equilibrium	

	Name:	
________________________	

Class:	
________________________	

Date:	
________________________	

 

 	
Time:	128 minutes	
Marks:	120 marks	
Comments:		

Q1.Which change would alter the value of the equilibrium constant (Kp) for this reaction?
2SO2(g) + O2(g)  2SO3(g)
A        Increasing the total pressure of the system.         
B        Increasing the concentration of sulfur trioxide.     
C        Increasing the concentration of sulfur dioxide.     
D       Increasing the temperature.                                  
(Total 1 mark)

Q2.The following information concerns the equilibrium gas-phase synthesis of methanol.
CO(g) + 2H2(g)  CH3OH(g)
At equilibrium, when the temperature is 68 °C, the total pressure is 1.70 MPa.
The number of moles of CO, H2 and CH3OH present are 0.160, 0.320 and 0.180, respectively.
Thermodynamic data are given below.
 
 	Substance	ΔH / kJ mol−1	S / J K-1 mol-1	
 	CO(g)	−110	198	
 	H2(g)	0	131	
 	CH3OH(g)	−201	240	

The mole fraction of hydrogen in the equilibrium mixture is
A       0.242
B       0.485
C       0.653
D       0.970
(Total 1 mark)

Q3.The following information concerns the equilibrium gas-phase synthesis of methanol.
CO(g) + 2H2(g)  CH3OH(g)
At equilibrium, when the temperature is 68 °C, the total pressure is 1.70 MPa.
The number of moles of CO, H2 and CH3OH present are 0.160, 0.320 and 0.180, respectively.
Thermodynamic data are given below.
 
 	Substance	ΔH / kJ mol−1	S / J K-1 mol-1	
 	CO(g)	−110	198	
 	H2(g)	0	131	
 	CH3OH(g)	−201	240	

Which one of the following statements applies to this equilibrium?
A       The value of Kp increases if the temperature is raised.
B       The value of Kp increases if the pressure is raised.
C       The yield of methanol decreases if the temperature is lowered.
D       The yield of methanol decreases if the pressure is lowered.
(Total 1 mark)

Q4.The following information concerns the equilibrium gas-phase synthesis of methanol.
CO(g) + 2H2(g)  CH3OH(g)
At equilibrium, when the temperature is 68 °C, the total pressure is 1.70 MPa.
The number of moles of CO, H2 and CH3OH present are 0.160, 0.320 and 0.180, respectively.
Thermodynamic data are given below.
 
 	Substance	ΔH / kJ mol−1	S / J K-1 mol-1	
 	CO(g)	−110	198	
 	H2(g)	0	131	
 	CH3OH(g)	−201	240	

With pressures expressed in MPa units, the value of the equilibrium constant, Kp, under these conditions is
A       1.37
B       1.66
C       2.82
D       4.80
(Total 1 mark)

Q5.This question relates to the equilibrium gas-phase synthesis of sulphur trioxide:
2SO2(g) + O2(g)  2SO3(g)
Thermodynamic data for the components of this equilibrium are:
 
 	Substance	ΔH / kJ mol−1	S / J K-1 mol-1	
 	SO3(g)	−396	+257	
 	SO2(g)	–297	+248	
 	O2(g)	0	+204	

This equilibrium, at a temperature of 585 K and a total pressure of 540 kPa, occurs in a vessel of volume 1.80 dm3. At equilibrium, the vessel contains 0.0500 mol of SO2(g), 0.0800 mol of O2(g) and 0.0700 mol of SO3(g).
At equilibrium in the same vessel of volume 1.80 dm3 under altered conditions, the reaction mixture contains 0.0700 mol of SO3(g), 0.0500 mol of SO2(g) and 0.0900 mol of O2(g) at a total pressure of 623 kPa. The temperature in the equilibrium vessel is
A       307 °C
B       596 K
C       337 °C
D       642 K
(Total 1 mark)

Q6.The following information concerns the equilibrium gas-phase synthesis of methanol.
CO(g) + 2H2(g)  CH3OH(g)
At equilibrium, when the temperature is 68 °C, the total pressure is 1.70 MPa.
The number of moles of CO, H2 and CH3OH present are 0.160, 0.320 and 0.180, respectively.
Thermodynamic data are given below.
 
 	Substance	ΔH / kJ mol−1	S / J K-1 mol-1	
 	CO(g)	−110	198	
 	H2(g)	0	131	
 	CH3OH(g)	−201	240	

Possible units for the equilibrium constant, Kp, for this reaction are
A       no units
B       kPa
C       MPa−1
D       kPa−2
(Total 1 mark)

Q7.This question relates to the equilibrium gas-phase synthesis of sulphur trioxide:
2SO2(g) + O2(g)  2SO3(g)
Thermodynamic data for the components of this equilibrium are:
 
 	Substance	ΔH / kJ mol−1	S / J K-1 mol-1	
 	SO3(g)	−396	+257	
 	SO2(g)	–297	+248	
 	O2(g)	0	+204	

