TOPIC 5 ANSWERS & MARK SCHEMES QUESTIONSHEET 1 AS Level ENTHALPY OF DISPLACEMENT a) (i) 1 × 20 × 10 -3 = 0.020 (1) (ii) 1.95/65 = 0.030 (1) b) 0.02(1) Zn in excess/equivalent mol in reaction(1) c) (i) to ensure heat equilibrium/balance/steady temperature(1) (ii) regression to point of addition(1) (iii) to predict the value without heat loss(1) d) (i) 20 × 4.2 × 39.5 (1) = 3318 J(1) (ii) 3318/0.02 (1) = 165900 J(1) = -165.9 (1) kJ mol -1 (1) e) more exothermic/greater (1) zinc more electropositive/reactive/higher in Reactivity Series (1) larger release of energy (1)
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= -5576/0.1 = -55,760 J mol-1 or –55.76 kJ mol-1 (1)Maximum 2 marks if negative sign omitted* mark consequentially from graph.
b) Quantity of energy produced by the coil = 30 × 145 = 4350 J (1)Amount of water formed = 40 × 2/1000 = 0.08 mol (1)(Note: NaOH is in excess ∴ calculation based on acid)∴ the quantity of energy produced by the formation of 1 mole of water = 4350/0.08 = 54,375 J mol-1 (1)∴ ∆H
neut = -54,375 J mol-1 or –54.4 kJ mol-1 (1)
= -5576 J per 0.1 mol of H2O formed (1)
TOPIC 5 ANSWERS & MARK SCHEMES
QUESTIONSHEET 3
AS Level
ENTHALPY OF FORMATION
a) The heat change (not ‘energy change’(-1) but allow ‘enthalpy change’)when 1 mole (1)of a compound is formed from its elements (1)under standard conditions / 1 atm (105 Pa) and 298 K (1)
a) The heat evolved (accept ‘enthalpy change’, but not ‘energy change’ or ‘heat absorbed or evolved’ (-1))when 1 mole (1)of a substance undergoes complete combustion / is burned in excess oxygen (1)under standard conditions / 1 atm (105 Pa) and 298 K (1)
b) For 2C → 2CO2, ∆Hθ = -222 + (-566) = -788 kJ mol-1 (1)
Reason ∆H depends only on the initial and final states of the system / is independent of the route taken (1)
TOPIC 5 ANSWERS & MARK SCHEMES
QUESTIONSHEET 5
AS Level
BOND DISSOCIATION ENTHALPY
a) The energy required (½)to break the covalent bonds in 1 mole (½)of gaseous hydrogen molecules (½)to give gaseous hydrogen atoms at 298 K (½)
b) Bond dissociation enthalpy is the energy required to break one O—H bond (1)in each molecule in 1 mole of gaseous water molecules (1)whereas mean bond dissociation enthalpy is the average energy required to break all the O—H bonds (1)
c) (i) C—C < C=C < C≡C (1)
(ii) C—I < C—Br < C—Cl (1)
d) Bond dissociation enthalpy is directly related to bond strength (1)but inversely related to bond length (1)
a) The enthalpy change (not ‘energy change’) (1)when one chemical system is converted into another (1)is independent of the route taken (1)but depends only on the initial and final states of the system (1)
b) (i) Energy can be neither created nor destroyed (1)but can be converted from one form to another (1)
(ii) Consider a conversion and reconversion by two routes (1)If Hess’s law were obeyed, conversion by route 1 and reconversion by route 2 would give overall ∆H of 0,consistent with the first law (1)If Hess’s law were not obeyed, conversion by route 1 and reconversion by route 2 would give overall ∆H > 0 or< 0, which would violate the first law (1)
c) (i) C3H
8(g) + 5O
2(g) → 3CO
2(g) + 4H
2O(l) (2)
Award (1) for balance and (1) for state symbols.
(ii)
(2 for diagram, 2 for start & finish labels, 2 for route labels)
(3) For either scheme. Deduct 1 mark for each error
Finish
(-196.2)/2 = - 98.1 kJ mol-1
Start
-297.2 kJ mol-1
(2)
(2)
OrS(monoclinic) + 1½O2(g) Start
Finish
SO2(g) + 1½O
2(g)
SO3(g)
(-196.2)/2= -98.1 kJ mol-1
-297.2 kJ mol-1
∆Hêf(SO
3)
2H2(g) + 3S(monoclinic) + O
2(g)
2H2S(g) + S(monoclinic) + O
2(g)
2H2S(g) + SO
2(g)
2H2O(g) + 3S(monoclinic)
2H2O(l) + 3S(monoclinic)
Start
Finish
Finish
2(-20.2) = -40.4 kJ mol-1
- 297.2 kJ mol-1
∆Hêreaction
2(+41.1)
= +82.2 kJ mol-1
2(-286)= -575 kJ mol-1
Start
Finish
S(rhombic) + O2(g)
SO2(g)
-296.9 kJ mol-1
-297.2kJ mol-1
∆Hconversion
(2)
TOPIC 5 ANSWERS & MARK SCHEMES
QUESTIONSHEET 8
AS Level
HESS’S LAW WITH CALORIMETRY
a) Graph:axes appropriately labelled (½ each)suitable scales to match size of graph paper - at least 2/
3 of each side used (½ each)
points correctly plotted (1 but -½ for each error)two extended straight lines drawn (½ each)Maximum temp. change (taken as the vertical distance at 3½ min) between the two straight lines = 25.0 ± 0.2 0C (1)
b) q = 50.24 × 4.2 × (-25)* *Or value from graph= -5275 J (1)