enthalpy of displacement

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)


QUESTIONSHEET 2

AS Level

THERMOMETRIC TITRATION

a) (i) Axes: sensible choice of scales (1)labelling (1)

Points correctly plotted (1)Two intersecting straight lines of best fit (1)Volume of acid = 22.5 ± 0.5 cm3 (1)

(ii) n (NaOH) = 2.0 × 50/1000 = 0.1 mol (1)∴ n (HCl) = 0.1 mol (1)∴ concentration of HCl = 0.1/0.0225* = 4.44 M (1)* mark consequentially from graph

(iii) q = m.θ.∆Τ= (50 + 22.5*) × 4.18 × (17.6 – 36.0*) (1)

∴ ∆Hneut

= -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)


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)

b) ∆Hθreaction

= ∑ ∆Hθf (products) - ∑ ∆Hθ

f (reactants) (1)

= [(+90 × 4) + (-242 × 6)] - [-46 × 4] (1)= -1092 + 184 = -908 kJ mol-1 (1)

(Or accept a Hess’s law solution)

c) ∆Hθreaction

= [(-394 × 4) + (-286 × 2)] - [(+228 × 2)] (1)= -2148 – 456 = -2604 kJ mol-1 (1)

∴ ∆Hθc (C

2H

2) = -2604/2 = -1302 kJ mol-1 (1)


d) ∆Hθf(MgO) = -602 kJ mol-1; ∆Hθ

f(CO

2) = -394 kJ mol-1 (1) (Stated or implied)

For MgCO3(s) → MgO(s) + CO

2(g)

∆Hθreaction

= +100 = [(-602) + (-394)] - ∆Hθf (MgCO

3) (1)

∴ ∆Hθf(MgCO

3) = -602 – 394 – 100 = - 1096 kJ mol-1 (1)



QUESTIONSHEET 4

AS Level

ENTHALPY OF COMBUSTION

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)

∴ for 1C → 1CO2, ∆Hθ = -788/2 = -394 kJ mol-1 (1)

c) ∆Hθreaction

= ∑∆Hθc (reactants) - ∑∆Hθ

c (products) (1)

= [-6778] - [(-5470) + (-1411)] (1)= -6778 – (-6881) = + 103 kJ mol-1 (1)


d) (i) ∆Hθreaction

= [(-2718) + (-286)] - [-2877] (1)= - 3004 –(-2877) = -127 kJ mol-1(1)

(ii) Expected change Nil (1)

Reason ∆H depends only on the initial and final states of the system / is independent of the route taken (1)


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)

e) ∆H for bond breaking

6 C—H = 6 × 414 = + 2484 (½) 6 C—H = + 2484 (½)2 C—C = 2 × 346 = + 692 (½) 2 C—C = + 692 (½)1 C=O = 804 = + 804 (½)4 O=O = 4 × 498 = + 1992 (½) 4 O=O = +1992 (1)

Total + 5972 kJ mol-1 (1) + 5168 kJ mol-1 (1)

∆H for bond making

6 C=O = 6 × (-804) = -4824 (½) 5 C=O = -4020 (½)6 H—O = 6 × (-463) = -2778 (½) 6 H—O = -2778 (½)

Total -7602 kJ mol-1 (1) -6798 kJ mol-1 (1)

∴ ∆Hc

= (+5972 – 7602) (1) ∴ ∆Hc

= (+5168 – 6798) (1)= -1630 kJ mol-1 (1) = -1630 kJ mol-1 (1)

Or


QUESTIONSHEET 6

AS Level

HESS’S LAW AND ENTHALPY DIAGRAMS

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)

(iii) By Hess’s law, ∆H(Route 1) = ∆H(Route 2) (1)∴ -1182 + (-1144) = ∆Hθ

f (C

3H

8) + (-2222) (1)

∴ ∆Hθf (C

3H

8) = -1182 + (-1144) + 2222

= -104 kJ mol-1 (1)

3C(s) + 4H2(g) + 5O2(g)

3CO2(g) + 4H2(g) + 2O2(g)

3CO2(g) + 4H2O(l)

3C3H8(g) + 5O2(g)3(-394)= -1182 kJ mol-1

∆Hθf (C3H8)

4(-286)= -1144 kJ mol-1

-2222 kJ mol-1

Finish (1)

