# AS 2.5 - Physical Chemistry - BoP-Waikato-Gisborne-Science

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```					 AS 2.5 - Physical
Chemistry

Formative Assessments in
Energy, Rates of Reaction,
Equilibrium and Acid/Base
Reactions
Overall Outline
Success Criteria
Can classify    Can classify    Can classify
a reaction as   several         reactions as
exothermic or   reactions       exothermic or
endothermic     correctly as    endothermic
exothermic or   from either
endothermic     descriptions,
equations or
energy
profiles
Energy
State whether heat is taken in or given out in
the following state changes.

Change of State        Heat in or out?
Evaporating                  in
Melting                      in
Freezing                    out
Condensing                  out
Sublimating                   in
Energy
State whether the following are exothermic or
endothermic reactions.

Example                  Exothermic or
Achieved
Merit          Endothermic
Clothes drying outside    Endothermic
Dehydrating CuSO4         Endothermic
Combustion                Exothermic
Photosynthesis            Endothermic
State whether the following are
exothermic or endothermic reactions
Process                   Exothermic or
Merit
Achieved
Excellence               Endothermic
Burning Wood                              Exothermic
Some ammonium chloride was dissolved
in water. The temperature of the water   Endothermic
changed from 19 oC to 10 oC
CH4 (g) + H2O (g)     CO (g) + 3H2 (g)
Endothermic
DrH = + 206 kJ mol -1

Exothermic
Success Criteria
Can carry out Can carry out   Can carry out
a single step a multi-step    a multi-step
calculation   calculation     calculation
and give the    and give the
correct         correct
appropriate
sig.figures
and unit
Octane is a key component in petrol, and burns
according to the following equation:

C8H18(l) + 121/2 O2(g) ® 8 CO2 (g) + 9 H2O(l)
ΔrH = -5500 kJ mol-1

1.00 litre of Octane contains 6.12 moles of the fuel.
Calculate the energy released when 1.00 litre of fuel is
burnt.

Energy Released             = n x ΔrH
= 6.12 x 5500
Achieved
= 33 660 kJ
Using hydrogen gas (H2) as a fuel for cars, rather than
octane, is often viewed as better for the environment.
Calculate the mass of H2 required to produce the same
amount of energy as 1.00 litre of octane ( 33 660 kJ ).

H2(g) +    1/
2 O2(g) ® H2O(g)   ΔrH = -286 kJ mol-1
n (H2) = 33 660 / 286
= 118 mol          Achieved
m (H2) = 117.7 x 2
= 235.4          Merit
= 235 g         Excellence
29.6 g of sodium hydroxide was dissolved in water and
excess hydrochloric acid was added. Using the temperature
increase and the heat capacity of water, it was calculated that
43.5 kJ of heat was released.

Determine the enthalpy change,   DrH, for the following
reaction:

NaOH(aq)     +    HCl(aq)      NaCl(aq)    +   H2O(l)
mol NaOH      =    29.6 g
40.0 g mol-1      Achieved
= 0.740 mol
0.740 mol reacts to produce 43.5 kJ
1 mol reacts to produce 43.5 = 58.82 kJ  Merit
0.740

DrH = -58.8 kJ mol-1               Excellence
29.6 g of sodium hydroxide was dissolved in water and
excess hydrochloric acid was added. Using the
temperature increase and the heat capacity of water, it
was calculated that 43.5 kJ of heat was released.

NaOH(aq)     +    HCl(aq)       NaCl(aq)   +   H2O(l)
ΔrH = -58.8 kJ mol-1

What mass of sodium hydroxide is required to produce
150 kJ of energy?

