas will no longer
decrease, because the heat energy is being used to
form the intermolecular bonds between the liquid
molecules. This plateau is the change of state from
gas to liquid known as condensation.
Cooling Curve
Once all of the liquid
molecules have changed into solid molecules, the
heat energy will continue to be removed from the
solid molecules causing them to lose kinetic energy
and to decrease their temperature. This loss in
kinetic energy will cause the molecules to move even
more slowly and to come together even more and their
temperature will continue to decrease.
Cooling Curve
Once all of the liquid
molecules have changed into solid molecules, the
heat energy will be removed from the solid molecules
causing them to lose kinetic energy and to decrease
their temperature.
The above sketch is the
graph of the cooling process.
Problem 18
How do I sketch and label a potential energy diagram for the
reaction:
2NOBr(g) → 2NO(g) + Br2 + 50kJ, in which Ea(f) = 30kJ
Determine the value of Ea(r)
Answer
Sketch and label a
potential energy diagram for the reaction:
2NOBr(g) → 2NO(g) +
Br_{2} + 50kJ, in which E_{a(f)} = 30kJ
Determine the value of E_{a(r)}
Thought process
When solving this problem, you
need to know the equation for enthalpy
of a reaction:
E_{a(f)} = E_{a(r)}
+ H
You also need to know how to
determine if the reaction is endothermic
or exothermic from the given
thermochemical equation.

Because the 50kJ is written
as a product (on the right side of the equation), we
know that energy is being given off; therefore, the
reaction is exothermic
 Exothermic
reactions have negative enthalpies
 Now, you can substitute in
your values and rearrange the equation to solve for the
unknown, E_{a(r)}
Ea_{a(f)} = E_{a(r)}
+ H
Ea_{a(r)} = Ea_{a(f)}  H
Ea_{a(r)} = 30kJ – (50kJ)
Ea_{a(r)} = 80kJ
Now we know the value of the
forward and reverse activation energies and the value of the
enthalpy; therefore, we can sketch our graph.