CHEM1001 2014-J-11 June 2014 22/01(a)

• Combustion of 15.0 g of coal provided sufficient heat to increase the temperature of 7.5 kg of water from 286 K to 298 K. Calculate the amount of heat (in kJ) absorbed by the water. The heat capacity of water, Cp° = 4.2 J K–1 g–1.

Marks 3

The temperature rise of water is: ΔT = (298 – 286) K = 12 K.

As the heat capacity Cp° = 4.2 J K–1 g–1, the heat absorbed by 7.5 kg of water is:

q = mCp°ΔT = (7.5 × 103 g) × (4.2 J K-1 g-1) × (12 K) = 380 kJ

Answer: 380 kJ

Assuming that coal is pure carbon, calculate the heat of combustion (in kJ mol–1) of the coal.

The number of moles of carbon in 15.0 g is: number of moles = mass / molar mass = 15.0 g / 12.01 g mol-1 = 1.25 mol

As this quantity generates 380 kJ, the amount generated by 1 mol is: q = 380 kJ / 1.25 mol = 300 kJ

As energy is released by the combustion reaction, it is exothermic. Hence: ΔcombustionH = -300 kJ mol-1

Answer: ΔcombustionH = -300 kJ mol-1

CHEM1001 2014-J-11 June 2014 22/01(a)

• The standard enthalpy of formation of SO2(g) is the enthalpy change that accompanies which reaction?

3

S(s) + O2(g) → SO2(g)

Calculate the standard enthalpy of formation of SO2(g) given the following data.

SO2(g) + 1/2O2(g) → SO3(g) ΔH = –99 kJ mol–1

S(s) + 3/2O2(g) → SO3(g) ΔH = –396 kJ mol–1

Using ΔrxnH = ΣmΔfH°(products) - ΣnΔfH°(reactants), the enthalpy of these two reactions are: (1) ΔrxnH = ΔfH°(SO3(g)) - ΔfH°(SO2(g)) = -99 kJ mol-1

(2) ΔrxnH = ΔfH°(SO3(g)) = -369 kJ mol-1 as ΔfH°(O2(g)) and ΔfH°(S(s)) are both zero as O2(g) and S(s) are elements in their standard states. Substituting (2) into (1) gives:

(-396 kJ mol-1) - ΔfH°(SO2(g))= -99 kJ mol-1

Hence:

ΔfH°(SO3(g)) = -297 kJ mol-1

Answer: -297 kJ mol-1

CHEM1001 2014-J-12 2201(a)

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• The standard heat of formation of ClF3(g) is –159 kJ mol–1. Use the bond enthalpies below to calculate the average Cl–F bond enthalpy in ClF3(g).

Marks 4

Bond Cl–Cl F–F

Bond enthalpy / kJ mol–1 243 158

The heat of formation of ClF3(g) corresponds to the reaction: 1/2 Cl2(g) + 3/2 F2(g) → ClF3(g) In terms of bond enthalpies, this corresponds to (i) breaking 1/2 mol of Cl-Cl bonds and 3/2 mol of F-F bonds and (ii) forming 3 mol of Cl-F bonds: ΔfHo = (1/2 × 243 + 3/2 × 158) – 3 × ΔbondH(Cl-F) = -159 kJ mol-1 ΔbondH(Cl-F) = +173 kJ mol-1

Answer: +173 kJ mol-1

Explain why this number is different from the average Cl–F bond enthalpy estimated for ClF5(g) of 151 kJ mol–1.

Bond strength is dependent on the interaction between the bonding electrons and the two nuclei of the atoms involved. The bonds in ClF3 and ClF5 are similar but not identical. For example, the oxidation numbers of Cl are +3 and +5 and it has 2 and 1 lone pair respectively. This leads to the Cl-F bonds being a little different in strength.

• Explain the observation that the boiling point of ethanol is much higher than that of dimethyl ether despite these compounds having the same molar mass.

1

compound formula boiling point / °C

ethanol CH3CH2OH 78.3

dimethyl ether CH3OCH3 –22.0

Ethanol can from strong intermolecular hydrogen bonds. Dimethyl ether cannot, it can only form much weaker dispersion forces and dipole-dipole interactions.

CHEM1001 2014-J-11 June 2014 22/01(a)

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CHEM1001 2013-J-12 June 2013 22/01(a)

• Write the equation whose enthalpy change represents the standard enthalpy of formation of NO(g).

