Chemical Reactions and Energy Changes

Classified in Chemistry

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Relative Molecular Mass and Composition

60.The relative molecular mass of aluminium chloride is 267 and its composition by mass is 20.3%....

moles of Na == 0.05;

moles of NaOH = 0.05; same as moles of Na.concentration == 0.20 (mol dm–3)3

Reaction of Sodium with Water

61.Sodium reacts with water as follows.2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)

1.15 g of sodium....

AlCl .empirical formula AlCl3;molecular formula: n = = 2;Al2Cl6

Reaction of Calcium Carbonate with Acids

62.(i)Calcium carbonate is added to separate solutions of hydrochloric acid and ethanoic.....bubbling/effervescence/dissolving of CaCO3/gas given off 

more vigorous reaction with HCl/OWTTE;2

(ii)2HCl(aq) + CaCO3(s)  CaCl2(aq) + CO2(g) + H2O(1); 

(iii)amount of CaCO3.amount of HCl = 2×0.0125 = 0.0250 mol .volume of HCl = 0.0167 dm3/16.7 cm3

(iv)1:1 ratio of CaCO3 to CO2 /use 0.0125 moles CO2
(0.0125×22.4) = 0.28 dm3/280 cm3/2.8×10–4 m3 .pV=nRT

Analysis of an Organic Compound

.63.An organic compound, A, containing only the elements carbon, hydrogen and oxygen was analysed.

(a)% of oxygen = 36.4;    C2H4O;

12, 1 and 16 are used.C2H4O;3

(b)pV = nRT/pV = /correct rearrangement;

..368..Mr = 87.8;


Empirical Formula of an Organic Compound

64.An organic compound A contains 62.0 by mass of carbon, 24.1 by mass..


Correct empirical formula scores 

(ii)the average mass of a molecule;

compared to 1/12 of (the mass of) one atom of 12C/compared to
C-12 taken as 12;



Award [2] for the equation above.


Empirical Formula of an Organic Compound

65.An organic compound A contains 62.0 by mass of carbon, 24.1 by mass of nitrogen, the remainder ....


(ii)the average mass of a molecule;
compared to 1/12 of (the mass of) one atom of 12C/compared to
C-12 taken as 12;



Award [2] for the equation above.


Combustion of Propane

66.Propane and oxygen react according to the following equation.

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g).....

80.0 dm3 H2O;

20.0 dm3 O2;

Effect of Temperature and Volume on Pressure

67.State and explain what would happen to the pressure of a given mass of gas when its absolute temperature and volume are both doubled.

overall there will be no change to the pressure;

double absolute temperature and the pressure doubles;

double volume and the pressure halves;

Empirical and Molecular Formula of Crocetin

68.(i)Crocetin consists of the elements carbon, hydrogen and oxygen. Determine the empirical formula of crocetin, if 1.00 g of crocetin ...

n(C)(= n(CO2) = 2.68 g÷44.01 g mol1) = 0.0609 mol;

n(H)(= 2×n(H2O) = 0.657 g÷18.02 g mol1) = 0.0729 mol;

m(C) = 0.0609 mol×12.01 g mol1 = 0.731 g

and m(H) = 0.0729 mol×1.01 g mol1 = 0.0736 g;

m(O) = (1.00  0.731  0.0736)g = 0.195g;








empirical formula: C5H6O;or C5H6 

(ii)M(crocetin) = 98.5 g÷0.300 mol = 328 (g mol1); molecular formula: C20H24O4;2

Titration of Ammonia

69.A solution containing ammonia requires 25.0 cm3 of 0.100 mol dm–3 hydrochloric acid to reach the equivalence point of a titration.