This equilibrium, at a temperature of 585 K and a total pressure of 540 kPa, occurs in a vessel of volume 1.80 dm3. At equilibrium, the vessel contains 0.0500 mol of SO2(g), 0.0800 mol of O2(g) and 0.0700 mol of SO3(g).
With pressures expressed in MPa units, the value of the equilibrium constant, Kp, is
A       4.90
B       6.48
C       9.07
D       16.8
(Total 1 mark)

Q8.This question relates to the equilibrium gas-phase synthesis of sulphur trioxide:
2SO2(g) + O2(g)  2SO3(g)
Thermodynamic data for the components of this equilibrium are:
 
 	Substance	ΔH / kJ mol−1	S / J K-1 mol-1	
 	SO3(g)	−396	+257	
 	SO2(g)	–297	+248	
 	O2(g)	0	+204	

This equilibrium, at a temperature of 585 K and a total pressure of 540 kPa, occurs in a vessel of volume 1.80 dm3. At equilibrium, the vessel contains 0.0500 mol of SO2(g), 0.0800 mol of O2(g) and 0.0700 mol of SO3(g).
Possible units for the equilibrium constant Kp include
A       no units
B       kPa
C       Mpa−1
D       kPa−2
(Total 1 mark)

Q9.The equation for the combustion of butane in oxygen is
C4H10 + 6 O2 → 4CO2 + 5H2O
The mole fraction of butane in a mixture of butane and oxygen with the minimum amount of oxygen required for complete combustion is
A       0.133
B       0.153
C       0.167
C       0.200
(Total 1 mark)

Q10.This question relates to the equilibrium gas-phase synthesis of sulphur trioxide:
2SO2(g) + O2(g)  2SO3(g)
Thermodynamic data for the components of this equilibrium are:
 
 	Substance	ΔH / kJ mol−1	S / J K-1 mol-1	
 	SO3(g)	−396	+257	
 	SO2(g)	–297	+248	
 	O2(g)	0	+204	

This equilibrium, at a temperature of 585 K and a total pressure of 540 kPa, occurs in a vessel of volume 1.80 dm3. At equilibrium, the vessel contains 0.0500 mol of SO2(g), 0.0800 mol of O2(g) and 0.0700 mol of SO3(g).
The mole fraction of SO3 in the equilibrium mixture is
A       0.250
B       0.350
C       0.440
D       0.700
(Total 1 mark)

Q11. 
 	Summarised directions for recording responses to multiple completion questions	
 	A
(i), (ii) and (iii)
only	B
(i) and (iii) only	C
(ii) and (iv) only	D
(iv) alone	

Which of the following statements about a catalyst is / are true?
(i)      It speeds up the forward reaction and slows down the reversere action.
(iii)    It increases the proportion of molecules with higher energies.
(iii)    A homogeneous catalyst usually acts in the solid state.
(iv)    It does not alter the value of the equilibrium constant.
(Total 1 mark)

Q12.This question is about the reaction given below.
    CO(g) + H2O(g)   CO2(g) + H2(g)
Enthalpy data for the reacting species are given in the table below.
 
 	Substance	CO(g)	H2O(g)	CO2(g)	H2(g)	
 	ΔH / kJ mol−1	−110	−242	−394	0	

Which one of the following statements is not correct?
A       The value of Kp changes when the temperature changes.
B       The activation energy decreases when the temperature is increased.
C       The entropy change is more positive when the water is liquid rather than gaseous.
D       The enthalpy change is more positive when the water is liquid rather than gaseous.
(Total 1 mark)

Q13.          (a)     The gaseous reactants W and X were sealed in a flask and the mixture left until the following equilibrium had been established.
2W(g)  +  X(g)    3Y(g)  +  2Z(g)        ΔH = –200 kJ mol–1
Write an expression for the equilibrium constant, Kp, for this reaction.
State one change in the conditions which would both increase the rate of reaction and decrease the value of Kp. Explain your answers.
(7)

(b)     Ethyl ethanoate can be prepared by the reactions shown below.
          Reaction 1
CH3COOH(l) + C2H5OH(l)    CH3COOC2H5(l) + H2O(l)      ∆H = –2.0 kJ mol–1
          Reaction 2
CH3COCl(l) + C2H5OH(l) → CH3COOC2H5(l) + HCl(g)            ∆H = –21.6 kJ mol–1
(i)      Give one advantage and one disadvantage of preparing ethyl ethanoate by Reaction 1 rather than by Reaction 2.

(ii)     Use the information given above and the data below to calculate values for the standard entropy change, ∆S, and the standard free-energy change, ∆G, for Reaction 2 at 298 K.
 
 	CH3COCl(l)	C2H5OH(l)	CH3COOC2H5(l)	HCl(g)	
S/JK1mol1	201	161	259	187	

(8)
(Total 15 marks)

 

Q14.          Hydrogen and carbon monoxide were mixed in a 2:1 mole ratio. The mixture was allowed to reach equilibrium according to the following equation at a fixed temperature and a total pressure of 1.75 × 104 kPa.
2H2(g)   +   CO(g)      CH3OH(g)
(a)     The equilibrium mixture contained 0.430 mol of carbon monoxide and 0.0850 mol of methanol.
(i)      Calculate the number of moles of hydrogen present in the equilibrium mixture.
.............................................................................................................

