NCEA Level 2 Chemistry (90310) 2009 — page 1 of 5
Assessment Schedule – 2009
Chemistry: Describe thermochemical and equilibrium principles (90310)
Evidence Statement
Q
ONE
(a)
(b)(i)
Evidence
Achievement
Achievement with
Merit
(i) H3O+ + Cl–
THREE OF
TWO OF:
(ii) H3O+ + CH3COO–
ONE equation
completed a
ONE equation
completed a
ONE equation
completed a
OR
ONE correct pH a
OR
ONE correct pH a
AND
ONE correct pH a
OR
TWO concentrations
correctly calculated a
OR
TWO
concentrations
correctly
calculated a
AND
TWO
concentrations
correctly
calculated a
(accept rounding
from 2 to 4sf)
(accept rounding
up to 4sf)
AND
AND
Explains that
EITHER
[H3O+]=[HCl]
because HCl is a
strong acid so it
completely
dissociates /
ionises
OR
[H3O+]<[CH3COO
H] because
CH3COOH is a
weak acid so it
does not
completely
dissociate /
ionisation.
m
Discusses the idea
that because HCl
is a strong acid
and completely
dissociates; this
means that [HCl]
= [H3O+].
AND ethanoic
acid, a weak acid,
hardly dissociates,
and therefore the
[H3O+] is less than
[CH3COOH].
[H3O+] /
mol L–1
0.0376
[OH–] / mol L–1
pH
2.66  10–13
1.42
[H3O+] /
mol L–1
4.03 
10–11
[OH–] / mol L–1
pH
2.48  10–4
10.4
(accept rounding
from 2 to 4sf – do not
accept only 1 sf )
Achievement with
Excellence
OR
(ii)
For HCl, the [H3O+] is equal to [HCl].
HCl is a strong acid and completely
dissociates in solution. Consequently the
[H3O+] is equal to [HCl].
(iii)
For CH3COOH, [H3O+] is less than
[CH3COOH].
CH3COOH is a weak acid and only partially
dissociates in solution. Consequently, the
[H3O+] would be less than the concentration
of the ethanoic acid.
EITHER
Identifies HCl as a
strong acid /
completely
dissociates
OR
ethanoic acid, as a
weak acid / partially
dissociates
a
OR
(c)
Conductivity of NaOH and HCl are both
high. Conductivity of CH3COOH is low.
HCl is a strong acid and it completely
dissociates into H3O+ and Cl– ions. This
means the conductivity will be high, as there
will be a large concentration of ions in
solution. NaOH is a strong base, and in
solution is completely ionised to Na+ ions and
OH– ions. Because there are a large
concentration of ions in solution, the
conductivity will be high. The conductivity of
both these will be the same, as they have the
same concentration of ions in solution.
Conductivity of CH3COOH is low.
CH3COOH is a weak acid and only partially
dissociates. Since there is a low concentration
of ions, it will have low conductivity.
TWO correct
conductivities
described.
AND
a
Conductivities
correct for TWO
solutions linked to
two correct
reasons.
(must include a
link to charged
particles / ions)
m
e
AND
Compares the
conductivity of all
solutions with
reasons that link to
concentration /
amount.
[ ] / n can be
implied.
Must also include
the need for ions
in the solution to
allow
conductivity.
e
NCEA Level 2 Chemistry (90310) 2009 — page 2 of 5
TWO
(a)
(i), (ii),
(iii)
Diagram One – exothermic and forward
THREE of:
Diagram Two – endothermic and reverse
SIX out of eight
correct for
exothermic,
endothermic,
forward,
reverse reactions.
∆rH (twice)
Ea (twice).
Enthalpies of reaction and activation energies
correctly labelled.
(Need to explain both equal and opposite i.e.
gains instead of releases energy.)
TWO of:
Needs 2a, 1m and
1e
SIX out of eight
correct for
exothermic,
endothermic,
forward,
reverse reactions.
∆rH (twice)
Ea (twice).
a
a
(iv)
∆rH = + 950 kJ mol–1
Because the same amount of energy is being
gained / absorbed / taken in (it is an
endothermic reaction) so the ∆rH is positive.
