AS past paper S3 Energetics

AS past paper Section 3 Energetics

94B 9(a)

95A 2.

(a)

(b)

95B 9(a)

The table below lists the enthalpy changes H, in kJ mol-1 , of reactions (1), (2), (3) and (4) respectively at room temperature and atmospheric pressure. Reaction ∆H/kJ mol-1 (1) CaCO3 (s) + 2 HCl (aq)  CaCl2 (aq) + H2O (l) + CO2 (g) - 25.1 (2) Ca (s) + 2 HCl (aq)  CaCl2 (aq) + H2 (g) -484.0 (3) C(graphite) + O2 (g)  CO2 (g) - 393.5 (4) 2 H2 (g) + O2 (g)  2 H2O (l) -571.8 (i) Outline an experiment to determine the enthalpy change of reaction (1) in the school laboratory.

(ii)

Using the given thermochemical data, construct an enthalpy level diagram and thus calculate the molar enthalpy change of formation of calcium carbonate, at room temperature and atmospheric pressure.

(8 marks)

The standard enthalpy change of formation of NH4Cl (s) cannot be directly determined by an experiment. It can be obtained, however, via an enthalpy level diagram constructed using the following thermochemical data.

Process

Hθ / kJ mol-1 Formation of NH3 (g) from its elements

- 46 Formation of HCl (g) from its elements - 92

NH3 (g) + HCl (g)  NH4Cl (s)

- 177

Explain why the standard enthalpy change of formation of NH4Cl (s) cannot be directly determined.

(1 mark)

Determine the standard enthalpy change of formation of NH4Cl (s), via an enthalpy level diagram constructed using the above data.

AL test_ (4 marks)

The following experiment is used to determine the strength of the hydrogen bond formed between ethyl ethanoate and trichloromethane.

14.0 cm3 of ethyl ethanoate and 10.0 cm3 of trichloromethane were measured separately into two measuring cylinders. The temperature of each liquid was determined. The liquids were then poured into an insulated test tube. The mixture was stirred with a thermometer and the maximum temperature was noted. The results were as follows:

Initial temperature of ethyl ethanoate = 19.5C Initial temperature of trichloromethane = 19.0C Maximum temperature of mixture = 34.0C

96B 8(a)

AS past paper / Section 3 Energetics / Page 2

You are provided with the following data: Compound Density / g cm- 3 Specific heat capacity / J g- 1K- 1 ethyl ethanoate 0.90 1.92 trichloromethane 1.48 0.96 (i) Give the structures of ethyl ethanoate and trichloromethane.

(ii)

Draw a diagram to show the hydrogen bond formed between ethyl ethanoate and trichloromethane.

(iii) Deduce which compound is in excess. Why should an excess of this compound be

used?

(iv) Calculate the enthalpy change per mole of hydrogen bond formed between ethyl

ethanoate and trichloromethane, assuming that the heat capacity of the test tube is negligible.

(9 marks)

In an experiment to determine the enthalpy change of neutralization at room temperature and pressure, a student mixed different volumes of 1.0 M hydrochloric acid with different volumes of 1.0 M sodium hydroxide solution. He measured the rise in temperature of each reaction mixture. The results are tabulated below: Volume of HCl used Volume of NaOH Rise in temperature of the / cm3 used / cm3 reaction mixture / K 16 84 2.2 30 70 4.1 44 56 6.0 58 42 5.7 72 28 3.8 86 14 1.9 (i) By plotting a graph, deduce the rise in temperature of the reaction mixture when 50 cm3 of 1.0 M hydrochloric acid and 50 cm3 of 1.0 M sodium hydroxide solution are mixed.

(ii) Calculate the enthalpy change of neutralization of hydrochloric acid with sodium hydroxide, in kJ mol-1, at room temperature and pressure.

(Assuming that the specific heat capacity and the density of each reaction mixture are 4.2 J g-1 K-1 and 1.0 g cm-3 respectively. Also disregard heat loss to the container and surroundings.)