NH3(aq) + HCl(aq)  NH4Cl(aq);1

(ii)n(HCl) = cV = 0.100 mol dm3×0.0250 dm3 = 0.00250 mol;
n(NH3) = n(HCl) = 0.00250 mol;

(iii)(M (NH3) = 14.01 + 3(1.01) =) 17.04/17.0 (g mol1);
m(NH3) = 0.00250 mol×17.04g mol 1 = 0.0426g/0.0425g;2


70.A toxic gas, A, consists of 53.8% nitrogen and 46.2% carbon by mass. At 273 K and
1.01×105 Pa, 1.048 g of A occupies 462 ...

empirical formula = CN;

Working must be shown to get point.

Mr = 51.9 (g mol–1);

71.An oxide of copper was reduced in a stream of hydrogen as shown below.

After heating, the stream of hydrogen gas was maintained until the apparatus had cooled.

The following results were obtained.

Mass of empty dish = 13.80 g
Mass of dish and contents before heating = 21.75 g
Mass of dish and contents after heating and leaving to cool = 20.15 g

(a)to prevent (re)oxidation of the copper/OWTTE;1

(b)number of moles of oxygen == 0.10;
number of moles of copper == 0.10;
empirical formula = Cu (0.10) : O (0.10) = CuO;3

Alternate solution

= 79.8%  = 20.2%

= 1.25 = 1.29

(c)H2 + CuO → Cu + H2O; 1

Allow ECF.

(d)(black copper oxide) solid turns red/brown;
condensation/water vapour (on sides of test tube);2

Accept change colour.

72.Copper metal may be produced by the reaction of copper(I) oxide and copper(I) sulfide according to the below equation.

2Cu2O + Cu2S  6Cu + SO2

A mixture of 10.0 kg of copper(I) oxide and 5.00 kg of copper(I) sulfide was heated until no further reaction occurred.

(a)Determine the limiting reagent in this reaction, showing your working.

(a)n(Cu2O) = 10.0×103÷143.1 = 69.9 mol;

n(Cu2S) = 5.00×103÷159.16 = 31.4 mol;

Penalise failure to convert kg g once only.

Cu2S is the limiting reagent;3

ECF from above answers.

(b)n(Cu) = 6×n(Cu2S) = 6×31.4 = 188 mol;

m(Cu) = 188×63.55 = 11900  12000 g/11.9  12.0 kg;2

If Cu2O given in (a), allow 3×n(Cu2O) and 3×n(Cu2O)×63.55.

Allow ECF from (a).

73.The reaction below represents the reduction of iron ore to produce iron.

2Fe2O3 + 3C  4Fe + 3CO2

A mixture of 30 kg of Fe2O3 and 5.0 kg of C was heated until no further reaction occurred.

Calculate the maximum mass of iron that can be obtained from these masses of reactants.

n(Fe2O3) = 30×103÷159.7/n(Fe2O3) = 188 mol;

n(C) = 5.0×103÷12.01/n(C) = 416 mol;

Fe2O3 is the limiting reagent or implicit in calculation;

n(Fe) = 2×n(Fe2O3) = 2×188 = 376 mol;

m(Fe) = 376×55.85 = 21 kg;

Accept 2 sig. fig. or 3 sig. fig., otherwise use  1(SF).

Correct final answers score [5].

Allow ECF.

74.0.502 g of an alkali metal sulfate is dissolved in water and excess barium chloride solution, BaCl2(aq) is added to ....

(a)Calculate the amount (in mol) of barium sulfate formed.

(a)M(BaSO4) (= 137.34 + 32.06 + 4(16.00)) = 233.40 (g mol1);

Accept 233.4 but not 233

n(BaSO4)= 0.00288 / 2.88×103(mol);2

ECF from M value

(b)n (alkali metal sulfate) = 0.00288 / 2.88×103(mol);1

(c) 174.31 / 174.3 / 174;

units: g mol1;2

(d)(2(Ar) + 32 + 4(16) = 174, thus) Ar = 39 / Ar ==39;



(e)K2SO4(aq) + BaCl2(aq)  BaSO4(s) + 2KCl(aq)2

A alkali metal arrived at in (d)

Accept net ionic equation

75.0.600 mol of aluminium hydroxide is mixed with 0.600 mol of sulfuric acid, and the following reaction occurs:

