NSS Chemistry Part 8 Chemical Reactions and Energy

Part 8 Chemical Reactions and Energy/Page.1

HKAL Past Paper Questions: Part 8 Chemical Reactions and Energy 1. [HKAL 1985 II Q2a ] A form 6 student carried out the following experiments to determine the standard enthalpy of dissociation of

RCO2H.

Experiment (I) Experiment (II)

50 cm3 of 2.0M NaOH solution were added to

50 cm3 of 2.0 M HCl solution in a polystyrene

cup. The maximum temperature rise was found

to be 13.0oC.

50 cm3 of 2.0M NaOH solution were added to

50 cm3 of 2.0 M RCO2H solution in a

polystyrene cup. The maximum temperature rise

was found to be 10.5oC.

equation:

H+(aq) + OH-(aq) = H2O(l)

equation:

RCO2H(aq) + OH-(aq) = RCO2-(aq) + H2O(l)

Given at 298K and 1 atmosphere pressure: Specific heat capacity of the solution = 4.2 Jg-1K-1; density of the solution = 1.0 g cm-3. (i) Calculate the standard enthalpy of neutralization for experiment (I) and experiment (II) respectively. Explain the difference in values obtained. (ii) Calculate the standard enthalpy of dissociation of RCO2H.

(8 marks) (i) In Experiment (I) No. of moles of NaOH = (0.05 dm3)(2.0M) = 0.1 mol No. of moles of HCl = (0.05 dm3)(2.0M) = 0.1 mol Mass of the mixture = (50 cm3 + 50 cm3)( 1.0 g cm-3) = 100.0 g Heat produced by the reaction between 0.1 mol of NaOH and 0.1 mol of HCl = mcΔT = (100.0)(4.2)(13) = 5460J Enthalpy of neutralization (ΔH1) = -(5460J)÷0.1 mol = -54.6kJmol-1

In Experiment (II) No. of moles of NaOH = (0.05 dm3)(2.0M) = 0.1 mol No. of moles of RCO2H = (0.05 dm3)(2.0M) = 0.1 mol Mass of the mixture = (50 cm3 + 50 cm3)( 1.0 g cm-3) = 100.0 g Heat produced by the reaction between 0.1 mol of NaOH and 0.1 mol of HCl = mcΔT = (100.0)(4.2)(10.5) = 4410J Enthalpy of neutralization (ΔH2) = -(4410J)÷0.1 mol = -44.1kJmol-1

In the experiment (II), RCO2H only partially ionizes, therefore, some energy are absorbed to ionize the RCO2H

for the neutralization reaction. (ii) standard enthalpy of dissociation of RCO2H = ΔH2 - ΔH1

= -44.1 – (-54.6) = +10.5 kJmol-1


2. [HKAL 1987 II Q2a] Given the following thermochemical data at 298K: standard enthalpy of formation of CO2 (g) -394.80 kJ mol-1

standard enthalpy of formation of CH3COOH(l) -488.88 kJ mol-1

enthalpy of vaporization of H2O(l) to H2O(g) +39.48 kJ mol-1

enthalpy of reaction: H2(g) + 1/2 O2(g) → H2O(g) -242.76 kJ mol-1

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) -809.34 kJ mol-1

Calculate (i) the standard enthalpy of formation of H2O(l) at 298K, and (ii) the enthalpy change for the following reaction at 298K: CH3COOH(l) → CH4(g) + CO2(g)

(8 marks) (i) Since H2(g) + 1/2 O2(g) → H2O(g) ΔH = -242.76 kJ mol-1

H2O(g) → H2O(l) ΔH = -39.48 kJ mol-1

Add up these two equations, H2(g) + 1/2 O2(g) → H2O(g) ΔH = -242.76 kJ mol-1

+ H2O(g) → H2O(l) ΔH = -39.48 kJ mol-1

--------------------------------------------------------------------------------- H2(g) + 1/2 O2(g) → H2O(l) ΔH = -282.24 kJmol-1

(ii) Since CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔH = -809.34 kJ mol-1

-809.34 kJmol-1 = ΔHf[CO2(g)] + 2ΔHf[H2O(g)] - ΔHf[CH4(g)] -809.34 = -394.80 + 2(-242.76) - ΔHf[CH4(g)] ΔHf[CH4(g)] = -70.98 kJmol-1

For CH3COOH(l) → CH4(g) + CO2(g) ΔH = ΔHf[CH4(g)] + ΔHf[CO2(g)] – ΔHf[CH3COOH(l)] ΔH = -70.98 + (-394.80 kJ mol-1) – (-488.88 kJ mol-1) ΔH = +23.1 kJmol-1


3. [HKAL 1988 I Q1a] From calorimetric measurements, the standard molar enthalpies of combustion at 298K for graphite, hydrogen,

and ethanol are –393.1 kJ mol-1, –285.8 kJ mol-1, and –1367 kJ mol-1 respectively. Calculate the standard molar enthalpy of formation for ethanol at 298K.