OR
∆rH = + 950 kJ
mol–1. Explains that
energy is gained /
absorbed.
(must have units
correct)
m
EITHER
∆rH = + 950 kJ
mol–1
OR
Explains that
energy is gained /
absorbed.
∆rH = + 950 kJ
mol–1. Explains that
energy is gained /
absorbed.
(must have units
correct)
m
a
(b)(i)
(ii)
Diagram redrawn showing the reactants and
products having a higher energy, ∆rH remains
the same, with a smaller Ea.
Explanation:
The energy of the reactants and products is
higher because with increased temperature the
particles have more (kinetic) energy.
The activation energy gap is reduced because
the particles have more energy to start with so
require less energy for effective / successful
collisions.
The ∆rH will remain the same, this is still the
same reaction so regardless of what
temperature / how much energy the reactants
start with the same amount of energy is
released.
(Could acknowledge that the ∆rH changes
when temperature changes but don’t know
how it changes.)
OR
OR
Redraws diagram
with ONE correct
aspect or describes
one aspect correctly
a
Redraws diagram
with 2 aspects and
explains 1 correctly
OR explains 2
aspects correctly.
m
OR
OR
ONE step in
calculation correct.
a
Correct answer
with correct units
(Incorrect units
negates back to
achieved).
n(CaO) = 287 / 82.0 = 3.50
m(CaO) = 3.50  56.0 = 196 g
e
Correct answer
with correct units.
(Incorrect units
negates back to
achieved).
m
(c)
Redraws diagram
with 3 aspects
correct and fully
explains 3 aspects
correctly.
m
NCEA Level 2 Chemistry (90310) 2009 — page 3 of 5
THREE
(a)(i)
Kc 
 NO2 
2
2
 NO  ΚO
 2 
TWO of:
Needs 1a, 1m and
1e
Kc expression
correct.
Kc expression
correct.
a
a
OR
At 230°C a brown colour would be observed.
(ii)
(b)
NO2 is in the highest concentration. The value
of Kc is greater than 1. This means that the
amount of the products will be greater than
the reactants. Therefore the concentration of
NO2(g) will be greater than NO(g) or O2(g).
This explains the dark brown colour, as there
is more NO2(g), which is a brown colour
compared with the other two, which are both
colourless.
HI(g) added.
(i) Colour will become more purple.
(ii) As the concentration of HI is increased,
the equilibrium will shift to oppose the
change, ie decrease the concentration of HI.
This will favour the reverse reaction
producing more I2, so more purple.
Mixture is cooled.
(i) The purple colour fades.
(ii) Decreased temperature causes the
equilibrium to shift to favour the reaction that
releases energy / heat, to replace the heat that
has been lost. ie the exothermic direction.
This will favour the forward reaction
resulting in less I2 so less purple.
Increase in pressure.
(i) No change.
(ii) Increase in pressure causes the
equilibrium to shift to reduce the number of
gaseous particles, ie shifts equilibrium to the
side with the least number of moles. Since
each side of the equilibrium equation has two
moles there will be no equilibrium shift. The
colour will remain the same.
Students need to explain terms like Le
Chatelier’s principle, ie they can’t use Le
Chatelier’s principle as a reason.
AND
EITHER
NO2 in highest
concentration.
OR
Brown colour seen.
a
[ ] / n can be
implied
EITHER
Explains that Kc
value is greater
than 1 / large,
meaning the
amount /
concentration of
products is greater
than the amount /
concentration of
reactants.
OR
Explains that the
appearance will be
brown because the
concentration /
amount of NO2(g)
is greater than the
amount /
concentration of
NO(g) and O2(g).
(Links to Kc not
made).
[ ] / n can be
implied
m
OR
AND
EITHER
Describes TWO
observations.
or
TWO shifts
correctly described.
a
Explains TWO
observations by
linking them to the
appropriate shifts or
by linking the two
observations to the
species causing the
colour change (I2)
or link to gaseous
moles as
appropriate.
Relates the idea
that a large Kc value
means there is a
higher
concentration /
amount of products
than reactants (all
but gone to
completion) and
that the colour will
be brown as there
are so many NO2(g)
molecules.
