1.
(a) amount of energy required to break bonds of reactants
–1 –1
3 × 413 + 358 + 464 + 1.5 × 498 (kJ mol ) / 2808 (kJ mol );
amount of energy released during bond formation of products
–1 –1
4 × 464 + 2 × 746 (kJ mol ) / 3348 (kJ mol );
–1
∆H = –540 (kJ mol ); 3
Award [3] for correct final answer.
Award [2] for (+)540.
–1
If old Data Booklet is used accept answer: –535 (kJ mol ) or
award [2] for (+)535.
(b) (i) m(methanol) = (80.557 – 80.034) = 0.523 (g);
0.523 g
32.05 g mol 1
n(methanol) = = 0.0163 (mol); 2
Award [2] for correct final answer.
(ii) ∆T = (26.4 – 21.5) = 4.9 (K);
–3
q = (mc∆T =) 20.000 × 4.18 × 4.9 (J) / 20.000 × 4.18 × 4.9 × 10 (kJ);
0.41 (kJ); 3
Award [3] for correct final answer.
0.41 (kJ)
(iii)
O
∆Hc = 0.0163 (mol) / –25153 (J mol–1);
–1
= –25 (kJ mol ); 2
Award [2] for correct final answer.
–1
Award [1] for (+)25 (kJ mol ).
(c) (i) bond enthalpies are average values/differ (slightly) from
one compound to another (depending on the neighbouring
atoms) / methanol is liquid not gas in the reaction; 1
(ii) not all heat produced transferred to water / heat lost to
surroundings/environment / OWTTE / incomplete combustion
(of methanol) / water forms as H2O(l) instead of H2O(g) ; 1
Do not allow just “heat lost”.
[12]
3.99
2. (a) (i) amount = 159.61 = 0.0250 (mol); 1
IB Questionbank Chemistry 1
(ii) 26.1 (°C);
Accept answers between 26.0 and 26.2 (°C).
temperature rise = 26.1 – 19.1 = 7.0 (°C); 2
Accept answers between 6.9 °C and (7.1 °C).
Award [2] for the correct final answer.
No ECF if both initial and final temperatures incorrect.
50.0
(iii) heat change = 1000 × 4.18 × 7.0 / 50.0 × 4.18 × 7.0;
Accept 53.99 instead of 50.0 for mass.
= 1.5 (kJ); 2
Allow 1.6 (kJ) if mass of 53.99 is used.
Ignore sign.
1.5
–1
(iv) ∆H1 = 0.0250 = –60 (kJ mol ); 1
Value must be negative to award mark.
Accept answers in range –58.0 to –60.0.
–1
Allow –63 (kJ mol ) if 53.99 g is used in (iii).
6.24
(b) (i) (amount of CuSO4•5H2O = 249 .71 =) 0.0250 (mol);
(amount of H2O in 0.0250 mol of CuSO4•5H2O
= 5 × 0.0250 =) 0.125 (mol). 2
(ii) (50.0 × 4.18 × 1.10 =) 230 (J);
IB Questionbank Chemistry 2
229.9
(1000 0.0250 + 9.20 (kJ); 2
Accept mass of 47.75 or 53.99 instead of 50.00 giving answers
of +8.78 or +9.9.
Do not penalize missing + sign but penalize – sign unless charge
already penalized in (a) (iv).
–1
(iii) (∆Hx = ∆H1 – ∆H2 = –58.4 – (+9.20) =) –67.6 (kJ mol ); 1
[78.0 (67.6)]
(c) (i) 78.0 × 100 = 13.3 %; 1
–1
If 70.0 kJ mol is used accept 10.3 %.
(ii) the anhydrous copper(II) sulfate had already absorbed
some water from the air / OWTTE;
the value would be less exothermic/less negative than
expected as the temperature increase would be lower /
less heat will be evolved when the anhydrous salt is
dissolved in water / OWTTE; 2
Do not accept less without a reason.
[14]
3. (a) N2H4(g) + 2F2(g) → N2(g) + 4HF(g)
Award [1] for reactants and products.
Award [1] if this equation is correctly balanced.
Ignore state symbols. 2
(b) Hydrazine:
Nitrogen:
N
N
;
Accept lines, dots and crosses to show electron pairs.
Penalize missing lone pairs once only. 2
(c) ΣBE (bonds broken) = (4 × 391) + 158 + 2(158) / 2038(kJ);
ΣBE (bonds formed) = (945) + 4(568) / 3217 (kJ);
O
∆H = 2038 – 3217 = –1179 (kJ);
Award [3] for correct final answer.
Award [2] for (+)1179 (kJ). 3
(d) (N2H4 / F2) better rocket fuel;
ECF: answer must be consistent with equation in (a) and ∆H in (c).
IB Questionbank Chemistry 3
5 vol/mol (g) > 3 vol/mol (g) / more moles/greater amount of gas produced;
O O
∆H (N2H4 / F2) > ∆H (N2H4 / O2) (per mole) / (N2H4 / F2) reaction more
exothermic; 2 max
[9]
O O O
4. (a) ∆H reaction = Σ∆Hf (products) – Σ∆Hf (reactants)
= [(1)(–85) + (2)(–242)] – [(2)(–201)];
–1
= –167 (kJ/kJ mol );
Award [1] for (+) 167. 2
O O O
(b) ∆S reaction = ΣS (products) – ΣS (reactants)
= [(1)(230) + (2)(189)] – [(2)(238) + (1)(131)];
–1 –1 –1
= 1 (J K /J K mol ); 2
O O O
(c) ∆G reaction = (∆H – T∆S ) = (–167) – (298)(0.001);
Award [1] for correct substitution of values.
= –167 kJ/–167000 J;
Units needed for mark in (c) only.
–1 –1
Accept –167 kJ mol or –167000 J mol .
spontaneous;
Award marks for final correct answers throughout in each of (a), (b) and (c). 3
[7]
5. (i) lattice enthalpy for a particular ionic compound is defined as ΔH for the
+ –
process, MX(s) → M (g) + X (g);
Accept definition for exothermic process
electron affinity is the energy change that occurs when an electron is added
to a gaseous atom or ion; 2
IB Questionbank Chemistry 4
(ii)
H f = – 4 11 k J m o l – 1
N a(s) + 1
2 C l 2 (g ) N aC l(s)
+ 1 0 8 k J m o l –1 + 1 2 1 k J m o l –1
N a(g ) C l(g )
+ 4 9 4 k J m o l –1 – 3 6 4 k J m o l –1
–
N a + (g ) + C l (g )
lattice enthalpy = –[(–411) – (+108) – (+494) – (+121) – (–364)]
–1
= 770 (kJ mol )
Award [2] for all correct formulas in correct positions on cycle
diagram.
1 incorrect or missing label award [1].
Award [1] for all correct values in correct positions on cycle
diagram.
–1
calculation of lattice enthalpy of NaCl(s) = 770 (kJ mol ); 4
Allow ECF.
Accept alternative method e.g. energy level diagram.
(iii) lattice/network/regular structure;
each chloride ion is surrounded by six sodium ions and each sodium ion is
surrounded by six chloride ions/6:6 coordination; 2
[8]
IB Questionbank Chemistry 5