(3 marks)

2C(s) + 3H2(g) + 0.5O2(g) CH3CH2OH(l)

2CO2(g) + 3H2O(l)

+ 3O2(g) + 3O2(g)

ΔH

ΔH2ΔH1

ΔH1 = 2ΔHc[C(s)] + 3ΔHc[H2(g)] = 2(-393.1) + 3(-285.8) = -1643.6 kJmol-1

ΔH2 = ΔHc[CH3CH2OH(l)] = -1367 kJmol-1

Since ΔH1 = ΔH + ΔH2

Therefore ΔH = -1643.6 - (-1367) = -276.6 kJmol-1

i.e. standard enthalpy of formation for ethanol at 298K is -276.6 kJmol-1

4. [HKAL 1990 I Q2b] Show how ΔHf for CuSO4

. 5H2O(s) can be determined using the following data: ΔHsol, 298 CuSO4

. 5H2O(s) = +8 kJ mol-1; ΔHf CuSO4(s) = -773 kJ mol-1

ΔHsol, 298 CuSO4(s) = -66 kJ mol-1; ΔHf H2O(l) = -286 kJ mol-1

CuSO4(s) + 5H2O(l) CuSO4.5H2O(s)

CuSO4(aq)

+ aq+ aq

ΔH2

ΔH

ΔH1

ΔH1 = ΔHsol, 298 CuSO4(s) = -66 kJ mol-1

ΔH2 = ΔHsol, 298 CuSO4. 5H2O(s) = +8 kJ mol-1

Since ΔH1 = ΔH + ΔH2

Therefore, ΔH = -66 – (+8) = -74 kJmol-1

CuSO4(s) + 5H2O(l) CuSO4.5H2O(s)

ΔH

ΔH = ΔHf for CuSO4. 5H2O(s) – {ΔHf CuSO4(s) + 5 ΔHf H2O(l)}

ΔHf for CuSO4

. 5H2O(s) = -74 + {-773 + 5(-286)} = -2277 kJmol-1


5. [HKAL 1991 I Q4b] You are required to determine indirectly the enthalpy of formation of calcium carbonate(s). You are given the reagents: calcium, calcium carbonate, a strong acid, and water; the enthalpies of formation of

carbon dioxide(g) and water(l) are available. (i) Give equations of suitable reactions for this determination. (ii) Name the major pieces of apparatus needed for this determination. (iii) Suggest one experimental difficulty in the direct determination of the enthalpy of formation of

calcium carbonate from its elements. (i) CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l) Ca(s) + 2HCl(aq) → CaCl2(aq) + H2 (g) (ii) Equipments include: balance, burette, pipette, thermometer and calorimeter. (iii) For the reaction Ca(s) + C(s) + 1.5 O2(g) → CaCO3(s) side reactions may occur: 2Ca(s) + O2(g) → 2CaO(s) C(s) + O2(g) → CO2(g) Besides, CaCO3 may decompose to CaO. 6. [HKAL 1991 II Q1b ] (i) Calculate the enthalpy of formation of NaCl(s) from the following data: Reaction ΔH / kJ mol-1

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) -57.3 H2(g) + 1/2 O2(g) → H2O(l) -285.9 1/2 H2(g) + 1/2 Cl2(g) → HCl(g) -92.3 HCl(g) + aq → HCl(aq) -71.9 Na(s) + 1/2 O2(g) + 1/2 H2(g) + aq → NaOH(aq) -425.6 NaCl(s) + aq → NaCl(aq) +3.9 (ii) The enthalpy of neutralization of ethanoic acid with aqueous sodium hydroxide is –55.2 kJmol-1

while that of hydrochloric acid is –57.3 kJ mol-1. Account for the difference in these two values. (6 marks)