(ii)     Hence calculate the mole fraction of hydrogen in the equilibrium mixture.
.............................................................................................................
.............................................................................................................
.............................................................................................................

(iii)     Calculate the partial pressure of hydrogen in the equilibrium mixture.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(5)

(b)     In a different mixture of the three gases at equilibrium, the partial pressure of carbon monoxide was 7550 kPa, the partial pressure of hydrogen was 12300 kPa and the partial pressure of methanol was 2710 kPa.
(i)      Write an expression for the equilibrium constant, Kp, for this reaction.
.............................................................................................................

(ii)     Calculate the value of the equilibrium constant, Kp, for the reaction under these conditions and state its units.
Kp ........................................................................................................
.............................................................................................................
Units ....................................................................................................
(3)

(c)     Two isomeric esters E and F formed from methanol have the molecular formula C6H12O2
Isomer E has only 2 singlet peaks in its proton n.m.r. spectrum.
Isomer F is optically active.
Draw the structures of these two isomers.
Isomer E

 

Isomer F

 

(2)
(Total 10 marks)

 

Q15.          When a mixture of 0.345 mol of PCl3 and 0.268 mol of Cl2 was heated in a vessel of fixed volume to a constant temperature, the following reaction reached equilibrium.
PCl3(g)   +   Cl2(g)      PCl5(g)           DH = –93 kJ mol–1
At equilibrium, 0.166 mol of PCl5 had been formed and the total pressure was 225 kPa.
(a)     (i)      Calculate the number of moles of PCl3 and of Cl2 in the equilibrium mixture.
Moles of PCl3 ......................................................................................
Moles of Cl2 .........................................................................................

(ii)     Calculate the total number of moles of gas in the equilibrium mixture.
.............................................................................................................
(3)

(b)     Calculate the mole fraction and the partial pressure of PCl3 in the equilibrium mixture.
Mole fraction of PCl3 .....................................................................................
......................................................................................................................
Partial pressure of PCl3 ................................................................................
......................................................................................................................
(3)

(c)     (i)      Write an expression for the equilibrium constant, Kp, for this equilibrium.
.............................................................................................................
.............................................................................................................

(ii)     The partial pressures of Cl2 and PCl5 in the equilibrium mixture were 51.3 kPa and 83.6 kPa, respectively, and the total pressure remained at 225 kPa. Calculate the value of Kp at this temperature and state its units.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)

(d)     State the effect on the mole fraction of PCl3 in the equilibrium mixture if
(i)      the volume of the vessel were to be increased at a constant temperature,
.............................................................................................................
(ii)     the temperature were to be increased at constant volume.
.............................................................................................................
(2)
(Total 12 marks)

 

Q16.          At high temperatures, SO2Cl2 dissociates according to the following equation.
SO2Cl2(g)    SO2(g)  +  Cl2(g)               ΔH = +93 kJ mol–1
When 1.00 mol of SO2Cl2 dissociates, the equilibrium mixture contains 0.75 mol of Cl2 at 673 K and a total pressure of 125 kPa.
(a)     Write an expression for the equilibrium constant, Kp, for this reaction.
......................................................................................................................
......................................................................................................................
(1)

(b)     Calculate the total number of moles of gas present in the equilibrium mixture.
......................................................................................................................
(2)

(c)     (i)      Write a general expression for the partial pressure of a gas in a mixture of gases in terms of the total pressure.
.............................................................................................................
.............................................................................................................

(ii)     Calculate the partial pressure of SO2Cl2 and the partial pressure of Cl2 in the equilibrium mixture.
Partial pressure of SO2Cl2 ...................................................................
.............................................................................................................
Partial pressure of Cl2 ..........................................................................
.............................................................................................................
(5)

(d)     Calculate a value for the equilibrium constant, Kp, for this reaction and give its units.
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)

(e)     State the effect, if any, of an increase in temperature on the value of Kp for this reaction.
Explain your answer.
Effect on Kp ..................................................................................................
Explanation ..................................................................................................
......................................................................................................................
(2)

(f)      State the effect, if any, of an increase in the total pressure on the value of Kp for this reaction.
......................................................................................................................
(1)
(Total 14 marks)

 

Q17.          Sulphur dioxide and oxygen were mixed in a 2:1 mol ratio and sealed in a flask with a catalyst.
The following equilibrium was established at temperature T1
2SO2(g) + O2(g)    2SO3(g)              ΔH =  –196 kJ mol–1
The partial pressure of sulphur dioxide in the equilibrium mixture was 24 kPa and the total pressure in the flask was 104 kPa.
(a)     Deduce the partial pressure of oxygen and hence calculate the mole fraction of oxygen in the equilibrium mixture.
Partial pressure of oxygen ...........................................................................
Mole fraction of oxygen ................................................................................
......................................................................................................................
......................................................................................................................
(3)

