AS AQA Chemistry Paper 1 2017 Exam
Questions With New Update Solutions
/. Write the full electron configuration for Cl - and Fe2+ - Answer-Cl - = 1s2 2s2 2p6 3s2
3p6
Fe2+ = 1s2 2s2 2p6 3s2 3p6 3d6
/.Write an equation, including state symbols, to represent the process that occurs when
the third ionisation energy of manganese is measured. - Answer-Mn2+ (g) → Mn3+ (g)
+ e−
/.State which of the elements magnesium and aluminium has the lower first ionisation
energy. Explain your answer. - Answer-Al
Outer electron in (3)p orbital
Is higher in energy
Due to being further from the nucleus so easier to remove
/.Write an equation, including state symbols, for the reaction with an enthalpy change
equal to the enthalpy of formation for iron(III) oxide. - Answer-2Fe(s) + 3 /2 O2(g) →
Fe2O3(s)
/.CO(g) ΔfH o / kJ mol−1 = −111
Fe2O3(s) ΔfH o / kJ mol−1 = −822
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔH = −19 kJ mol−1
Use these data and the equation for the reaction of iron(III) oxide with carbon monoxide
to calculate a value for the standard enthalpy of formation for carbon dioxide. - Answer-
(3x∆fHCO2) = −19 + (−822) + 3(−111) - 0
(3x∆fHCO2) = −1174
∆fHCO2 = −391 kJmol-1
/.N2(g) + 3H2(g) → 2NH3(g) ΔH / kJ mol−1 = −92
N2(g) → 2N(g) ΔH / kJ mol−1 = +944
H2(g) → 2H(g) ΔH / kJ mol−1 = +436
Use the data from TABLE 3 to calculate the bond enthalpy for N-H in ammonia. -
Answer-(6(N-H)) = 944 + 3(+436) + 92
(6(N-H)) = 2344
N-H = (+)391kJmol−1
/.Give one reason why the bond enthalpy that you calculated in Question 2.3 is different
from the mean bond enthalpy quoted in a data book (388 kJ mol−1). - Answer-Data
, book value derived from (a number of) different compounds (not just different NH3
molecules)
/.Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s) The student used this method: • A measuring
cylinder was used to transfer 50 cm3 of a 1.00 mol dm−3 aqueous solution of zinc
sulfate into a glass beaker.
• A thermometer was placed in the beaker.
• 2.08g of magnesium metal powder were added to the beaker.
• The mixture was stirred and the maximum temperature recorded. The student
recorded a starting temperature of 23.9 o C and maximum temperature of 61.2 o C.
Show by calculation which reactant was in excess. Use the data to calculate the
experimental value for enthalpy of reaction in kJ mol−1
Assume that the specific heat capacity of the solution is 4.18 J K−1g−1and the density
of the solution is 1.00 g cm−3. - Answer-Amount ZnSO4 = 0.050 mol
Amount Mg= 0.0856 mol (Hence Mg in excess)
Q=mc∆T
Q=50.0 x 4.18 x 37.3
Q=7795.7J
(Energy released per mole) = 7.796/0.05kJmol−1
∆H= − 156 kJmol−1
/.Another student used the same method and obtained a value for the enthalpy of
reaction of −142 kJ mol−1 A data book value for the enthalpy of reaction is −310 kJ
mol−1
Suggest the most likely reason for the large difference between the student's
experimental value and the data book value. - Answer-Heat loss (from the apparatus
would mean the experimental value is smaller than the data source)
/.Suggest how the students' method, and the analysis of the results, could be improved
in order to determine a more accurate value for the enthalpy of reaction. - Answer-Stage
1
Improved insulation
Insulate the beaker or use a polystyrene cup or a lid
To reduce heat loss
Stage 2 Improved temperature recording
Record the temperature for a suitable time before adding the metal
To establish an accurate initial temperature OR 2c Record temperature values at
regular time intervals 2d To plot the temperature results against time on a graph
Stage 3 Improved analysis of results
Extrapolate the cooling back to the point of addition
To establish a (theoretical) maximum temperature