n (NaOH) =        150 kJ
-1
Achieved
58.8 kJ mol
=     2.55 mol

m (NaOH)     = 2.55 mol x 40.0 g mol-1        Merit
= 102.3
Excellence
m = 102 g
Carbohydrates are an important source of energy in our diet. Two
common carbohydrates are glucose (C6H12O6) and sucrose
(C12H22O11).
The equation below shows the combustion of glucose to form carbon
dioxide and water.
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(ℓ)
∆rH = –2820 kJ mol–1
Calculate the enthalpy change, ∆rH, when 100 g of glucose reacts to
form carbon dioxide and water. M(C6H12O6) = 180 g mol–1

n (glucose) = 150 / 180 = 0.556 mol
Achieved
∆rH = 0.556 x -2820
= -1567
Merit

∆rH = -1567 kJ mol-1              Excellence
The equation below shows the combustion of sucrose to form
carbon dioxide and water.
C12H22O11(s)    +   12O2(g)   →        12CO2(g)    +   11H2O(ℓ)

When 150 g of sucrose, C12H22O11, undergoes combustion,
2478 kJ of energy is released.
Calculate the enthalpy change when 1 mole of sucrose
undergoes complete combustion.
M(C12H22O11 ) = 342 g mol–1

n (sucrose) = 150 / 342 = 0.439 mol

∆r H =       1     x -2478                 Achieved
0.439
= - 5650
Merit

-1        Excellence
∆rH = -5650 kJ mol
Success Criteria
Can describe    Can explain    Can fully
what            how the        explain the
happens to      change in      change in
the particles   condition      rate due to
due to          affects the    the frequency
change in       frequency of   of collisions
observation
seen
The reaction between 20.0 mL of 0.500 mol L–1 hydrochloric acid
and 20.0 mL of 0.250 mol L–1 sodium thiosulfate solution at room
temperature (25°C) produces a precipitate of sulfur that makes
the solution go cloudy after about 5 minutes.
With reference to the collisions of particles, explain how and why
the reaction is affected, if the reaction is carried out in a water
bath at a temperature of 50°C.

An increase in temperature means the particles have
more kinetic energy. There will be an increase in the
frequency of collisions. And when the particles
collide there is more chance that they will reach the
activation energy required for the reaction to take
place. Therefore the frequency of successful
collisions will increase and hence the reaction rate
will increase, so lowering the time taken for the
reaction to take place.

Achieved                 Merit         Excellence
Hydrogen peroxide decomposes at room temperature (25 oC)
according to the following equation.
2H2O2(aq)      2H2O(l)   +    O2(g)
On addition of a very small amount of solid manganese dioxide, the
rate at which the bubbles of gas are produced is increased so that
rapid fizzing is observed. Further observation indicates that
manganese dioxide remains after reaction has stopped.
With reference to the collisions of particles, explain why the
reaction rate has increased.

MnO2 is a catalyst for the reaction. The catalyst
provides an alternative pathway of lower activation
energy for the reaction. So molecules that previously
have enough energy to react, can now reach the lowered
activation energy. Therefore the successful collision
rate is increased and the reaction rate is increased. So
more rapid fizzing is observed.

Achieved                Merit              Excellence
Hydrogen peroxide decomposes at room temperature (25 oC)
according to the following equation.
2H2O2(aq)     2H2O(l)    +   O2(g)
Hydrogen peroxide is stored at a low temperature. Discuss this
statement in terms of reaction rate.

The low temperature means that the molecules have
less energy. So there is a decrease in the frequency of
collisions. Less energy means when the molecules
collide, they have less chance of reaching the
activation energy for the reaction. Therefore there
are fewer successful collisions in the same time, so
the reaction rate decreases and the rate of
decomposition is decreased.

Achieved                 Merit          Excellence
Success Criteria
Can place      Can correctly   Can discuss
products on    write Kc        the
top and        expressions     relationship
reactants on   using square    shown by Kc
the bottom     brackets and
subscripts
The following equilibrium system is established when
thiocyanate ions (SCN-) are added to iron (III) ions
(Fe3+). The resulting aqueous solution is a dark red
colour. The equation representing the equilibrium
system and the colours of each species involved are
given below.

Fe3+ (aq) + SCN– (aq)            FeSCN2+ (aq)
pale orange colourless           dark red

Write the equilibrium constant expression for the above
reaction.