(b)     Calculate the partial pressure of sulphur trioxide in the equilibrium mixture.
......................................................................................................................
(1)

(c)     Write an expression for the equilibrium constant, Kp, for this reaction. Use this expression to calculate the value of Kp at temperature T1 and state its units.
Expression for Kp .........................................................................….............
......................................................................................................................
Calculation .................................................................................…...............
......................................................................................................................
......................................................................................................................
Units .............................................................................................................
(4)

(d)     When equilibrium was established at a different temperature, T2, the value of Kp was found to have increased. State which of T1 and T2 is the lower temperature and explain your answer.
Lower temperature........................................................................................
Explanation ...................................................................................................
......................................................................................................................
(3)

(e)     In a further experiment, the amounts of sulphur dioxide and oxygen used, the catalyst and the temperature, T1, were all unchanged, but a flask of smaller volume was used.
Deduce the effect of this change on the yield of sulphur trioxide and on the value of Kp.
Effect on yield of SO3 ...................................................…............................
Effect on Kp ...................................................................................................
......................................................................................................................
(2)
(Total 13 marks)

 

Q18.Many chemical processes release waste products into the atmosphere. Scientists are developing new solid catalysts to convert more efficiently these emissions into useful products, such as fuels. One example is a catalyst to convert these emissions into methanol. The catalyst is thought to work by breaking a H–H bond.
An equation for this formation of methanol is given below.
CO2(g) + 3H2(g)  CH3OH(g) + H2O(g)    ∆H = −49 kJ mol−1
Some mean bond enthalpies are shown in the following table.
 
 	Bond	C=O	C–H	C–O	O–H	
 	Mean bond enthalpy / kJ mol−1	743	412	360	463	

(a)     Use the enthalpy change for the reaction and data from the table to calculate a value for the H–H bond enthalpy.

 
 
H–H bond enthalpy = ............................... kJ mol−1
(3)

(b)     A data book value for the H–H bond enthalpy is 436 kJ mol−1.
Suggest one reason why this value is different from your answer to part (a).
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)

(c)     Suggest one environmental advantage of manufacturing methanol fuel by this reaction.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)

(d)     Use Le Chatelier's principle to justify why the reaction is carried out at a high pressure rather than at atmospheric pressure.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)

(e)     Suggest why the catalyst used in this process may become less efficient if the carbon dioxide and hydrogen contain impurities.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)

(f)      In a laboratory experiment to investigate the reaction shown in the equation below, 1.0 mol of carbon dioxide and 3.0 mol of hydrogen were sealed into a container. After the mixture had reached equilibrium, at a pressure of 500 kPa, the yield of methanol was 0.86 mol.
CO2(g) + 3H2(g)  CH3OH(g) + H2O(g)
Calculate a value for Kp
Give your answer to the appropriate number of significant figures.
Give units with your answer.

 

 

 

Kp = ............................... Units = ...............................
(7)
(Total 16 marks)

Q19.          The manufacture of methanol can be achieved in two stages.
(a)     In the first stage, methane and steam react according to the following equation. 
CH4(g) + H2O(g)  CO(g) + 3H2(g)                           ∆Hο = +210 kJ mol–1
Discuss, with reasons, the effects of increasing separately the temperature and the pressure on the yield of the products and on the rate of this reaction.
(6)

(b)     In the second stage, carbon monoxide and hydrogen react according to the following equation. 
CO(g) + 2H2(g)  CH3OH(g)
A 62.8 mol sample of carbon monoxide was added to 146 mol of hydrogen. When equilibrium was reached at a given temperature, the mixture contained 26.2 mol of methanol at a total pressure of 9.50 MPa. 
Write an expression for the equilibrium constant, Kp, for this reaction. Calculate a value for Kp at this temperature and give its units.
(8)
(Total 14 marks)

 

Q20.          A sealed flask containing gases X and Y in the mole ratio 1:3 was maintained at 600 K until the following equilibrium was established.
X(g) + 3Y(g)  2Z(g)
The partial pressure of Z in the equilibrium mixture was 6.0 MPa when the total pressure was 22.0 MPa.
(a)     (i)      Write an expression for the equilibrium constant, Kp, for this reaction.
.............................................................................................................
.............................................................................................................

(ii)     Calculate the partial pressure of X and the partial pressure of Y in the equilibrium mixture.
Partial pressure of X ...........................................................................
.............................................................................................................
Partial pressure of Y ............................................................................

