Equilibria (With Solution)
Equilibria (With Solution)
Equilibria (With Solution)
sg
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Q20 AJC/P2/Q2b
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[ Kc =656; 0.364mol]
Q21 SRJC/P3
3c The table below lists the equilibrium constants, Kc, for the reversible reaction
(i) Based on the above information, deduce whether the forward reaction is
exothermic or endothermic.
(ii) 2.0 moles each of H2 (g) and CO2 (g) are allowed to react in a 4.0 dm3
closed vessel. Calculate the concentration of CO (g), in mol dm-3, in the
equilibrium mixture at 700 K. [ 0.130 ]
(iii) State the effect of an increase in pressure on the percentage yield of CO.
Q22 SAJC/P2/Q3
The reaction between NO and F2 was studied by mixing the two gases. At different times
during the experiment, various single changes were made to the conditions inside the
reaction vessel.
2NO(g) + F2(g) 2ONF(g)
The change in concentrations of the three compounds in the equilibrium mixture with time
is given by the graph below:
1.4
1.3
1.2
1.1 F2
1
/ mol dm / mol dm-3
0.9
0.8
NO
Concentration Concentration
0.7
-3
0.6
0.5
0.4 ONF
0.3
0.2
0.1
0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Time / min
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(ii) Calculate the equilibrium constant, Kc, for the system at the 7th minute. Include units in
your answer. [0.123 mol-1 dm3 ]
(iii) The changes in concentration during the period 7.5 to 10 minutes was produced by
heating the system. Is this equilibrium reaction an endothermic or exothermic reaction?
Explain your answer.
(iv) Suggest the change in condition that caused the sudden increase in concentrations at
the 10th minute.
(v) Using your answer in (iv), explain the changes in concentrations from the 10th to 15th
minutes.
(vi)To investigate the equilibrium concentrations of the species at the 7th minute, the
reaction was quenched by rapid cooling. Explain why cooling must be done rapidly.
Q23 TJC/P3
5(a) Sulphuric acid is manufactured by the contact process which involves the following
reaction:
In order to increase the yield of sulphur trioxide, a student suggested increasing the
temperature and adding a catalyst to the reaction.
(b) An equimolar mixture of sulphur dioxide and oxygen was injected into a reaction vessel
of fixed volume and maintained at a fixed temperature. When equilibrium was reached,
it was found that one-third of the oxygen had reacted and that the total pressure in the
vessel was 227.5 kPa.
(i)Write an expression for the equilibrium constant, Kp, for this reaction.
The volume of the reaction mixture was compressed to half its original volume to
increase the yield of sulphur trioxide before a large portion of the sulphur trioxide was
extracted. After extraction of sulphur trioxide the mixture was allowed to reach
equilibrium and the partial pressure of sulphur dioxide was found to be 13 kPa.
(The temperature of the reaction was maintained throughout the entire process)
(iii) Determine the partial pressures of SO2, O2 and SO3 at the instant when the
volume of the reaction vessel was halved.`[ 91 , 182, 182 kPa]
(iv) Determine the new equilibrium partial pressure of sulphur trioxide. [ 32.6 kPa]
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Q24. MJC/P2
[ 99.9%]
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Q25 HCI/P2/Q3
[ 20:1; pH=7.28 ]
Q26 MJC/P3/Q5
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Q27 SRJC/P3
5(b) Morphine is a weak monoacidic alkali and is used in small quantities as a pain relief.
When 30 cm3 of 0.30 mol dm-3 morphine solution was titrated against a 0.20 mol
dm-3 hydrochloric acid, the end-point was determined using a suitable indicator.
You may use Mor and MorH+ to denote morphine and its conjugate acid
respectively.
(ii) Determine the concentration of the salt that was formed at the equivalence
point. [0.120 mol dm-3 ]
(iii) Hence, determine the pH at the equivalence point. Explain why the end
point of this titration is not neutral. [ 6.30]
(iv) Sketch and label the graph of pH against volume of hydrochloric acid used,
indicating the region of maximum buffering capacity.[8]
(c) Given 25.0 cm3 of 0.10 mol dm-3 of aqueous ammonia (pKb = 4.74), what is the
mass of ammonium nitrate that must be added to produce a solution with maximum
buffering capacity? In addition, state the pH of this buffer solution.
[pH= 9.26; mass= 0.200g ] [2]
Q28 TJC/P3
4(a) (i) Given that the Ka of benzoic acid, C6H5CO2H, is 6.46 105 mol dm3,
calculate the pH of 0.100 mol dm3 benzoic acid. [2.59]
(ii)Calculate the Kb of benzoate ion, C6H5CO2, and hence calculate the pH of
0.0500 mol dm3 sodium benzoate. [ 1.55 X 10-10 ; 8.44]
(iii)Hence sketch the pH curve when 20 cm3 of 0.100 mol dm3 benzoic acid is
titrated against 0.100 mol dm3 sodium hydroxide, showing all relevant
working.