Achieved
Kc =         FeSCN 2+
[FeSCN2+ ]
[Fe3+] .. [SCN- ]
Fe3+ SCN-                       Merit
Ammonia is produced industrially according to the
Haber Process as shown below:
N2 (g) + 3H2 (g)          2NH3 (g)

Complete the equilibrium constant expression for the
above reaction.

2             Achieved
=
KcKc =          [NH3]
NH3
[NN2 .. [H2]3
2]    H2
Merit
Ag+(aq) + 2NH3(aq)                Ag(NH3)2+(aq)

At 25°C the value of Kc is 1.70 ´ 107. Circle the species that would be
present in the higher concentration in the equilibrium mixture at this
temperature.
Ag+(aq)         or         Ag(NH3)2+(aq)
Achieved
Kc =
Kc =          [Ag(NH3)2+(aq)]
Ag(NH (aq)
[Ag++(aq) . NH33(aq)]2
Ag (aq)] [NH (aq)                    Merit
Excellence
Ag(NH3)2+(aq) Kc is very large, so the
concentration of product is high compared to that of
reactants (as the product concentration is on top of
the ratio).
2NO2(g)                 2NO(g) + O2(g)

At 200°C the value of Kc is 1.10 ´ 10–5. Circle the species that
would be present in the higher concentration in the equilibrium
mixture at this temperature.
NO2(g)         or        NO(g)
Achieved
2                        Merit
Kc = NO
Kc = [NO            (g)    . [O2
]. O2 (g) (g)]
Excellence
NO2(g) ]2
[NO2(g)
NO2(g). Kc is very small, so concentration of
products is low compared to that of reactants. So
the product concentration is the top of the ratio.
Success Criteria
State         Explain how     Relates
observation   the             equilibrium
seen          equilibrium     theory to
shifts due to   current
the change in   situation to
condition       explain
observation
seen
The following equilibrium system is established when thiocyanate
ions (SCN-) are added to iron (III) ions (Fe3+). The resulting aqueous
solution is a dark red colour.

Fe3+ (aq) +    SCN– (aq)           FeSCN2+ (aq)
pale orange     colourless         dark red

When iron (III) ions (Fe3+) are removed from the equilibrium
mixture (by adding sodium fluoride), a colour change is
observed. Describe the colour change you would expect to see
and explain why it occurs.

The red colour lightens and becomes more
orange. Removal of the Fe3+ causes the
equilibrium position to shift towards the
reactants in order to minimise the change.
This replaces some of the Fe3+ that has been
removed. Therefore there is less FeSCN2+
present which results in a lighter colour.
Ammonia is produced industrially according to the Haber
Process as shown below:

N2 (g)   +   3H2 (g)        2NH3 (g)

The pressure of the system at equilibrium is increased (by
decreasing the total volume of the system).

Describe the effect of this change on the amount of NH 3 in

The amount of NH3 will increase. Increasing
the pressure of the system causes a shift to
the right. This is to decrease the amount
of pressure by forming fewer gas moles of
gas.
The following reaction is exothermic:

2N2O5(g)         4NO2(g)    +   O2(g)

Both N2O5 and O2 are colourless gases and NO2 is a brown
gas. A mixture of these gases exists at equilibrium and is
observed as a brown colour.
The mixture of gases is heated (at constant pressure).
Describe the expected observation and explain why this
occurs.

The brown colour will lighten. When the
mixture is heated the endothermic reaction is
favoured. In this case, this is the reverse
reaction. So the amount of NO2 gas is
decreased.
The following reaction is exothermic:

2N2O5(g)        4NO2(g)     +    O2(g)

Both N2O5 and O2 are colourless gases and NO2 is a brown
gas. A mixture of these gases exists at equilibrium and is
observed as a brown colour.
The pressure is increased, by decreasing the volume of the
container. Describe the expected observation and explain
why this occurs.

The brown colour will become lighter. As the
pressure is increased the formation of fewer
moles of gas is favoured. This favours the
reverse reaction as the ratio is 5:2 moles of
gas. Therefore the amount of brown gas
will decrease.
An equilibrium system involving different species of cobalt(II) is shown
in the equation below.