(iii) At room temperature and pressure, the enthalpy change for

NaOH (s) water NaOH (aq, 0.5 M) was found to be - 42.3 kJ mol-1. Using your answer in (ii), calculate the enthalpy change for the following reaction:

NaOH (s) + HCl (aq, 0.5M)  NaCl (aq, 0.5M) + H2O (l)

(iv) Under the same conditions, the enthalpy change of neutralization of hydrochloric

acid with sodium hydroxide differs from that of ethanoic acid with sodium hydroxide. Explain.

(9 marks)

AS past paper / Section 3 Energetics / Page 3

97B 8(a) (i) You are provided with the following materials and apparatus:

98B 8(a)

99A 2.

methanol,

an aluminium can containing water as a calorimeter, an electronic balance a thermometer, and

a spirit lam (see diagram below). Outline an experimental procedure to determine the enthalpy change of combustion of methanol, Hc[CH3OH (l)]. Write an expression to show how Hc[CH3OH (l)] can be calculated from the experimental results.

(You may assume that the heat capacity of the calorimeter is C JK-1

.) (ii) You are provided with the following thermochemical data at 298K: Standard enthalpy change of combustion of CH-13OH (l) = - 726 kJ mol Standard enthalpy change of formation of CO2 (g) = - 394 kJ mol-1 Standard enthalpy change of formation of H2O (l) = - 286 kJ mol-1

With the help of an enthalpy level diagram, calculate the standard enthalpy

change of formation of CH3OH(l) at 298K.

AL test_ (9 marks)

The table below lists the enthalpy changes of formation of three substances under standard conditions.

Substance Hf, 298 /kJ mol-1 H2O (l) - 286 CO2 (g) - 393 C2H4 (g) + 52 (i) State the meaning of the terms below. (1) standard conditions

(2)

standard enthalpy change of formation of C2H4 (g)

(ii)

Calculate the standard enthalpy change of combustion of C2H4 (g) at 298K.

(5 marks)

Consider the standard enthalpy changes of combustion, ΔHcθ,298 of the alkanols listed in the table below:

Alkanol ΔHcθ,298 /kJ mol-1 CH3OH(l) -726 CH3CH2OH(l) -1367 CH3(CH2)2OH(l) -2017 CH3(CH2)3OH(l) X AS past paper / Section 3 Energetics / Page 4

(a) (b) (c)

Explain why the combustion of CH3OH(l) is exothermic.

(1 mark)

Estimate the value of x. Show how you arrive at your answer.

(2 marks)

At 298 K, the standard enthalpy changes of formation of CO2(g) and H2O(l) are –393 kJ mol-1 and –286 kJ mol-1 respectively. Calculate the standard enthalpy change of formation of CH3OH(l) at 298 K.

(3 marks)

99B 13(I)(c) The enthalpy change of combustion of T can be determined using the set-up shown

02B 8(c)

below :

When 2.30g of T was burnt, the temperature of water of mass 250g in the aluminium can was found to increase by 20.5oC.

(i) Calculate the enthalpy change of combustion of T, in kJmol-1

, under the conditions of the experiment.

(ii) Suggest TWO main sources of error in the experiment.

(Specific heat capacity water = 4.18Jg-1K-1, relative molecular mass of T = 878)

(5 marks)

The table below lists the standard enthalpy changes of neutralization of three acids with NaOH(aq).

Acid H1neutralization,298/kJmol HCI(aq) -57.3 HNO3(aq) -57.3 CH3CO2H(aq) -55.2 Account for the following statements :

(i)

The standard enthalpy change of neutralization of HCl(aq) with NaOH(aq) is the same as that of HNO3(aq) with NaOH(aq).

(ii) The standard enthalpy change of neutralization of HCI(aq) with NaOH(aq) is

more negative than that of CH3C02H(aq) with NaOH(aq).