(iii)     Calculate the value of Kp for this reaction under these conditions and state its units.
Value of Kp ..........................................................................................
.............................................................................................................
.............................................................................................................
Units of Kp ...........................................................................................
(6)

(b)     When this reaction is carried out at 300 K and a high pressure of 100 MPa, rather than at 600 K and 22.0 MPa, a higher equilibrium yield of gas Z is obtained.
Give two reasons why an industrialist is unlikely to choose these reaction conditions.
Reason 1 .....................................................................................................
Reason 2 .....................................................................................................
(2)
(Total 8 marks)

 

Q21.An experiment was carried out to determine the equilibrium constant, Kc, for the following reaction.
CH3CH2COOH   +   CH3CH2CH2OH      CH3CH2COOCH2CH2CH3   +   H2O
A student added measured volumes of propan-1-ol and propanoic acid to a conical flask. A measured volume of concentrated hydrochloric acid was added to the flask, which was then sealed.
After 1 week, the contents of the flask were poured into water and the solution was made up to a known volume.
This solution was titrated with standard sodium hydroxide solution.
(a)     Explain how the student could determine the amount, in moles, of propan-1-ol added to the flask.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)

(b)     The titration described above gives the total amount of acid in the equilibrium mixture.
Explain how, by carrying out a further experiment, the student could determine the amount of propanoic acid in the equilibrium mixture.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)

(c)     In a repeat experiment, the student failed to seal the flask that contained the equilibrium mixture.
Explain why this error would lead to the student obtaining an incorrect value for the equilibrium constant Kc
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 6 marks)

 

M1.D
[1]

M2.B
[1]

M3.D
[1]

M4.B
[1]

M5.D
[1]

M6.D
[1]

M7.C
[1]

M8.C
[1]

M9.A
[1]

M10.B
[1]

M11.D
[1]

M12.B
[1]

M13.          (a)     M1    Kp = (PY)3. (PZ)2/ (PW)2.( PX)      NB [ ] wrong
1
M2    temperature
1
M3    increase
1
M4    particles have more energy or greater velocity/speed
1
M5    more collisions with E > Ea or more successful collisions
1
M6    Reaction exothermic or converse
1
M7    Equilibrium moves in the left
1
          Marks for other answers
Increase in pressure or concentration   allow M1, M5, M6         Max 3
Addition of a catalyst;                             allow M1, M5, M6         Max 3
Decrease in temperature;                      allow M1, M2, M6         Max 3
Two or more changes made;                 allow M1, M6                Max 2

(b)     (i)      Advantage; reaction goes to completion, not reversible
or faster
                1
         Disadvantage; reaction vigorous/dangerous
(exothermic must be qualified)
          or HCl(g) evolved/toxic
or CH3COCl expensive
NB     Allow converse answers
     Do not allow reactions with other reagents e.g. water
     or ease of separation
1

(ii)     ΔS = ΣS products – ΣS reactants
1
         ΔS = (259 + 187) – (201 + 161)
1
         ΔS = 84 (JK–1 mol–1)     (Ignore units)
Allow – 84 to score (1) mark
1
         ΔG = ΔH – TΔS
1
      = – 21.6 – 298 × 84/1000
 = – 46.6 kJ mol–1 or – 46 600 J mol–1
1
Allow (2) for – 46.6 without units
(Mark ΔG consequentially to incorrect ΔS)
(e.g. ΔS = –84 gives ΔG = +3.4 kJ mol–1)
1
[15]

 

M14.          (a)     (i)      0.86 (1)

(ii)     total moles = 0.86 + 0.43 +0.085 = 1.375 (1)
 mole fraction of H2 =  = 0.625 (1)
                                                     (0.62 ‑ 0.63)
Conseq on (i)

(iii)     pp = mole fractn × total P (1)
     = 0.625 × 1.75 × 104
     = 1.09 × 104 (kPa) (1)
or 1.1(0)
Ignore units
Conseq on (ii)
5

(b)     (i)      Kp =  (1)
Penalise [  ]

(ii)     Kp =  = 2.37 (2.4) × 10–9  (1)
OR 2.37 × 10–15
         Units: kPa–2 (1)
or Pa–2
not conseq to wrong Kp expression
3

(c)     Isomer E: 
          Isomer F: 
2
[10]

 

M15.          (a)     (i)      Moles of PCl3: 0.345 – 0.166 = 0.179 (1)
Moles of Cl2: 0.268 – 0.166 = 0.102 (1)
3 sig figs
(ii)     0.447 (1)
allow 2 sig figs
conseq on (i)
3

(b)     Mole fraction of PCl3: 0.179/0.447 (1) = 0.4(00)
Partial pressure of PCl3: pp = mol fn × total P (1)
                                 = 0.400 × 225 = 90 (1) kPa (1)
3

(c)     (i)      Kp =  (1)
ignore brackets except [ ]
must show P
(ii)     Kp =  (1) = 1.8(1) × 10–2 (1) Kpa–1 (1) (or 1.81 × 10–5 Pa–1)
If 83.6 and 51.3 wrong way round, AE – 1,
answer = 6.81 × 10–3
If Kp × in (i) allow max 2 for substitution of numbers and conseq units
4

(d)     (i)      increased (1)
(ii)     increased (1)
2
[12]

          Organic points
(1)     Curly arrows: must show movement of a pair of electrons,
i.e. from bond to atom or from lp to atom / space
e.g.