Must mention [ ] or
n to get to e.
e
AND
m
Discusses THREE
observations by
linking them to the
correct shift in
equilibrium and
explaining why the
shift occurs and
either link to the
species causing the
colour change (I2)
or link to gaseous
moles as
appropriate.
e
NCEA Level 2 Chemistry (90310) 2009 — page 4 of 5
FOUR
(a)
Rate of reaction can be increased by
 increasing the temperature (heat the
reaction)
 increasing the surface area of the calcium
carbonate (use powdered calcium
carbonate)
 increasing the concentration of
hydrochloric acid. (not reactants).
TWO OF:
TWO correct
factors identified.
TWO correct
factors identified.
a
OR
(b)
Line A – Beaker 3
Line B – Beakers 1 and 2
Beakers 1 and 2 have the same concentration
of HCl and Beaker 3 has a lower
concentration of HCl.
Line B corresponds to beakers 1 and 2
because it is steeper showing a faster rate of
reaction. In Beakers 1 and 2, there are more
reactant particles per unit volume compared
to Beaker 3, so the concentration is greater.
As the concentration of reactants is increased,
the collision rate of reactant particles
increases, so there are more successful /
effective collisions and the reaction rate is
increased. Hence a steeper line on the graph.
Line A corresponds to beaker 3 because it is
less steep and shows a slower rate of reaction.
In beaker 3 there are fewer reactant particles
per unit volume compared to beakers 1 and 2,
so the concentration is lower. As the
concentration of the reactants is decreased,
the collision rate of the reactant particles
decreases, so there are fewer successful /
effective collisions and the rate of the reaction
is decreased.
(c)(i)
(ii)
States that as temperature increases the rate of
reaction increases / time taken for the solution
to decolourise decreases.
This is because as temperature increases, the
molecules have more kinetic energy / higher
energy and moving faster. There will be an
increase in the frequency of collisions
between particles. Particles also collide more
effectively when they actually do collide.
There are more effective / successful
collisions because more particles have
enough kinetic energy to overcome the
activation energy for the reaction. Leading to
an increased rate of reaction.
EITHER
Line B corresponds
to Beakers 1 and 2
OR
Line A corresponds
to Beaker 3
(could be implied
or stated in
explanation)
OR
Link concentration
and rate of reaction
for one of beakers
1, 2 or 3.
a
a
EITHER
Line B
Explains that
Beakers 1 and 2
have a higher
concentration of
reactants / more
particles than
Beaker 3 and
therefore there are
more collisions /
effective collisions.
OR
Line A
Explains that
Beakers 3 has a
lower concentration
of reactants / fewer
particles than
Beakers 1 and 2
and therefore there
are fewer collisions
/ effective
collisions
[ ] can be implied.
m
OR
AND
Discussion of rates
of reaction in terms
of effective particle
collision and
concentration of
particles for both
Lines A and B.
If Lines A and B
have been mixed
up, E can still be
achieved by
including a link to
the graph, i.e. the
steepness of the
gradient of the rate
of H2 gas evolution.
[ ] must be stated.
e
AND
AND
States the trend, ie
that reaction rate is
increased.
a
Links temperature
increase to
molecules which
are moving faster,
so increased
frequency of
collisions
AND that more of
these collisions will
be effective OR
have sufficient
energy to reach the
activation energy
required.
m
Discussion of rates
of reaction in terms
of an increase in
temperature results
in particles having
more kinetic energy
/ higher energy and
moving faster.
AND
Collisions are more
likely to have
sufficient energy to
overcome the
activation energy
barrier so more
successful /
effective collisions
AND an increase in
rate of reaction.
e
NCEA Level 2 Chemistry (90310) 2009 — page 5 of 5
Judgement Statement
Achievement
Achievement with Merit
Achievement with Excellence
3A
2M+1A
2E+1M
OR
OR
1M+1A
2E+2A
NOTE:
Lower case a, m, e may be used throughout the paper to indicate contributing evidence for overall grades for
questions.
Only upper case A, M and E grades shown at the end of each full question are used to make the final judgement.