2Al(OH)3(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 6H2O(l)

(a)0.600 mol Al(OH)3 ≡ (1.5)(0.600) mol H2SO4/0.900 mol H2SO4
needed, but only 0.600 mol used;    H2SO4 limiting reactant;2

(b)0.200 mol Al2(SO4)3;   68.4(g);2

(c)0.200 mol;1

(d)A Brønsted-Lowry acid is a proton/H+ donor;
A Lewis base is an electron-pair donor;2

(e)H2CO3 and carbonic acid/CH3COOH and ethanoic acid;1

37.The diagram shows the distribution of energy for the molecules in a sample of gas at a given temperature, T1.

(a)In the diagram Ea represents the activation energy for a reaction. Define this term.

activation energy = minimum energy required for a reaction to occur;1

(b)curve moved to the right;
peak lower,2

Deduct [1] if shaded area smaller at T2 or if T2 line touches the x-axis

(c)rate increased;
as more molecules with energy  Ea;2

38.(a)Define the term average bond enthalpy, illustrating your answer with an equation for methane, CH4.

(b)The equation for the reaction between methane and chlorine is

(a)energy for the conversion of a gaseous molecule into (gaseous) atoms;
(average values) obtained from a number of similar bonds/compounds/OWTTE;
CH4(g) → C(g) + 4H(g);3

State symbols needed.

(b)(bond breaking) = 1890/654;
(bond formation) = 2005/769;
enthalpy = –115(kJ mol–1)3

Award [3] for correct final answer.

Penalize [1] for correct answer with wrong sign.

(c)molecules have insufficient energy to react (at room temperature)/
wrong collision geometry/unsuccessful collisions;
extra energy needed to overcome the activation energy/Ea for the reaction;2

(d)Draw an enthalpy level diagram for this reaction.

exothermic shown;

activation energy/Ea shown;2

39.In aqueous solution, potassium hydroxide and hydrochloric acid react as follows.

KOH(aq) + HCl(aq) → KCl(aq)+ H2O(l)

The data below is from an experiment to determine the enthalpy change of this reaction.

50.0 cm3 of a 0.500 mol dm–3 solution of KOH was mixed rapidly in a glass beaker with 50.0 cm3 of a 0.500 mol dm–3 solution of HCl.

Initial temperature of each solution = 19.6°C
Final temperature of the mixture = 23.1°C

(a)exothermic because temperature rises/heat is released;1

(b)to make any heat loss as small as possible/so that all the heat will be
given out very quickly;1

Do not accept “to produce a faster reaction”.

(c)heat released = mass×specific heat capacity×temp increase/q = mcT =/
= 1463 J/1.463 kJ; (allow 1.47 kJ if specific heat = 4.2)
amount of KOH/HCl used = 0.500×0.050 = 0.025 mol;
H = (1.463÷0.025) = –58.5 (kJ mol–1); (minus sign needed for mark)4

Use ECF for values of q and amount used.

Award [4] for correct final answer.

Final answer of 58.5 or +58.5 scores [3].

Accept 2,3 or 4 significant figures.

(d)heat loss (to the surroundings);
insulate the reaction vessel/use a lid/draw a temperature versus time graph;2

(e)3.5°C/temperature change would be the same;
amount of base reacted would be the same/excess acid would not react/
KOH is the limiting reagent;2

40.The data below is from an experiment used to measure the enthalpy change for the combustion of 1 mole of sucrose (common table sugar), C12H22O11(s). The time-temperature data was taken from a data-logging software programme.

Mass of sample of sucrose, m = 0.4385 g

Heat capacity of the system, Csystem = 10.114 kJ K–1

(a)Calculate ΔT, for the water, surrounding the chamber in the calorimeter.

(a)ΔT = 23.70 – 23.03 = 0.67 (°C/K);1

(b) = 1.281×10–3;1

(c)(i)ΔHc = (C ΔT)/n == –5.3×103 kJ mol–1;1

Use ECF for values of T and n.