(2)     Structures
penalise sticks (i.e. ) once per paper

Penalise once per paper
          allow CH3– or –CH3 or  or CH3
    or   H3C–

 

M16.          (a)     Kp =  (1)
1

(b)     0.25 + 0.75 + 0.75 = 1.75 (1) (1)
2

(c)     (i)      p = Total pressure × mol fraction (1)
(ii)     Partial of SO2Cl2: 125 ×  = 17.9 kPa (1)
Partial pressure of Cl2: 125 ×  = 53.6 kPa (1) (1)
5

(d)     Kp =  (1) = 161 (1) kPa (1)
3

(e)     Effect on Kp: increase (1)
Explanation: increase T sends equilibrium in endothermic direction (1)
2
(f)      no effect (1)
1

          Notes
(a)     If Kp has [ ] lose mark in (a) but allow full marks in (d)
          If Kp wrong/upside down etc, allow max 2 in (d) for substitution of numbers (1) 
and consequential units (1)

(b)     Mark for moles of SO2Cl2 can be scored in part (c) (ii) if not gained in (b)
          1.75 get (2)
          If moles of SO2Cl2 = 1, this is a Chemical Error, hence a 2 mark penalty
•    If total moles given in (b) = 1.75, this scores [2] in (b); but if the no moles of 
SO2Cl2 = 1 in (c)(ii), lose both marks in (c)(ii) for pp of SO2Cl2 = (1/1.75) × 125, 
i.e. the 2 mark penalty is in (c)(ii).
•    If total moles given in (b) = 2.5, score zero in (b), but can gain full marks in (c)(ii) consequentially, i.e. the 2 mark penalty is in (b).
•    If moles of SO2Cl2 = 1 and total in (b) does not equal 2.5, still lose both in (b) 
but can get all 4 conseq in (c)(ii) for 1/x etc and 0.75/x etc

(c)     (i)      Allow “Total pressure = sum of partial pressures” for (1) or pA = xA × ptot
(ii)     First mark is for mole fraction.
If either number in either mole fraction is not consequential on (b), 
then lose both marks for that partial p.

(d)     If pCl2 is not equal to pSO2 or any number used in Kp is not conseq on (c)(ii), 
allow units only
          SIG FIGS; must be 3 sig figs in (b) but then allow 2 sig figs in (c) and (d); 
(ignore extra figs) but penalise incorrect rounding

(e)     If effect wrong, no marks for explanation.
If effect missing, e.g. answer states “equm shifts to right”, mark on.
In the explanation, the word “endothermic” (or its equivalent) is essential.
[14]

 

M17.          (a)     12 (kPa)
1
          pp = mole fraction × total pressure or mole fraction = 12/104
1
          = 0.115
(allow 0.12)
1

(b)     68 (kPa)
1

(c)     Kp = 
(If Kp wrong, allow consequential units only)
(penalise square brackets in expression but then mark on)
1
               = 
1
      = 0.669
(Allow 0.67)
(Allow full marks in calculation consequential on their 
values in (a) and (b))
1
          kPa–1
1

(d)     T2
(Must be correct to score any marks in this section)
1
          Exothermic
1
Reduce T to shift equilibrium to the right
or forward reaction favoured by low T
or Kp increases for low T
or low T favours exothermic reaction
1

(e)     Increase
1
          None
1
[13]

 

M18.(a)     Bonds broken = 2(C=O) + 3(H–H) = 2 × 743 + 3 × H–H
Bonds formed = 3(C–H) +(C–O) + 3(O–H) = 3 × 412 + 360 + 3 × 463
Both required
1

–49 = [2 × 743 + 3 × (H–H)] – [3 × 412 + 360 + 3 × 463]
3(H–H) = –49 – 2 × 743 + [3 × 412 + 360 + 3 × 463] = 1450
Both required
1

H–H = 483 (kJ mol–1)
Allow 483.3(3)
1

(b)     Mean bond enthalpies are not the same as the actual bond enthalpies in CO2 (and / or methanol and / or water)
1

(c)     The carbon dioxide (produced on burning methanol) is used up in this reaction
1

(d)     4 mol of gas form 2 mol
1

At high pressure the position of equilibrium moves to the right to lower the pressure / oppose the high pressure
1

This increases the yield of methanol
1

(e)     Impurities (or sulfur compounds) block the active sites
Allow catalyst poisoned
1

(f)     Stage 1: moles of components in the equilibrium mixture
Extended response question
         CO2(g)  +    3H2(g)   CH3OH(g) +   H2O(g)
 
 	Initial
moles	    1.0	   3.0	  0	  0	
 	Eqm
moles	(1–0.86)
  = 0.14	(3–3×0.86)
= 0.42	0.86	0.86	

1

Stage 2: Partial pressure calculations
Total moles of gas = 2.28
Partial pressures = mol fraction × ptotal
1

pCO2 = mol fraction × ptotal = 0.14 × 500 / 2.28 = 30.7 kPa
pH2 = mol fraction × ptotal = 0.42 × 500 / 2.28 = 92.1 kPa
M3 is for partial pressures of both reactants
Alternative M3 =
ppCO2 = 0.0614 × 500
ppH2 = 0.1842 × 500
1

pCH3OH = mol fraction × ptotal = 0.86 × 500 / 2.28 = 188.6 kPa
pH2O = mol fraction × ptotal = 0.86 × 500 / 2.28 = 188.6 kPa
M4 is for partial pressures of both products
Alternative M4 =
ppCH3OH = 0.3772 × 500
ppH2O = 0.3772 × 500
1