(b) In the identification of benzoic acid, a commonly used reagent is neutral
iron(III) chloride solution, in which a buff precipitate of iron(III) benzoate,
Fe(C6H5CO2)3 is formed. When 50.0 cm3 of iron(III) chloride solution was added to
50.0 cm3 of benzoic acid solution, 0.0532 g of the buff precipitate was formed in the
mixture.
(ii)Calculate the number of moles of benzoic acid that reacted with the neutral
iron(III) chloride solution. [3.81 X 10-4 ]
(iii)Given that the pH of the solution after the reaction is 2.33, calculate the number
of moles of H+ in the mixture. [4.68 X 10-4 ]
(iv)Assuming that the H+ ions in solution are formed only from the dissociation of
benzoic acid in solution, as well as the reaction of benzoic acid with iron(III)
chloride, calculate the concentration of benzoate ions in the mixture. [8.70X10-4]
C6H5CO2H C6H5CO2 + H+
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Q29 TPJC/P3
2(b)(i) Write ionic equation to explain why aqueous solution of ammonium chloride is
acidic.
(ii) With reference to your answer in (b)(i), write an expression of acid dissociation
constant, Ka, for ammonium ion.
(iii) Calculate the pH of a solution containing 30 cm3 of 0.1 mol dm-3 ammonia and 20
cm3 of 0.1 mol dm-3 ammonium chloride.[Ka of ammonium ion is 6.00x 10-10 mol
dm-3. [9.40]
(iv) Calculate the pH of the solution if 1.00 cm3 of 1.00 mol dm-3 hydrochloric acid is
added to the solution in (iii). [9.05]
(c) Hydrangeas are flowering plants which commonly have pink flowers. They produce blue
flowers only in soils which contain high concentrations of Mg2+(aq). The pH of well limed
soil is 9.25, whereas the pH of peat-based soil is 6.55.
The numerical value of the solubility product, Ksp of magnesium hydroxide is
1.8 x 10-12.
(i) Write an expression for the Ksp of magnesium hydroxide, stating its unit.
(ii) Calculate the theorectical maximum value of [Mg2+(aq)] in a well limed soil.
(iii) Calculate the theorectical maximum value of [Mg2+(aq)] in a peat-based soil.
(iv) Deduce which of these types of soil favours good cultivation of blue hydrangeas.
Q30 SAJC/P3
3(a) YOH is a weak base. 25.0 cm3 of aqueous YOH was titrated against 0.08 mol dm-3
hydrochloric acid and the following titration curve was obtained.
12
10
Z
8
pH 6
0
0 10 20 30 40 50
Vol. of HCl added / cm 3
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(i) Explain qualitatively with a balanced equation, how the pH would change when a
small amount of NaOH is added to the solution, marked at Z, on the graph.
(ii) Using the graph, determine the dissociation constant, Kb, of YOH. [1.58X10-5]
Q31 IJC/P2/Q5
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Q32 YJC/P3
5(a) When 100 cm3 of aqueous squaric acid containing 0.011 moles of the acid was
titrated against aqueous sodium hydroxide, the following graph was obtained with
the use of a pH meter.
How many acidic hydrogen atoms are contained in one molecule of squaric
acid? [2 acidic hydrogen atoms]
pH
12
pH 7
0.022
Moles of sodium hydroxide added
(b) Most common indicators are weak acids, the un-ionised molecule being one colour
whilst the anion is a different colour, e.g.
HIn(aq) H+(aq) + In-(aq)
(colour 1) (colour 2)
(i) Write an expression for the concentration equilibrium constant, KIn, for the above
indicator, Hln.
Two indicators of different KIn values are given in the table below:
Indicator
KIn values
Bromocresol green 1.5 x 10-5
Cresol red 1.7 x 10-8
(ii) Using the values given, determine the working pH ranges for both the indicators
where there is marked change in colour.
Hence deduce which of the above two indicators would be more suitable for the
titration in (a) (ii).
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Q33 VJC/P3/Q1c
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Solutions
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Q20 AJC/P2/Q2b
Q21 SRJC/P3
Change in no of -y -y +y +y
mol
Eqm no of mol 2.0-y 2.0-y y y
y 2
()
[CO][H 2O ] 4 . 0
Kc 1.23 10 1
[H 2 ][CO2 ] ( 2.0 y )2
4. 0
y = 0.5192
Q22 SAJC/P2/Q3
3. (i) Kc = ___[ONF]2___
[NO]2[F2]
(iii) When T increases, more ONF produced, i.e. eqm shifts to the right. Hence, forward
reaction must be endothermic to absorb the excess heat.
(v) When P increases, eqm shifts to the right where fewer molecules are present.