Marks 3

½ N2(g) + ½ O2(g) à NO(g)

Given the following data, calculate the standard enthalpy of formation of NO(g). N2(g) + 2O2(g) 2NO2(g) ΔH° = 66.6 kJ mol–1 2NO(g) + O2(g) 2NO2(g) ΔH° = –114.1 kJ mol–1

Using ΔrxnH° = ΣmΔfH°(products) - ΣnΔfH°(reactants), the enthalpy of these two reactions are:

(1) ΔH° = [2ΔfH°(NO2(g))] = 66.6 kJ mol-1

(2) ΔH° = [2ΔfH°(NO2(g))] – [2ΔfH°(NO(g))] = -114.1 kJ mol-1

where ΔfH°(N2(g)) = ΔfH°(O2(g)) = 0 has been used for elements already in their standard states. Substituting (1) into (2) gives: (66.6 kJ mol-1) – [2ΔfH°(NO(g))] = -114.1 kJ mol-1 ΔfH°(NO(g))] = 90.4 kJ mol-1

Answer: +90.4 kJ mol-1

• Hydrazine, N2H4, burns completely in oxygen to form N2(g) and H2O(g). Use the bond enthalpies given below to estimate the enthalpy change for this process.

3

Bond Bond enthalpy (kJ mol–1) Bond Bond enthalpy (kJ mol–1) N–H 391 O=O 498 N–N 158 O–O 144 N=N 470 O–H 463 N≡N 945 N–O 214

The chemical equation for the combustion is: N2H4(g) + O2(g) à N2(g) + 2H2O(g)

ANSWER CONTINUES ON THE NEXT PAGE

NN

H

H

H

H

+ O O N N OH H

+ OH H

+

CHEM1001 2013-J-12 June 2013 22/01(a)

This requires:

• breaking 1 N-N bond, 4 N-H bonds and 1 O=O bond • making 1 N≡N bond and 4 O-H bonds

Breaking bonds is endothermic: ΔbreakingH° = (158 + 4 × 391 + 498) kJ mol-1 = +2220 kJ mol-1

Making bonds is exothermic: ΔmakingH° = - (945 + 4 × 463) kJ mol-1 = -2797 kJ mol-1 Overall, ΔH° = ΔbreakingH° + ΔmakingH° = (2220 – 2797) kJ mol-1 = -577 kJ mol-1

Answer: -577 kJ mol-1

CHEM1001 2012-J-9 June 2012 22/01(a)

• A 120.0 g piece of copper is heated to 80.0 °C before being added to 150.0 mL of water at 25.0 °C. What is the final temperature of the mixture? The specific heat capacity of copper is 0.385 J g–1 K–1 and the specific heat capacity of water is 4.18 J g–1 K–1.

Marks 3

The copper will cool down and the water will heat up when the two are mixed. The final temperature, Tf, will be the same for both. For the copper, qcopper = m c ΔT = (120.0 g) × (0.385 J g-1 K-1) × ΔTcopper

For the water, qwater = m c ΔT = (150.0 g) × (4.18 J g-1 K-1) × ΔTwater

As the heat lost by the copper is gained by the water, qwater = -qcopper: (150.0 g) × (4.18 J g-1 K-1) × ΔTwater = - (120.0 g) × (0.385 J g-1 K-1) × ΔTcopper or 627 × ΔTwater = - 46.2 × ΔTcopper

Using ΔTwater = (Tf – 25.0) oC and ΔTcopper = (Tf – 80.0) oC gives: Tf = 28.8 oC

Answer: 28.8 oC

CHEM1001 2012-J-12 June 2012 22/01(a)

Use average bond dissociation enthalpies given below to calculate the molar enthalpy

change for the following chemical transformation:

N2(g) + 3H2(g) 2NH3(g)

Marks

6

Bond H–H N–H NN

H / kJ mol–1

436 391 945

The reaction requires:

breaking 1 NN and 3 H-H bonds and

formation of 6 N-H bonds

The enthalpy required to break these bonds is:

H (bond breaking) = [945 (NN) + 3 × 436 (H-H)] kJ mol-1

= +2253 kJ mol-1

Enthalpy is released by making the new bonds is:

H (bond making) = - [6 × 391 (N-H)] kJ mol-1

= -2346 kJ mol-1

The overall enthalpy change is therefore:

H = [(+2253) + (-2346)] kJ mol-1

= -93 kJ mol-1

Answer: -93 kJ mol-1

What is the standard enthalpy of formation, fH, of NH3(g)? -47 kJ mol-1

(Note: the reaction in the question produces 2 mol of NH3 so the enthalpy of

formation is half of the enthalpy change of this reaction.)