Stage 3: Equilibrium constant calculation
Kp = pCH3OH × pH2O / pCO2 × (pH2)3
1

Hence Kp = 188.6 × 188.6 / 30.7 × (92.1)3 = 1.483 × 10–3 = 1.5 × 10–3
Answer must be to 2 significant figures
1

Units = kPa–2
1
[16]

M19.          (a)     (must state correct effect on yield or rate to score the reason mark)
T       effect:  higher temp:    yield greater or shifts equilibrium to right;
1
effect:   higher temp:   rate increased;
1
reason:  endothermic
OR
more particles have E>Ea
1
OR
more successful/productive collisions;
1
P       effect:   higher pressure: yield less or shifts equilibrium to left;
1
         effect:   higher pressure: rate increased;
reason: increase in gas moles L to R
OR
greater collision frequency;
(Q of L mark)
1

(b)     M1     equilibrium moles of CO = 62.8 - 26.2 = 36.6
1
M2     equilibrium moles of H2 = 146 – 2(26.2) = 93.6
1
M3     total no moles = 36.6 + 93.3 + 26.2 = 156.4
1
M4     partial pressure = mole fraction x total pressure
1
M5            
1
M6    

1

M7     0.022(1)        2.2(l)×10–8                         2.2(l)×10–14
1
M8        MPa–2                      kPa–2                                 Pa–2
1
If no subtraction lose M1, M2 and M3)
(If ×2 missed in M2, lose both M2 and M3)
(If M1 gained but moles of H2 = 73.2 (i.e. double CO), M2 and M3 lost)
(If M1 gained but mol H2 = 2(146 – 26.2), M2 and M3 lost)
(If M1 and M2 correct but M3 lost for CE, penalise M6 also)
(M4 can be gained from the numbers in the expression for M6 even if these numbers are wrong)
(If Kp contains [ ] lose M5 but then mark on)
(If chemically wrong expression for Kp, lose M5, M6 and M7 (allow M8 conseq on their Kp))
(If divided by 9.5, or not used 9.5 at all, lose M6 and M7 (and M4))
(If tried to convert to kPa and is factor(s) of 10 out, penalise in M6 and allow M8 for kPa–2)
[14]

 

M20.          (a)     (i)      (Kp) = (pz)2/(px)(py)3
(penalise use of square brackets, allow ())
1
(ii)     X (22–6)/4 = 4 (MPa)
(mark is for value 4 only, ignore units)
1
Y obtained by multiplying value for X by 3
(allow conseq on wrong value for X)
1
Y 4.0 × 3 = 12 (MPa)
(mark is for value 12 only)
1

(iii)     Kp = 6.02/4.0 × 12.03 = 5.21 × 10–3
(allow conseq on wrong values for X and
Y e.g.62/3 × 93 = 0.165)
(if Kp wrong in (a)(i) CE)
1
MPa–2
(allow any unit of P–2 provided ties to P used for Kp value)
1

(b)     high pressure expensive (due to energy or plant costs)
1
(Rate is) slow (at lower temperatures)
1
[8]

 

M21.(a)     Multiply volume of propan−1−ol by density
Allow measure the mass of the volume added
Any reference to concentration of propan−1−ol CE = 0
1
Divide the mass by the Mr of propan−1−ol
1

(b)     Titrate a measured volume of the concentrated HCl added initially to determine moles of HCl used in the experiment
Allow addition of AgNO3 to form AgCl precipitate. Use mass of precipitate to calculate initial moles of HCl added.
1
Subtract this number of moles of HCl from the total moles of acid at equilibrium
1

(c)     M1 ester will evaporate / escape
Allow reactants / products will evaporate
1
M2 incorrect values used (to determine Kc)
Allow the system will no longer be at equilibrium
Do not allow references to equilibrium position shifting alone
1
[6]

 

E13.          Many candidates gave a correct expression for the equilibrium constant, Kp, for the reaction given in part (a). Some, who gave correct indices, lost a mark as they included square brackets, which are specific used to indicate concentration, in their expression. Identifying the one change which would increase the rate of reaction and decrease the value of Kp proved challenging but part marks were awarded for answers which addressed either of the two required changes. Candidates who gave more than one change of condition were able to score one mark for a correct Kp expression. There were many vague answers to part (b)(i), some of which assumed there were other reactants present with the most common example being water in Reaction 2. By contrast, (b)(ii) was very well answered with many candidates scoring full marks. Most candidates now remember to use either kJ mol–1 or J mol–1 throughout these calculations but a small number still failed to do this. Consequential marks were awarded in each part of the calculation.