Start (1)

Route 2 (1)

Route 1 (1)

↑Enthalpy

0


QUESTIONSHEET 7

AS Level

HESS’S LAW CALCULATIONS

a) (i)

By Hess’s law, ∆Hconversion

+ (-297.2) = -296.9 (1)∴ ∆H

conversion = -296.9 + 297.2 = +0.3 kJ mol-1 (1)

(ii) More stable form Rhombic (1)Reason The conversion of rhombic to monoclinic sulfur is endothermic / absorbs energy (1)

b)

By Hess’s law, - 40.4 + (-297.2) + ∆Hêreaction

= -572 + 82.2 (1)∴ ∆Hê

reaction= -572 + 82.2 + 40.4 + 297.2= -152.2 kJ mol-1 (1)

S(rhombic) ∆Hconversion

S(monoclinic) + O2(g)

SO2(g)

-297.2kJ mol-1

-296.9 kJ mol-1

Or

c)

By Hess’s law, ∆Hêf (SO

3) = -297.2 + (-98.1) (1)= -395.3 kJ mol-1 (1)

SO2(g) + 2H

2S(g)

∆Hêreaction

2H2O(g) + 3S(mono)

2(-20.2) = - 40.4 kJ mol-1

- 297.2 kJ mol-1

2H2(g) + O

2(g) + 3S(monoclinic)

S(monoclinic) + 1½ O2(g)

∆Hêf (SO

3)

SO3(g)

SO2(g) + ½O

2(g)

Start

+ O2(g) + O

2(g)

Finish

Or

Start

Finish

2(-286) = - 575 kJ mol-1

2(+41.1) = +82.2 kJ mol-1

(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)


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)

n (Mg) = 0.24/24 = 0.01 mol (1)∴∆H

reaction= -5275/0.01= -527,500 J mol-1 / -527.5 kJ mol-1 (1)

Maximum of 2 marks if –ve sign omitted

c) q = 50.84 × 4.2 × (-5.2) = -1110 J (1)∴ ∆H

reaction= -1110/0.01= -111,000 J mol-1 / -111 kJ mol-1 (1)

Maximum (1) if –ve sign omitted

d) Mg(s) + C(s) + 3/2 O2

(g) → MgCO3(s) (1 for formulae and balance; 1 for state symbols)

e)

Where ∆H1 = answer from b), ∆H

2 = answer from c), and

∆H3 = (-286 +(-394)) = - 680 kJ mol-1 for H

2O(l) + CO

2(g)

f) By Hess’s law, ∆Hf (MgCO

3) + ∆H

2 = ∆H

1 + ∆H

3 (1)

∴ ∆Hf (MgCO

3) = -527.5 – 680 + 111 = -1097 kJ mol-1 (1)*

*marked consequentially on answers in b) and c)

MgCl2(aq) + H2O(l) + CO2(g)

Mg(s) + C(s) + O2(g) + 2HCl(aq) MgCO3(s) + 2HCl(aq)

MgCl2(aq) + H2(g) + C(s) + O2(g)

∆H2 (1)∆H1 (1)

∆H3 (1)

∆Hf (MgCO3)

(1)

3

2

3

2

Start

Finish


QUESTIONSHEET 9

AS Level

HESS’S LAW WITH BOND DISSOCIATION ENTHALPY

a)

By Hess’s law, -278 = +1430 + 1308 + 248 + (-2264) + (-928) + ∆Hreaction

(1)∴ ∆H

reaction = -278 – 1430 – 1308 – 248 + 2264 + 928 = -72 kJ mol-1 (1)

Or

∆Hreaction

2C(s)

½O2(g)

3H2(g)

2(+715) = +1430 kJ mol-1 (1)

3(+436) = +1308 kJ mol-1 (1)

(+496)/2 = +248 kJ mol-1 (1)

2C(g)

6H(g)

O(g)

2CH2=CH2(g)

H2O(g)

+

-612 + 4(-413)= -2264 kJ mol-1 (1)

2(-464)

= -928 kJ mol-1 (1)

C2H5OH(l) (1) for cycle

∆Hθf (C2H2OH)

= -278 kJ mol-1

Finish

Start

-612 + 4(-413)

= -2264 kJ mol-1 (1)