(ii)Percentage experimental error == 5.4%;1

Use ECF for values of ΔHc.

(d)enthalpy change of combustion of sucrose > TNT, and therefore not important;
rate of reaction for TNT is greater than that of sucrose, so this is valid;
amount of gas generated (in mol) for sucrose > than that of TNT
(according to the given equation), so this is not important

41.(a)Define the term average bond enthalpy.

(b)Use the information from Table 10 of the Data Booklet to calculate the enthalpy change for the complete combustion of but-1-ene, according to the following equation.

C4H8(g) + 6O2(g) → 4CO2(g) + 4H2O(g)

(a)The amount of energy needed to break 1 mole of (covalent) bonds;
in the gaseous state;
average calculated from a range of compounds;2 max

Award [1] each for any two points above.

(b)Bonds broken
(612) + (2×348) + (8×412) + (6×496)/7580 (kJ mol–1);
Bonds made
(8×743) + (8×463) / 9648 (kJ mol–1);
H = –2068 (kJ mol–1);3

Award [3] for the correct answer.
Allow full ECF.
Allow kJ but no other incorrect units.
Even if the first two marks are lost, the candidate can score [1] for a clear correct subtraction for H.

42.Given the following data:

C(s) + F2(g) → CF4(g); ∆H1= –680 kJ mol–1
F2(g) → 2F(g); ∆H2= +158 kJ mol–1
C(s) → C(g); ∆H3= +715 kJ mol–1

calculate the average bond enthalpy (in kJ mol–1) for the C––F bond.

C(s) + 2F2(g) → CF4(g)∆H1 = –680 kJ;
4F(g) → 2F2(g)∆H2 = 2(–158) kJ;
C(g) → C(s)∆H3 = –715 kJ;

Accept reverse equations with +∆H values.

C(g) + 4F(g) → CF4(g)∆H = –1711 kJ,

so average bond enthalpy =

= –428 kJ mol–1;4

Accept + or – sign.

Lots of ways to do this! The correct answer is very different from the value in the Data Booklet, so award [4] for final answer with/without sign units not needed, but deduct [1] if incorrect units. Accept answer in range of 427 to 428 without penalty for sig figs.

43.Two reactions occurring in the manufacture of sulfuric acid are shown below:

reaction IS(s) +O2(g)  SO2(g)HӨ = –297 kJ

reaction IISO2(g) + O2(g)SO3(g)HӨ = –92 kJ

(i)State the name of the term HӨ. State, with a reason, whether reaction I would be accompanied by a decrease or increase in temperature.

(a)(i)standard enthalpy (change) of reaction;

(temperature) increase;

reaction is exothermic/sign of H is negative;3

(ii)more (negative);

heat given out when gas changes to solid/solid has less enthalpy than

(iii)–389 kJ;1

44.(i)Define the term average bond enthalpy.


(ii)Explain why Br2 is not suitable as an example to illustrate the term average bond enthalpy.

(i)the energy needed to break one bond;
(in a molecule in the) gaseous state;
value averaged using those from similar compounds;3

(ii)it is an element/no other species with just a Br-Br bond/OWTTE;1

(iii)(sum bonds broken =) 412 + 193 = 605;

(sum bonds formed =) 276 + 366 = 642;

(H =) –37 kJ;3

Award [3] for correct final answer.

Award [2] for “+ 37”.

Accept answer based on breaking and making extra C-H bonds.

(iv)Sketch an enthalpy level diagram for the reaction in part (iii).


CH4 + Br2 

CH3Br + HBr


Award [1] for enthalpy label and two horizontal lines, [1] for reactants higher than products.

ECF from sign in (iii), ignore any higher energy level involving atoms.

(v)(about) the same/similar;

same (number and type of) bonds being broken and formed;2

45.But–1–ene gas, burns in oxygen to produce carbon dioxide and water vapour according to the following equation.