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(vi) Cooling was done rapidly so as to prevent the equilibrium from shifting to the left such
that accurate equilibrium concentrations at the seventh minute can be found.
Q23 TJC/P3
5a) The reaction is an exothermic equation. When temperature increases, by Le
Chatlier’s Principle the position of equilibrium will shift left to absorb heat and
reduce the temperature. This decreases the yield of the SO3.
On addition of a catalyst, the position of equilibrium does not shift as the
catalyst increases the rate of the forward and backward reaction to the same
extent. Thus the yield of SO3 is also not increased.
Hence, the suggestion is not feasible.
( PSO3 ) 2
bi) Kp
( PSO2 ) 2 ( PO2 )
x 2x 2x
Total no. of moles of gas = + +
3 3 3
5x
=
3
1
PSO2 227.5 = 45.5 kPa
5
2
PO2 227.5 = 91.0 kPa
5
2
PSO3 227.5 = 91.0 kPa
5
( PSO3 ) 2
Kp =
( PSO2 ) 2 ( PO2 )
(91) 2
=
(45.5) 2 (91)
= 0.0440 kPa-1 (No units deduct 1 mark)
biii) When the volume is halved, the partial pressures are doubled
Partial pressure of SO2 = 91 kPa
Partial pressure of O2 = 182 kPa
Partial pressure of SO3 = 182 kPa
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(260 y ) 2
0.0440
(13) 2 (143)
y2 – 520y + 66537.71 = 0
y = 227.4
Q24 MJC/P2
(c) Dynamic equilibrium occurs when forward rate = backward rate and concentrations of
reactants and products are constant
(e)
Q25 HCI/P2/Q3
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Q26 MJC/P3/Q5
i.
ii.
iii.
iv.
v.
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b.
Q27 SRJC/P3
[2]
(i) Let [OH-] at eqm be x mol dm-3
Initial [ ]: 0.30 0 0
in [ ]: -x +x +x
Eqm [ ]: 0.30 – x +x +x
[OH ][MorH ]
Kb
[ Mor ]
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pH = 14 – 1.42 = 12.6
9 x103
Volume of HCl added = = 0.045 dm3
0.20
no of mol 9 x10 3
Concentration of salt = =
total vol (0.045 0.030)
(iii)
At equivalence point, salt (MorH+) undergoes hydrolysis. [1/2] or eqn
[ H ][Mor ]
Ka
[ MorH ]
[ H ]2
2.09x10-12 =
0.12
[H+] = 5.01 x 10-7 mol dm-3
pH = -log10(5.09 x 10-7) = 6.30
(iv)
[8]
pH
Maximum buffer region
12.6
6.3
45 vol HCl
(c)
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25
No. of mol of NH4+ = 0.1 X = 2.50 x 10-3 mol
1000
Mass of NH4NO3 to be added = 2.50 x 10-3 X 80.0 = 0.200 g
Q28 TJC/P3
4ai) ● pH = -log10 [H+]
= -log10 K ac
= -log10 (6.46 x 10-5 x 0.100)= 2.59
ii) ● Kb = Kw / Ka
= 1 x 10-14 / 6.46 x 10-5
= 1.55 x 10-10 moldm-3
● pH = 14 – 5.56
= 8.44
iii) ● Maximum buffering capacity occurs at 10.0 cm3 of NaOH, and the
pH = pKa = 4.19
● Correct profile and axes; students should label the sketch with calculated
answers from part (i) and (ii)
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pH
13
8.44
4.19
2.59
0 10 20 Volume of NaOH
added / cm3
C6H5CO2- ≡ C6H5CO2H
iv) Since the H+ comes from both dissociation of benzoic acid and reaction of
benzoic acid with iron (III) chloride,
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Q29 TPJC/P3
2b (i) soln is acidic due to hydrolysis of salt: presence of excess H3O+ resulted in acidic soln.
Q30 SAJC/P3
3(a)(i) At Z, equal proportion of YOH and YCl is present. Hence Z is at maximum buffer
capacity, pH does not change significantly when a small amount of NaOH is added
as Y+ will neutralize the excess OH- added.
Y+ + OH- YOH
(a)(ii) At Z, [Y+] = [YOH]
Hence, pOH = pKb
pH = 9.2
pOH = pKb = 14 – 9.2 = 4.8
Kb = 10-4.8 = 1.58 x 10-5 mol dm-3
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Q31 IJC/P2/Q5
Q32 YJC/P3
ai. From the graph, 0.011 moles of squaric acid reacts with 0.022 moles of NaOH solution.
Thus 1 mole of squaric acid contains 2 acidic hydrogen atoms
[H + ][In ]
KIn = [1m or zero]
[HIn]
Q33 VJC/P3/Q1c
i.
ii.
iii
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iv
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