AL test_(4 marks)

AS past paper / Section 3 Energetics / Page 5

Marking scheme Section 3 Energetics 94B 9(a)

95A 2(a)

(b)

(i) Experiment determination of ΔH of reaction (1) (ΔH1) Dissolve completely a known mass (W) of CaCO3 in a known volume of excess HCl in an polystyrene cup / vacuum flask / calorimeter. Measure the temperature of the acid in the calorimeter before the experiment and the highest temperature attained /

the maximum temperature change. ΔH1 = Heat capacity of calorimeter x rise in temp / ( w / Mr of CaCO3) OR ΔH1 = Heat capacity of calorimeter x rise in temp / ( no. of moles of CaCO3

used ) (ii)

Enthalpy level diagram (any correct form of enthalpy level diagram)

Ca (s) + C (graphite) + 1 1/2 O2 (g) + 2 HCl (aq) H2CaCl2 (aq) + H2 (g) + C (graphite) +1 1/2 O2 (g) HHf+3CaCO3 (s) + 2 HCl (aq)H4H1CaCl2 (aq) + H2O (l) + CO2 (g)

[deduct 1 mark if no state symbol in the enthalpy level diagram]

ΔH2 + ΔH3 + 1/2ΔH4 = ΔHf[CaCO3(s)] + ΔH1

So ΔHf[CaCO3(s)] = -484.0 – 393.5 – 1/2(571.8) – (-25.1)

= -1138.3 kJmol-1

N2, H2 and Cl2 do not combine directly under standard conditions

1/2 N2 (g) + 2 H2 (g) + 1/2 Cl2 (g)- 46NH3 (g) + 1/2 H2 (g) + 1/2 Cl2 (g)NH- 923 (g) + HCl (g)- 177NH4Cl (s)

Enthalpy level diagram

(Deduct 1 mark for missing the state symbols;

Deduct 1 mark for not listing the correct enthalpy values.)

12,12 12,12 12,

12

(1) 1

(1) 2

1 1/0 1

2

AS past paper / Section 3 Energetics / Page 6

ΔHf[NH4Cl(s)] = (-46) + (-92) + (-177) = -315 kJmol-1 (

95B 9(a)

121 1

mark for numerical answer;

12 mark for unit)

(i)

OH3CC ;OCH2CH3ClHCClCl

12,

12

(Accept also CH3COOCH2CH3; HCCl3)

96B 8(a)

(ii)

H3CClCOHCClCCH2CH3ClCHCOHCCl3CH2OClorH3OCl

(iii) no. of moles of ethyl ethanoate = 14 x 0.90 / 88.104 = 0.143 no. of moles of trichloromethane = 10 x 1.48 / 119.368

= 0.124

In H-bond formation, nCH3COOCH2CH3 : nCHCl3 = 1: 1

So ethyl ethanoate is in excess. The calculation that follows is based on the assumption that all trichloromethane molecules are involved in H-bond formation / CHCl3 is poisonous / anesthetic

which should be used as less as possible. (iv) Heat released in the reaction = 0.9 x 14 x 1.92 x (34 – 19.5) + 1.48 x 10 x 0.96 x (34 – 19) = 5 (J) ΔH of H-bond formed = -(5 / 0.124) x 10-3 (kJmol-1)

= -4.55 kJmol-1 / -4.6 kJmol-1

(Deduce 1/2 marks for missing out –ve sign / unit) (i) Plotting the graph (2 straight lines) Labelling the axes (1/2 marks each) Rise in temp. = 6.8(K)

(accept answers from 6.7 to 6.9K)

1 1 1 1

1

1 1* 1 1 1 1

AS past paper / Section 3 Energetics / Page 7

96B 8(a)

97B 8(a)

97B 8(a)

(ii) Heat evolved = 100 x 1.0 x 4.2 x 6.8 = 2856J

ΔH = -(2856 / 0.05)Jmol-1

= -57.1(2)kJmol-1 (Accept answer from 56.2 to 58.0 kJmol-1)

(iii) (1) NaOH(0.5M) + HCl(0.5M)  NaCl(0.5M) + H2O(l) ΔH= -57.1kJmol-1 (2) NaOH(s) + water  NaOH(0.5M) ΔH= -42.3kJmol-1

(2) + (1) NaOH(s) + HCl(0.5M)  NaCl(0.5M) + H2O(l) ΔH= -42.3 + (-57.1) = -99.4(kJmol-1)

(Accept answer from –98.5 to –100.3 kJmol-1) *step mark (0 marks for missing the –ve sign; deduct 12 marks for wrong / no unit)

