chem_sol

1. (i) rate = increase in

= slope of graph;time

volume

initially/to begin with steeper slope / fastest rate / volume of gas/CO2

produced faster/quickly as concentration of HCl highest / OWTTE;

as reaction progresses/with time, less steep slope / volume of gas

production slows / rate decreases due to less frequent collisions

as concentration (of HCl) decreases / OWTTE;

curve flattens/becomes horizontal when HCl used up/consumed

(as there are no more H+ ions to collide with the CaCO3 particles);

Each mark requires explanation. 3 max

(ii)

less steep curve;

same maximum volume at later time;

half/lower H+ /acid concentration less frequent collisions/slower rate;

same amount of HCl, same volume CO2 produced; 4

(iii) mass loss/of CO2 / mass of flask + content;

OR

OR

Do not penalize for missing x-axis label or for missing units on y-axis.

Accept if line meets time axis. 2

IB Questionbank Chemistry 1

(iv) minimum energy (of colliding particles) for a reaction to occur / OWTTE;

lower Ea / greater surface area/contact between CaCO3 and HCl / higher HCl

concentration / (sufficient) particles/molecules have activation energy; 2[11]

2. D[1]

3. C[1]

4. C[1]

5. B[1]

6. (i) by definition *HhO of elements (in their standard states) is zero / no

reaction involved / OWTTE; 1

(ii) *H = –104 – (+20.4);

= –124.4 (kJ mol–1); 2

Award [1 max] for 124.4 (kJ mol–1).

Award [2] for correct final answer.

(iii) *S = 270 – (267 + 131);

= –128 (J K mol–1); 2

Award [1 max] for +128 (J K–1 mol–1).


(iv) *G = *H – T*S = –124.4 –

;1000

)298128( ×−

= –86.3 kJ mol–1; 2

Units needed for the mark.


Allow ECF if only one error in first marking point.


(v) *G = *H – T*S = 0 / *H = T*S;

T =

= 972 K / 699 °C; 21000/128

4.124

−

−

Only penalize incorrect units for T and inconsistent ∆S value

once in (iv) and (v).[9]

7. C[1]

8. (a) (Kc =)

; 12

2

3

]CO][H[

OH]CH[

Do not award mark if incorrect brackets are used or brackets are missing.

(b) (i) amount (of methanol)/product decreases / less methanol;

(forward reaction) exothermic / reverse reaction endothermic / OWTTE; 2

(ii) amount (of methanol)/product increases / more methanol;

3 gas molecules/mol → 1 / decrease in volume / fewer gas

molecules on right hand side/products / more gas molecules

on left hand side/reactants; 2

(c) high pressure expensive / greater cost of operating at high pressure;

lower temperature – lower (reaction) rate; 2

(d) increases rate of forward and reverse reactions (equally) / lowers

activation energy/Ea (of both the forward and reverse reaction

equally) / provides alternative path with lower activation energy/Ea; 1

Accept reactants adsorb onto the catalyst surface and bonds

weaken resulting in a decrease in the activation energy.[8]

9. (a) (i) amount =

= 0.0250 (mol); 161.159

99.3


(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.

(iii) heat change =

× 4.18 × 7.0 / 50.0 × 4.18 × 7.0;1000

0.50

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.

(iv) *H1 =

= –60 (kJ mol–1); 10250.0

5.1−

Value must be negative to award mark.

Accept answers in range –58.0 to –60.0.

Allow –63 (kJ mol–1) if 53.99 g is used in (iii).

(b) (i) (amount of CuSO4•5H2O =

=) 0.0250 (mol);71.249

24.6

(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);

+ 9.20 (kJ); 2

=

× 0250.01000(

9.229

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).

(iii) (*Hx = *H1 – *H2 = –58.4 – (+9.20) =) –67.6 (kJ mol–1); 1

(c) (i)

× 100 = 13.3 %; 10.78

)]6.67(0.78[

−

−−−

If 70.0 kJ mol–1 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]

10. (i) 100 × 4.18 × 35.0;

14630 J / 14600 J / 14.6 kJ;


No ECF here if incorrect mass used. 2

(ii)

= 0.0386 mol;08.46

78.1

= (–)378 kJ mol–1;

0386.0

6.14

Accept (–)377 and (–)379 kJ mol–1.

Award [2] for correct final answer. 2

(iii) heat loss;

incomplete combustion;

heat absorbed by calorimeter not included;

Accept other sensible suggestions. 2 max[6]


11. (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:

;NN &&&&≡

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);

*HO = 2038 – 3217 = –1179 (kJ);


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).

5 vol/mol (g) > 3 vol/mol (g) / more moles/greater amount of gas produced;

*HO (N2H4 / F2) > *HO (N2H4 / O2) (per mole) / (N2H4 / F2) reaction more

exothermic; 2 max[9]

12. A[1]

13. B[1]

14. B[1]


15. C[1]

16. (i) reactants and products in same phase/state;

rate of forward reaction = rate of reverse reaction;

concentrations of reactants and products remain constant /

macroscopic properties remain constant; 2 max

Do not accept concentrations are equal.

(ii) (Kc) =

; 1]I][[H

]HI[

22

2

(iii) no change to position of equilibrium;

no change to value of Kc; 2

(iv) the reaction is exothermic/heat is given out/*H is negative; 1

(v) no effect (on the value of the equilibrium constant);

as it speeds up forward and reverse reaction / concentrations

of reactants and products do not change / position of equilibrium

does not change / no change in yield; 2[8]

17. B[1]

18. D[1]

19. (i) (Kc =) [SO3]2 /[O2] [SO2]

2; 1

(ii) yield (of SO3) decreases;

forward reaction is exothermic / reverse/backwards reaction is endothermic /

equilibrium shifts to absorb (some of) the heat;

Do not accept exothermic reaction or Le Chatelier’s Principle.

Do not allow ECF. 2


(iii) no effect; 1

(iv) no effect;

the rates of both the forward and reverse reactions increase equally; 2[6]

20. (i) exothermic;

Accept either of the following for the second mark.

increasing temperature favours endothermic/reverse reaction;

as yield decreases with increasing temperature; 2 max

(ii) yield increases / equilibrium moves to the right / more ammonia;

increase in pressure favours the reaction which has fewer

moles of gaseous products; 2

(iii) (rate increases because) increase in frequency (of collisions);

increase in energy (of collisions);

more colliding molecules with E ≥ Ea; 2 max

[6]


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