By Hess’s law, ∆Hêf (C

2H

4) = + 1430 + 872 + (-2264) (1) = + 38 kJ mol-1 (1)

By Hess’s law, ∆Hθf (H

2O) = +436 + 248 + (-928) (1)= -244 kJ mol-1 (1)

For the hydration of ethene,∆H

reaction= -278 – [+38 + (-244)] (1)= -72 kJ mol-1 (1)

b) ∆Hθreaction

= ∑∆Hθc (reactants) - ∑∆Hθ

c (products) (1)

= -1411 – (-1368) (1)= - 43 kJ mol-1 (1)


c) Mean/average bond dissociation enthalpies are used and not the dissociation enthalpies of the bonds in themolecules in the equation (1)

d) More or less negative ∆H is less negative (1)

Reason For the change C2H

5OH (l) → C

2H

5OH(g), enthalpy of vaporisation / latent heat of vaporisation (1)

is required / absorbed / positive (1)

2(-464)

= -928 kJ mol-1

(1)

++

C2H

4(g)

++

4H(g)

2C(g)2(+715) = +1430 kJ mol-1

2(+436) = +872 kJ mol-1

2H2(g)

2C(s)

∆Hêf (C

2H

4)

+436 kJ mol-1

(+496)/2 = + 248 kJ mol-1

∆Hêf (H

2O)

Start

Finish

H2(g)

½O2(g)

2H(g)

O(g)

H2O(g)Finish

Start


QUESTIONSHEET 10

AS Level

TEST QUESTION I

a) (i) Heat change / enthalpy change (½) when 1 mole of a compound (1)is dissolved in water (½) to form an infinitely dilute solution (1)

(ii) n (NH4Cl) = 5.35/53.5 = 0.1 mol (1)

Heat change = (100 + 5.35) × 4.18 × 10-3 × (20.55 - 17.10) = + 1.52 kJ (1)∴ ∆H

solution = +1.52/0.1 = + 15.2 kJ mol-1 (1)

(iii) endothermic (1)

b) Heat change = (50 + 50) × 4.18 × 10-3 × (20.00 - 23.12) = - 1.304 kJ (1)per 0.5 × 50.0/1000 = 0.025 mol (1)∴ ∆H

neut = -1.304/0.025 = -52.2 kJ mol-1 (1)

c) (i) ½N2(g) + 2H

2(g) + ½Cl

2(g) → NH

4Cl(s) (1)

(ii) Enthalpy cycle

Or accept an enthalpy diagram

Application of Hess’s law∆H

f (NH

3) + ∆H

f (HCl) + ∆H

solution (NH

3) + ∆H

solution (HCl) + ∆H

neut = ∆H

f (NH

4Cl) + ∆H

solution (NH

4Cl) (1)

Or -43.2 + (-92.3) + ∆Hsolution

(NH3) + (-187) + (-52.2*) = -315 + 15.2♦ (1)

* mark consequentially from b)♦ mark consequentially form a)

Molar enthalpy of solution of ammonia∆H

solution (NH

3) = -315 + 15.2 + 43.2 + 92.3 + 187 + 52.2 (1)

= +74.9 kJ mol-1 (1)


QUESTIONSHEET 11

AS Level

TEST QUESTION II

a) (i) 32.44-31.96 = 0.48g (1)

(ii) 33.1-16.4 = 16.7oC (1)

b) (i) 80x4.2x16.7 (1) = 5611(.2)J (1)

(ii) 5611 /0.48(1) = 11690 J (1)

(iii) Mr = 46(1) mol ethanol = 0.48/46 = 0.0104(1)

11690 = 1124 kJ(1)1000 × 0.0104enthalpy of combustion = -1124 kJ mol-1(1) (sign must be correct)

If different stages of figures, allow use of J not kJMark consequentially

(iv) standard conditions not stated (1) 298K(25oC)/1 atm P(most stable form)(1)

(v) Density = 1g cm-3 (1)or no account of container’s heat absorption (1)

c) Reason 1 Heat losses to atmosphere(1)Explanation Temp rise less than correct value(1)

Calculated J less than correct value(1)

Reason 2 Incomplete combustionExplanation Mol used less than from mass loss (1)

so energy/mol smaller than correct value (1)

enthalpy of displacement

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