C4H8 + 6O2  4CO2 + 4H2O

(a)Use the data below to calculate the value of HӨ for the combustion of but-1-ene.








Average bond
enthalpy / kJ mol–1







(a)(Amount of energy required to break bonds of reactants)

8×412 + 2×348 + 612 + 6×496/7580 (kJ mol1);

(Amount of energy released during bond formation)

4×2×743 + 4×2×463/9648 (kJ mol1);

H = 2068 (kJ or kJ mol1);3

ECF from above answers.

Correct answer scores [3].

Award [2] for (+)2068.

If any other units apply 1(U), but only once per paper.

(b)exothermic andHӨ is negative/energy is released;

46.Calculate the enthalpy change, H4 for the reaction

C + 2H2 + O2  CH3OHH4

using Hess’s Law and the following information.

CH3OH + O2  CO2 + 2H2OH1 = 676 kJ mol1

C + O2  CO2H2 = 394 kJ mol1

H2 + O2  H2OH3 = 242 kJ mol1



2×H3/– 484;

H4 = 202 (kJ mol1 );4

47.Methylamine can be manufactured by the following reaction.

CH3OH(g) + NH3(g)  CH3NH2(g) + H2O(g)

(a)energy needed to break (1 mol of) a bond in a gaseous molecule;

averaged over similar compounds;2

(b)bonds broken identified as CO and NH;

bonds formed identified as CN and OH;

H = 748  768 (kJ);

=  20 kJ/kJ mol1

48.(a)Define the term average bond enthalpy.

(b)Use the information from Table 10 in the Data Booklet to calculate the enthalpy change for the complete combustion of but-1-ene according to the following equation

C4H8(g)  4CO2(g) + 4H2O(g)

(a)amount of energy needed to break one mole of (covalent) bonds;

in the gaseous state;

average calculated from a range of compounds;2

Award [1] each for any two points above.

(b)bonds broken: 161 + 2×348 + 8×412 + 6×496/7580 kJ mol1;

bonds made: 8×743 + 8×463/9648 kJ mol1;

(bonds broken  bonds made =) H = 2068(kJ mol1);3

Award [3] for the correct answer.

Allow full ECF  1 mistake equals 1 penalty.

Allow kJ but not other wrong units.

(c)same/equal, because the same bonds are being broken and formed;1

(d)products more stable than reactants;

bonds are stronger in products than reactants/HP R/enthalpy/stored
energy of products less than reactants;

49.The reaction between ethene and hydrogen gas is exothermic.

(i)Write an equation for this reaction.

(a)(i)C2H4(g) + H2(g)  C2H6(g);1

State symbols not required for mark

(ii)products more stable than reactants/reactants less stable than products;

products lower in energy/reactants higher in energy;2

(iii)(overall) bonds in reactants weaker/(overall) bonds in product stronger
/all bonds in product are  bonds/weaker  bond broken and a
(stronger)  bond formed;

less energy needed to break weaker bonds/more energy produced
to make stronger bonds (thus reaction is exothermic)/OWTTE;


bond breaking is endothermic/requires energy and bond making is
exothermic/releases energy;

stronger bonds in product mean process is exothermic overall

50.(i)Define the term average bond enthalpy.


(ii)The equation for the reaction of ethyne and hydrogen is:

C2H2(g) + 2H2(g)  C2H6(g)

Use information from Table 10 of the Data Booklet to calculate the change in enthalpy for the reaction.

(i)energy required to break (a mole of) bonds in the gaseous state
/energy given out when (a mole of) bonds are made in the
gaseous state;
average value from a number of similar compounds;2

(ii)(HӨreaction = (∑BEbreak  BEmake))

= [(837) + 2(436)]  [(348 + 4(412)];

=  287(kJ/kJ mol1);2

Award [1 max] for 287 or + 287.

(iii)(BE): CCl > CBr > CI/CX bond becomes weaker;

halogen size/radius increases/bonding electrons further away from
the nucleus/bonds become longer

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