[CoCl4]2–(aq) +    6H2O(ℓ)           [Co(H2O)6]2+(aq)   + 4Cl–(aq)

[CoCl4]2–(aq) is blue and [Co(H2O)6]2+(aq) is pink.

At room temperature (25°C) the equilibrium mixture is pink.
Describe the expected observation when solid sodium chloride (NaCl)

The colour of the solution will turn blue. Adding
NaCl will increase the concentration of the Cl1-
ions. The equilibrium shift to decrease the
concentration of the chloride ion. In this case,
it will move in favour of the reactants so more
blue [CoCl4]2– is formed.
An equilibrium system involving different species of cobalt(II) is shown
in the equation below.
[CoCl4]2–(aq) +    6H2O(ℓ)           [Co(H2O)6]2+(aq)   + 4Cl–(aq)
[CoCl4]2–(aq) is blue and [Co(H2O)6]2+(aq) is pink.
At room temperature (25°C) the equilibrium mixture is pink.
The enthalpy change (∆rH) for this reaction as written above, has a
negative value. State the ion that would be present in the higher
concentration when the equilibrium mixture is heated.

[CoCl4]2– would be in present in the higher
concentration. As the temperature increases, the
equilibrium will shift to reduce the temperature
increase by moving in the endothermic direction.
As the reaction is exothermic, the equilibrium will
move in the reverse direction creating more [CoCl4]2– .
Success Criteria
Can identify   Can identify   Can identify
a conjugate    several        species that
acid-base      conjugate      can act as
pair           acid-base      acids or
pairs          bases and
their
conjugate
pair
Chickens make egg shell, CaCO3, using carbon dioxide gas from the air.
The carbon dioxide forms carbonic acid (H2CO3), which then reacts to form
the carbonate ions (CO32–) needed to make egg shell.
Two equations showing part of this process are given below.

Equation 1:      H2CO3(aq) + H2O(l)             HCO3-(aq) + H3O+(aq)

Equation 2:      HCO3-(aq) + H2O(l)            CO32-(aq) + H3O+(aq)

Identify three conjugate acid-base pairs in the equations above.

H2CO3 / HCO3-
HCO3- / CO32-
H 3O + / H 2O
Chickens make egg shell, CaCO3, using carbon dioxide gas from the air.
The carbon dioxide forms carbonic acid (H2CO3), which then reacts to form
the carbonate ions (CO32–) needed to make egg shell.
Two equations showing part of this process are given below.

Equation 1:        H2CO3(aq) + H2O(l)           HCO3-(aq) + H3O+(aq)

Equation 2:        HCO3-(aq) + H2O(l)           CO32-(aq) + H3O+(aq)

HCO3- can act as both an acid and a base.

Specify which equation above (1 or 2) shows HCO3- acting as an acid.

Equation two

1-                                           1+
HCO3             is donating a proton / H
Complete the table below to show the conjugate
acid-base pairs.

Conjugate        Conjugate
acid               base

NH4+
NH
3
H2PO4–        HPO4    2-

Cl–
HCl

H2SO            HSO4–
4
Which ion below can act as both an acid
and a base.
CH3COO–               HCO3–

HCO3–

It can donate an H + or it can accept
an H +
Success Criteria
Describe         Explain how     Fully explain
properties of    properties      how
weak or          seen are        properties
strong acids     linked to the   seen are
(conductivity,   strength of     linked to the
pH and           either weak     strength of
reactions with   or strong       weak and
metals and       acids           strong acids
carbonates)
Two acids of the same concentration, hydrochloric acid (HCl) and
propanoic acid (CH3CH2COOH), have properties as shown below:
Property                Hydrochloric acid   Propanoic acid
(0.100 mol L-1 )   (0.100 mol L-1 )
Relative conductivity         High               Low
of solution
pH of solution                1.00               2.93
Explain the differences in the conductivity and pH of the two acids.
In your explanation include reference to the species present in each
solution.
HCl has a low pH and a high conductivity as it is a strong
acid. This means it completely dissociates into its ions, so
there is a high [H3O+], which results in a low pH. The high
concentration of ions overall results in the high conductivity.
Propanoic acid is a weak acid. It only partially dissociates
in water, resulting in a low [H3O+]. This causes a high pH
and the low overall ion concentration results in low
conductivity.
The concentration and pH of three acids, HA, HB and HC, are shown in the table
below.
acid         concentration (mol       pH
L–1)
HA                  0.100            1.00
HB                  0.100            2.50
HC                 0.00100           3.00
A small piece of magnesium is added to a 20 mL sample of each of the acids.
State which acid that would be expected to react most rapidly with the magnesium.
Explain why this acid will react the fastest.
HA would react the most rapidly with the magnesium.
This is because HA has the lowest pH, which means it is the
strongest acid. It will completely dissociate into its ions,
producing a high [H3O+]. This will result in more particles
being available and a faster reaction rate.
The table below shows the pH of two acids, HA and HB, each with the
same concentration.
Acid                 pH
HA                  1.00
HB                  4.00