(i) Reaction ΔH / kJ mol-1

Na(s) + 1/2 O2(g) + 1/2 H2(g) + aq → NaOH(aq) -425.6 NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) -57.3 HCl(g) + aq → HCl(aq) -71.9 1/2 H2(g) + 1/2 Cl2(g) → HCl(g) -92.3 H2O(l) → H2(g) + 1/2 O2(g) +285.9 +) NaCl(aq) → NaCl(s) + aq -3.9 ___________________________________________________________________ Na(s) + 1/2 Cl2(g) → NaCl(s) -365.1 kJmol-1

i.e. the enthalpy of formation of NaCl(s) is -365.1 kJmol-1

(ii) As CH3COOH is a weak acid, it will only be slightly ionized. The difference in the two values is mainly due to

the enthalpy change of ionization, i.e. energy is absorbed for iuonization. CH3COOH(aq) → CH3COO-(aq) + H+(aq) ΔH = -55.2 – (-57.3) = +2.1 kJmol-1


7. [HKAL 1992 II Q1b ] (i) Define the standard enthalpy of formation of a compound, using CH3OH(l) as an illustration. (ii) Given the following thermochemical data at 298 K: standard enthalpy of combustion of CH3OH(l) -726.6 kJ mol-1

standard enthalpy of formation of CO2(g) -393.5 kJ mol-1

standard enthalpy of formation of H2O(l) -285.8 kJ mol-1

Calculate the standard enthalpy of formation of CH3OH(l) at 298 K.

(4 marks) (i) The standard enthalpy of formation of a compound is the standard enthalpy change that occurs when one

mole of the compound is made from its constituent elements under standard conditions (298 K and 1

atmospheric pressure)

e.g. C(s) + 2H2(g) + 0.5 O2(g) → CH3OH(l)

(ii) Given: CH3OH(l) + 1.5 O2(g) → CO2(g) + 2H2O(l)

Since ΔH =ΣΔHf[products] - ΣΔHf[products]

Therefore ΔH = {ΔHf[CO2(g)] + 2ΔHf[H2O(l)]} – {ΔHf[CH3OH(l)] + 1.5ΔHf[O2(g)]}

= -726.6 kJ mol-1

Since ΔHf[O2(g)] = 0

Therefore -726.6 = (-395.5) + 2(-285.8) - ΔHf[CH3OH(l)]

Therefore ΔHf[CH3OH(l)]= -238.5 kJ mol-1


8. [HKAL 1993 II Q3b] Given the following thermochemical data at 298 K:

Compound ΔHc, 298 / kJ mol-1 ΔHf, 298 / kJ mol-1

cyclopropane(g) -2091 -

propene(g) -2058 -

propane(g) -2220 -

water(l) - -285.8

(i) Calculate the enthalpy change involved in the hydrogenation of cyclopropane to propane. (ii) Calculate the enthalpy change involved in the conversion of cyclopropane to propene. Comment on the relative stabilities of cyclopropane and propene.

(8 marks) (i)

+ H2 (g) CH3CH2CH3(g)(g)

3CO2(g) + 4H2O(l)

+5O2(g)+5O2(g)

ΔH

ΔH1 ΔH2

ΔH1 = ΔHc[cyclopropane(g)] + ΔHc[H2(g)] = ΔHc[cyclopropane(g)] + ΔHf[H2O(l)] = -2091 + (-285.8) =-2376.8 kJmol-1

ΔH2 = ΔHc[propane(g)] = -2220 kJmol-1

Since ΔH = ΔH1 - ΔH2

Therefore, ΔH = -2376.8 + 2220 = -156.8 kJmol-1

(ii)

(g)

3CO2(g) + 3H2O(l)

+4.5O2(g)

ΔH

ΔH1 ΔH2

CH3CH=CH2 (g)

+4.5O2(g)

ΔH1 = ΔHc[cyclopropane(g)] = -2091 kJmol-1

ΔH2 = ΔHc[propene(g)] = -2058 kJmol-1

ΔH = ΔH1 - ΔH2 = -2091 + 2058 = -33 kJmol-1

The conversion of propene from cyclopropane is exothermic, hence propene is more stable. It is because in the structure of cyclopropane, there is a squeezing of bond angles / ring is highly

strained.