E14.          The equilibrium parts of this question were generally answered well. In part (a)(i), the correct value of 0.86 was the most common answer but this was the least well-answered section of part (a). Full marks were awarded in parts (ii) and (iii) if the number of moles of hydrogen given in part (i) was used correctly. Even if the numbers were not used correctly, a mark was given in part (iii) for the correct statement that partial pressure equals mole fraction multiplied by the total pressure. In part (b) common errors were to include square brackets in the expression for Kp or to forget that the partial pressure of hydrogen should be squared. The units of Kp expected were kPa–2 but Pa–2 were accepted as long as the value given for Kp was 2.37 × 10–15. Part (c) was poorly answered. Many structures were not esters at all and of those which were, few were methyl esters of C5 acids. F was answered better than E but only the better candidates scored any marks here.

E15.          The first three parts of this question were answered well even by some of the weaker candidates. Credit was given throughout for correct working consequential to errors in a previous part. Many candidates gained full marks in parts (a) and (b). In part (c)(i), most answers were correct except where square brackets were used. These are essential in Kc expressions but should not be used in expressions for Kp. Wrong substitution of the partial pressures values were sadly common but the correct units kPa–1 were usually given. After well-answered numerical sections, it was less common for the answers to both parts of (d) to be correct. Part (d)(i), which involved an increase in the volume of the vessel thereby causing a reduction in the total pressure, discriminated particularly well.

E16.          This question was well answered by many candidates and full marks were common for the most able.  In part (a) a few candidates were penalised for using square brackets, e.g. [SO2] meaning concentration of SO2 instead of using p for partial pressure, and a very few wrote the expression for Kp upside down. Part (b) caused the most problems. Many candidates forgot that, in order to produce 0.75 mol of Cl2 (and 0.75 mol of SO2), 0.75 mol of SO2Cl2 would be used up leaving only 0.25 mol in the equilibrium mixture. Both of the marks in part (b) were lost for this error, but full marks could be gained subsequently in parts (c)(ii) and (d) for correct working in the calculations. In part (e) most were able to deduce and explain that an increase in temperature would cause Kp to increase.  However there were many who thought, incorrectly, that pressure had a similar effect. 

E17.          This question was answered well by most candidates, particularly part (a). The 2:1 ratio or moles of sulphur dioxide to moles of oxygen was sometimes reversed, but marks were allowed in this part and in part (b) for correct use of their incorrect number of moles of oxygen. Most were able to write the expression for Kp correctly in part (c), although some were penalised for the use of square brackets. Full marks were available for correct substitution of wrong partial pressure values from parts (a) and (b). No marks were allowed for the calculation if the expression for Kp was wrong, but a mark was allowed for units correctly consequential on their expression. In part (d), better candidates were able both to predict and explain that T2 was the lower temperature using the information given that the reaction was exothermic.
In part (e) most correctly stated that the yield of SO3 would increase, but only the better candidates remembered that Kp is independent of pressure changes.

E19.          Part (a) was answered well as expected as the topic overlaps with qualitative equilibria in CHM2. Some candidates lost marks by discussing only the effect of increased temperature on the yield and not on the rate also. Several also suggested that the rate was unaffected or decreased by an increase of pressure. Part (b) was answered impressively by the better candidates, who were able to set out the answer clearly. Most were able to score marks for the expression for Kp or the definition of partial pressure (stated or inferred from a calculation). Too many, however, failed to realise that the amounts of the reagents at equilibrium were less than the initial amounts given. Even those who realised that the equilibrium amount of carbon monoxide was 62.8 – 26.2 mol did not always realise that, because of the 1:2 ratio in the equation, the equilibrium amount of hydrogen was 62.8 – (2 × 26.2) mol. However if this was their only error, six marks were still available. The pressure was given in MPa and the answer for Kp was expected with units MPa–2. Many attempted unnecessarily to convert the pressure into kPa or even into Pa and got into difficulties with the powers often.

E20.          Good candidates were able to score high marks for this question. In part (a)(i) some weaker candidates gave an expression for Kc rather than for Kv. In part (a)(ii) a very common error was to give the partial pressure of X as 3.0 MPa, half of Z. The partial pressure of Y was then given as 9 MPa. Such answers were awarded only one out of three marks for recognising that the pressure of Y should be three times that of X. Consequential marking was adopted for part (a)(iii) and answers usually gained both marks though it was surprising to find that a candidate sometimes only squared the partial pressure of Y even though the same candidate had shown the pressure of Y cubed in part (a)(i). In part (b) the extra cost of using a high pressure was more often given than the disadvantage of a lower reaction rate.

E21.The vast majority of students were unaware that propan-1-ol cannot have a concentration and so immediately lost both marks in part (a) due to the chemical error of using a concentration.
Answers to part (b) were often vague and, whilst a lot of students did score the second mark for deducting one value from the other, most did not score the first mark. Common incorrect answers saw students stating that sodium carbonate would only react with the carboxylic acid present, and so measuring the volume of carbon dioxide released, or measuring the pH of the mixture in the hope that they could use that to determine the amount of propanoic acid, despite the presence of concentrated hydrochloric acid.
Answers to part (c) often started well, with students giving valid explanations of how failing to seal the container could result in evaporation. They frequently then incorrectly went on to describe how this would change the actual value of Kc for the equilibrium, rather than just leading to the calculation being incorrect.