Questions With New Update Solutions
/. Write the full electron configuration for Cl - and Fe2+ - Answer-Cl - = 1s2 2s2 2p6 3s2
3p6
Fe2+ = 1s2 2s2 2p6 3s2 3p6 3d6
/.Write an equation, including state symbols, to represent the process that occurs when
the third ionisation energy of manganese is measured. - Answer-Mn2+ (g) → Mn3+ (g)
+ e−
/.State which of the elements magnesium and aluminium has the lower first ionisation
energy. Explain your answer. - Answer-Al
Outer electron in (3)p orbital
Is higher in energy
Due to being further from the nucleus so easier to remove
/.Write an equation, including state symbols, for the reaction with an enthalpy change
equal to the enthalpy of formation for iron(III) oxide. - Answer-2Fe(s) + 3 /2 O2(g) →
Fe2O3(s)
/.CO(g) ΔfH o / kJ mol−1 = −111
Fe2O3(s) ΔfH o / kJ mol−1 = −822
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔH = −19 kJ mol−1
Use these data and the equation for the reaction of iron(III) oxide with carbon monoxide
to calculate a value for the standard enthalpy of formation for carbon dioxide. - Answer-
(3x∆fHCO2) = −19 + (−822) + 3(−111) - 0
(3x∆fHCO2) = −1174
∆fHCO2 = −391 kJmol-1
/.N2(g) + 3H2(g) → 2NH3(g) ΔH / kJ mol−1 = −92
N2(g) → 2N(g) ΔH / kJ mol−1 = +944
H2(g) → 2H(g) ΔH / kJ mol−1 = +436
Use the data from TABLE 3 to calculate the bond enthalpy for N-H in ammonia. -
Answer-(6(N-H)) = 944 + 3(+436) + 92
(6(N-H)) = 2344
N-H = (+)391kJmol−1
/.Give one reason why the bond enthalpy that you calculated in Question 2.3 is different
from the mean bond enthalpy quoted in a data book (388 kJ mol−1). - Answer-Data
, book value derived from (a number of) different compounds (not just different NH3
molecules)
/.Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s) The student used this method: • A measuring
cylinder was used to transfer 50 cm3 of a 1.00 mol dm−3 aqueous solution of zinc
sulfate into a glass beaker.
• A thermometer was placed in the beaker.
• 2.08g of magnesium metal powder were added to the beaker.
• The mixture was stirred and the maximum temperature recorded. The student
recorded a starting temperature of 23.9 o C and maximum temperature of 61.2 o C.
Show by calculation which reactant was in excess. Use the data to calculate the
experimental value for enthalpy of reaction in kJ mol−1
Assume that the specific heat capacity of the solution is 4.18 J K−1g−1and the density
of the solution is 1.00 g cm−3. - Answer-Amount ZnSO4 = 0.050 mol
Amount Mg= 0.0856 mol (Hence Mg in excess)
Q=mc∆T
Q=50.0 x 4.18 x 37.3
Q=7795.7J
(Energy released per mole) = 7.796/0.05kJmol−1
∆H= − 156 kJmol−1
/.Another student used the same method and obtained a value for the enthalpy of
reaction of −142 kJ mol−1 A data book value for the enthalpy of reaction is −310 kJ
mol−1
Suggest the most likely reason for the large difference between the student's
experimental value and the data book value. - Answer-Heat loss (from the apparatus
would mean the experimental value is smaller than the data source)
/.Suggest how the students' method, and the analysis of the results, could be improved
in order to determine a more accurate value for the enthalpy of reaction. - Answer-Stage
1
Improved insulation
Insulate the beaker or use a polystyrene cup or a lid
To reduce heat loss
Stage 2 Improved temperature recording
Record the temperature for a suitable time before adding the metal
To establish an accurate initial temperature OR 2c Record temperature values at
regular time intervals 2d To plot the temperature results against time on a graph
Stage 3 Improved analysis of results
Extrapolate the cooling back to the point of addition
To establish a (theoretical) maximum temperature