When these acids react with magnesium metal, hydrogen gas (H2)
Is produced.
Discuss the reactions with both acids, HA and HB, with
magnesium metal when the same volume of each acid is used.

HA and HB will react with the magnesium metal and produce
the same amount of hydrogen gas. Although HA will produce
it faster. This is because HA is a strong acid as it has a low pH.
This means it fully dissociates into its ions and so has a high
[H3O+]. Therefore there are more particles present to react.
HB is a weak acid as it has a high pH. This means it only
partially dissociates into its ions, so has a low [H3O+].
Therefore there are less particles present to react with initially,
so a slower rate of reaction is seen.
Success Criteria
Can use Kw     Can use Kw    Can use Kw
or pH          or pH         and pH
expressions    expressions   expressions
to calculate   to solve      to solve
an unknown     several       unknowns
unknowns
Complete the following table showing hydronium ion
concentration, hydroxide ion concentration and pH for
some solutions.

Kw = 1.00 ´ 10–14

Solution       [H3O+]           [OH–]           pH
1           0.0350         2.86 x 10-13      1.46

2          1.58 x 10-11    6.31 x 10-4       10.8
3         1.77 x 10-9     5.66 × 10–6        8.75
Complete the table below to show the hydronium ion concentration,
hydroxide ion concentration, and pH for the three solutions shown.

Kw = 1.00 × 10–14

Solution            [H3O+]         [OH–]            pH
mol L–1        mol L–1

hydrochloric acid
(HCl)              0.0720      1.39 x 10-13       1.14

sodium hypochlorite
(NaOCl)          3.98 x 10-12   2.51 x 10-3
11.4

hypochlorous acid
4.46 x 10-4                       3.35
(HOCl)                           2.24 × 10   –11
If a solution of sodium hydrogen carbonate has
a pH of 9.20, calculate the concentration of
hydroxide ions, [OH-], present in the solution.

pH = -log [H3O+]

[H3O+] = inv log -9.20
= 6.31 x 10-10 mol L-1

[OH-] =    1 x 10-14
6.31 x 10-10
= 1.58 x 10-5 mol L-1
Success Criteria
the species    the species    products of
as either an   as either      the equation
acid or a      acid or base   to the
base, due to   using          increase or
its reaction   equations      decrease in
with water                    pH
A solution of sodium ethanoate (NaCH3COO) is tested and
found to have a pH of 8.50.

Discuss why the pH of the solution is greater than 7.

Sodium ethanoate solution contains both Na1+ and
CH3COO1- ions. Ethanoate ions react with water to
accept H1+ since ethanoic acid is a weak acid.
CH3COO1- + H2O                     CH3COOH        +    OH1-

So [OH1-] is increased. Therefore the
[OH1-] > [H3O1+], which results in a pH greater
than 7.
A solution of sodium hypochlorite, NaOCl, is basic.

Discuss the above statement, including appropriate

The hypochlorite ion is basic.
Therefore it accepts a proton from
water to form hydroxide ions.

OCl1- +       H2O           HOCl      +    OH1-

The [OH1-] is now greater than the
[H3O1+], producing a basic solution.

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