9. [HKAL 1994 II Q2a] Given the following thermochemical data. Reaction ΔH / kJ mol-1

C(graphite) + 2H2(g) → CH4(g) -75.0 C(graphite) + O2(g) → CO2(g) -393.5 H2(g) + 1/2 O2(g) → H2O(l) -285.9 a. Calculate the enthalpy change for the reaction CH4(g) + 2O2(g) → 2H2O(l) + CO2(g) b. The enthalpy change is –801.7 kJ mol-1 for the following reaction. CH4(g) + 2O2(g) → 2H2O(g) + CO2(g) Calculate the enthalpy change of vaporization of water at 298 K. c. The enthalpy change of combustion of diamond at 298 K is –395.4 kJ mol-1. Which allotrope of carbon,

diamond or graphite, is energetically more stable? Explain why carbon does not convert from the less stable allotrope to the more stable one at room temperature.

(6 marks)


10. [HKAL 1995 II Q1b] You are provided with the following thermochemical data: Reaction ΔH / kJ mol-1

Ag(s) + aq → Ag+(aq) + e- +105.56 1/2 N2(g) + 3/2 O2(g) + aq + e- → NO3

-(aq) -207.36 1/2 Cl2(g) + aq + e- → Cl-(aq) -167.15 Ag(s) + 1/2 Cl2(g) → AgCl(s) -127.07 Calculate the standard enthalpy change for the reaction AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)

(4 marks)

11. [HKAL 1997 II Q2c] Given the following thermochemical date at 298K: standard enthalpy change of formation of CO2(g) = -393.5 kJ mol-1

standard enthalpy change of formation of H2O(l) = -285.8 kJ mol-1

standard enthalpy change of combustion of CH3CH2OH(l) = -1336.9 kJ mol-1

calculate the standard enthalpy change of formation of CH3CH2OH(l) at 298 K.

(4 marks)


12. [HKAL 1998 II Q2c] Both H2(g) and CH3OH(l) are possible fuels for powering rockets. Their combustion reactions are shown

below. H2(g) + 1/2 O2(g) → H2O(g) CH3OH(l) + 3/2 O2(g) → CO2(g) + 2H2O(g) (i) For each of the above reactions, calculate the enthalpy change at 298 K per kg of the fuel-oxygen mixture in

the mole ratio as indicated in the stoichiometric equation. (ii) The effectiveness of a fuel can be estimated by dividing the enthalpy change per kg of the fuel-oxygen mixture

in its combustion reaction by the average molar mass of the product(s) in g. Deduce which of the above two fuels is more effective in powering rockets. Note: You are provided with the following data at 298K: Compound Molar mass /g ΔHf, 298 / kJ mol-1

CO2(g) 44 -394 H2O(g) 18 -242 CH3OH(l) 32 -239

(6 marks)


13. [HKAL 1999 I] In an experiment to determine the enthalpy change of neutralization, a polystyrene foam cup was used as a

calorimeter. When a solution of an acid was poured into a solution of an alkali in the calorimeter, the temperature rise was recorded by a thermometer which also served as a stirrer.

State THREE sources of error in the result obtained in such an experiment.

(3 marks)

14. [HKAL 2003 I Q3a] Chlorine dioxide, ClO2, can be prepared in the laboratory according to the following equation: 2AgClO3(s) + Cl2(g) → 2AgCl(s) + 2ClO2(g) + O2(g) ΔH = +10 kJ mol-1

Given that the standard enthalpy changes of formation at 298K of AgClO3(s) and AgCl(s) are –30 kJ mol-1

and –127 kJ mol-1 respectively, calculate ΔHf, 298 [ClO2(g)]. Hence, comment on the stability of ClO2(g) under standard conditions.

(3 marks)


15. [HKAL 2003 I Q7a] Outline an experimental procedure and the data treatment to determine the enthalpy change for the reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

(6 marks)


16. [HKAL 2004 II Q2a] (i) What is the standard enthalpy change of formation of a compound? (ii) 0.10 g of magnesium was added to an excess of dilute hydrochloric acid in a polystyrene foam cup with

negligible heat capacity. The maximum rise in temperature of the mixture was found to be 4.3oC. Given that the heat capacity of the acid used is 494 Jk-1, calculate the molar enthalpy change of the reaction: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) (iii) In a similar experiment, the molar enthalpy change for the reaction of magnesium carbonate with

hydrochloric acid was found to be -90 kJ. In addition, given that the standard enthalpy change of formation of H2O(l) is -285 kJmol-1 and that of CO2(g)

is -393 kJmol-1, estimate the enthalpy change of formation of MgCO3(s) under the conditions of the experiment.

(7 marks)


17. [HKAL 2005 II Q5a] Most of the petroleum stock located on Earth is likely to be used up in 50 to 100 years if petroleum

consumption is maintained at the current rate. With a view to cutting down petroleum consumption, some countries have adopted an alternative fuel for motor vehicles – gasoline which contains ethanol.