(iv) HCl is a strong acid / fully ionized in water CH3COOH is a weak acid / not completely ionized. Extra energy is required for the breaking of the O-H bond, so less heat is evolved

(DO NOT accept incomplete neutralization)

(i) Fill the spirit lamp with methanol and use it to heat up the calorimeter. Note the change in mass of the spirit lamp/weigh the spirit lamp before and after burning/note the mass of methanol burnt. (Δm)

Measure the change in temperature of the calorimeter /the temperature of the calorimeter before the experiment and the highest temperature attained (ΔT)

ΔHc = -Heat capacity of calorimeter x ΔT / (Δm / Mr(CH3OH))

(Deduce

12 marks for each mistake e.g. no negative sign, no unit)

(ii)

Enthalpy level diagram.

C(s) + 2 H2(g) + 2 O2 (g)Hf [CH3OH(l)]DHf [CO2(g)] + 2 DHf [H2O (l)]CH3OH(l) + 1 1/2 O2(g)Hc [CH3OH(l)]CO2(g) + 2 H2O (g)

(Deduce 1 mark if no state symbol in the enthalpy level diagram) ΔHf[CH3OH(l)] + ΔHc[CH3OH(l)] = ΔHf[CO2(g)] + 2ΔHf[H2O(l)]

ΔHf[CH3OH(l)] = -394 + 2 x (-286) – (-726)

= -240kJmol-1

(Deduct

12 marks for wrong unit)

1* 1

1*

1 1 1

12,

121

1 2

2

1* 1

AS past paper / Section 3 Energetics / Page 8

98B 8(a)

(i) (1)

Standard conditions : gases at 1 atm pressure; solutions at unit concentration; substances at their normal physical states at the specified temperature

(ii)

99A 2 (a)

During combustion, the energy required to break the bonds in oxygen (O=O) and in alkanol is less tan the energy released when CO2 and H2O are being formed.

or,

the products of combustion (CO2 and H2O) have lower enthalpy/energy than the reactants.

(b)

-1

Accepts answers from – 2658 to – 2676 kJ mol

1 1 1* (1) 2

(2) The enthalpy change when one mole of ethene is formed from its elements in their standard states ∆Hcø,298[C2H4(g)]

= 2∆Hfø,298[CO2(g)] + 2∆Hfø,298[H2O(l)] - ∆Hfø,298[C2H4(g)] = 2(-393) + 2(-286) – (52) = -1410kJmol-1

or ∆Hcø,298[C2H4(g)]

(1 mark for numerical answer; 1 mark for unit and sign)

1

1

(Award 0 mark for omitting of the –ve sign)

Descending a homologous series, each member has one –CH2- group more than the preceding one. Combustion of 1 mole of –CH2- group would release 1 to 650 kJ.

1 2

(c)

(1 mark for the answer of H[C4H9OH(l)]; 1 mark for the explanation.) cHc[CH3OH(l)] = Hf[CO2(g)] + 2 Hf[H2O(l)] –Hf[CH3OH(l)] Hf[CH3OH(l)] = (-393) + 2 (-286) – (-726)

= - 239 kJ mol

-1

1

(Award 1 mark for giving a balanced equation for the combustion of CH3OH.)

Heat produced = 250 4.18 20.5 10-3 H = -

= 21.4 kJ

1 1,1

99B 13(I)(c) (i)

21.4 = - 8178 kJ mol-1 (-1 mark for omitting of –ve sign.)

2.30878

(ii) (i) (ii)

(Accept answers form – 8165 to – 8185 kJ mol-1) Heat loss to the surrounding

The heat capacity of the aluminium can has been neglected Both HC1 and HNO3 are strong acids and NaOH is a strong alkali.

1 1 1 1

02B 8(c)

Hneutralization refer to the standard enthalpy change of the reaction:

H+(aq) + OH-(aq)  H2O(1)

CH3CO2H is a weak acid / does not undergo complete ionisation in water. The 1 reaction of CH3CO2H(aq) and NaOH(aq) involves breaking of O-H bonds in 1 CH3CO2H less energy is released.

4