(i) Based on the standard enthalpy changes of formation given below, calculate the standard enthalpy

changes for the complete combustion of octane and ethanol respectively.

(ii) Assuming that gasoline contains only octane, compare the enthalpy change of combustion values, in kJ g-1 , of gasoline and an alternative fuel containing gasoline and 10% ethanol by mass.

(iii) Besides cutting down petroleum consumption, suggest one additional advantage of using the alternative fuel over using gasoline.

(9 marks)


18. [HKAL 2008 II Q1c]

The table below lists the standard enthalpy change of formation of four compounds:

Compound -1 o 298f mol kj/ ,HΔ

)O(H2 l 286− HCl(g) 92−

)s(SiO2 910− )(SiCl4 l 640−

(i) State the meaning of the term ‘standard enthalpy change of formation of a compound’. (ii) undergoes hydrolysis to give )(SiCl4 l s)(SiO2

(I) Write the chemical equation for the hydrolysis. (II) Using the above data, calculate the standard enthalpy change for the hydrolysis. State ONE assumption made in your calculation. (III) Does the hydrolysis have a positive, negative or zero entropy change? Explain.

(7 marks)


19. [HKAL 2009 II Q1a]

Hydrolysis of proteins gives a variety of amino acids, and alanine (CH3CH(NH2)CO2H)) is one of the amino acids commonly obtained.

(i) Explain why alanine exists as a crystalline solid under room temperature and atmospheric

pressure, whereas amines and carboxylic acids with a comparable molar mass are liquids. (ii) Suggest why proteins in blood can function as a buffer. (iii) In the human body, alanine undergoes biological oxidation to give carbon dioxide, water and urea

(CO(NH2)2). Write the chemical equation for this reaction. (iv) When nitrogen-containing organic compounds are burnt in calorimetric experiments, the nitrogen

they contained is transformed to nitrogen molecules. Write the chemical equation for the combustion of each of the following compounds in a calorimetric

experiment: (I) alanine (II) urea (v) Using the equations that you have given in (iii) and (iv), as well as the standard enthalpy changes of

combustion given in the table below:

calculate the energy, in kJ, that can be obtained from the biological oxidation of 1,00 g of alanine at

298 K. (12 marks)


20. [HKAL 2010 I Q3a]

Answer the following multiple-choice questions. (i) Which of the following processes is endothermic ? A. freezing of water B. condensation of steam C. reaction of H+(aq) with OH－(aq) to give H2O(l) D. electrolysis of water

D

21. [HKAL 2010 II Q2a]

A flight of space shuttle requires the use of three propellants: A solid propellant, which is a mixture of powdered Al(s) and NH4ClO4(s), is used to power the rockets

carrying the shuttle. Upon ignition, the solid propellant reacts to give Al2O3(s), AlCl3(s), NO(g) and H2O(g). This reaction provides energy for launching the rockets and the shuttle up to the upper atmosphere.

After the shuttle separates from the rockets, the shuttle is propelled into its designated orbit by a cryogenic

propellant, which is a mixture of H2(l) and O2(l). When the shuffle is in its orbit, a hypergolic propellant, of which the fuel is CH3NHNH2(l) and the oxidant

is N2O4(l), will provide energy for manoeuvring the shuttle. The fuel and oxidant react upon mixing, without ignition, to give CO2(g), H2O(g) and N2(g).

(i) Write the chemical equation for the reactions of (I) Al(s) with NH4ClO4(s), and (II) CH3NHNH2(l) with N2O4(l).


(ii) Given the following standard enthalpy changes of formation, calculate the standard enthalpy change, at 298 K, of reaction (I) and that of reaction (II).

Compound θfHΔ ,298 /kJ mol-1

Al2O3(s) -1676 AlCl3(s) -704

CH3NHNH2(l) +53 CO2(g) -394 H2O(g) -242

NH4ClO4(s) -295 NO(g) +90 N2O4(l) -20

(iii) Suggest an advantage of using the solid propellant in powering the rockets. (iv) The cryogenic propellant is also used to produce electricity for use in the shuttle. Briefly describe the

electrochemical processes involved. (v) State an advantage of using the hypergolic propellant in manoeuvring the shuttle.

(10 marks)

NSS Chemistry Part 8 Chemical Reactions and Energy

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Transcript of NSS Chemistry Part 8 Chemical Reactions and Energy