The standard enthalpy of formation of hydrazine, N2H4(g) is +96 kJ mol–1

. Calculate

the strength of the N–N single bond in hydrazine.

The standard enthalpy of formation of N2H4(g) corresponds to the reaction:

N2(g) + 2H2(g) N2H4(g)

This reaction requires:

breaking 1 NN and 2 H-H bonds and

formation of 4 N-H and 1 N-N bond

The enthalpy required to break these bonds is:

H (bond breaking) = [945 (NN) + 2 × 436 (H-H)] kJ mol-1

= +1817 kJ mol-1

ANSWER CONTINUES ON THE NEXT PAGE

CHEM1001 2012-J-12 June 2012 22/01(a)

The standard enthalpy of formation of N2H4(g) corresponds to the reaction:

N2(g) + 2H2(g) N2H4(g)

This reaction requires:

breaking 1 NN and 2 H-H bonds and

formation of 4 N-H and 1 N-N bond

The enthalpy required to break these bonds is:

H (bond breaking) = [945 (NN) + 2 × 436 (H-H)] kJ mol-1

= +1817 kJ mol-1

Enthalpy is released by making the new bonds is:

H (bond making) = - [4 × 391 (N-H) + x (N-N)] kJ mol-1

= - (1564 + x) kJ mol-1

The overall enthalpy change is equal to the enthalpy of formation of N2H4:

fH = [(+1817) + -(1564 + x)] kJ mol-1

= +96 kJ mol-1

So

x = +157 kJ mol-1

Answer: +157 kJ mol

-1

Suggest why the N–N single bond in hydrazine is much weaker than the N–H and

H–H bonds. Hint: Draw its Lewis structure.

Each N atom in hydrazine has a lone pair of electrons and is the negative end of a

dipole formed with the H atoms.

These lone pairs repel each other, weakening the bond.

CHEM1001 2010-J-9 June 2010 22/01(a)

• How much energy is needed to convert 15 g of ice at 0.0 °C to water at 60.0 °C? The molar heat of fusion of water is 6.009 kJ mol–1 and the specific heat capacity of water is 4.18 J g–1 K–1.

Marks 3

The molar mass of H2O is (2 × 1.008 + 16.00) g mol-1 = 18.018 g mol-1. Hence, 15 g contains: number of moles = mass / molar mass = 15 g / 18.018 g mol-1 = 0.83 mol. Energy is required to (i) melt the ice and (ii) heat it; (i) 6.009 kJ is required to melt 1 mol. Hence, to melt 0.83 mol would require: qmelt = (0.83 mol) × (6.009 kJ mol-1) = 5.0 kJ (ii) The heat required to heat water from 0.0 °C to 60.0 °C is given by: qheat = mCΔT = (15 g) × (4.18 J g-1 K-1) × ((60.0 – 0.0) K) = 3800 J Overall; qtotal = qmelt + qheat = (5.0 × 103 J) + (3800 J) = 8800 J = 8.8 J

Answer: 8.8 J

CHEM1001 2010-J-9 June 2010 22/01(a)

• Calculate the standard enthalpy change for the combustion of 1.00 mol of propane gas, C3H8(g), to CO2(g) and H2O(l).

3

compound C3H8(g) CO2(g) H2O(l)

∆fH° / kJ mol–1 –105 –394 –286

The chemical equation for the combustion reaction is: C3H8(g) + 5O2(g) ���� 3CO2(g) + 4H2O(l) Using ∆rxnH° = Σm∆fH°(products) - Σn∆fH°(reactants), ∆rxnH° = [3 × ∆fH°(CO2(g)) + 4 × ∆fH°(H2O(l)) – [∆fH°(C3H8(g)] = ([(3 × -394) + (4 × -286)] – [-105]) kJ mol-1

= –2220 kJ mol–1

Answer: -2220 kJ mol-1

CHEM1001 2008-J-11 June 2008 22/01(a)

• The conversion of SO2 to SO3 can occur in the catalytic converters of cars using gasoline containing traces of sulfur compounds. Calculate the enthalpy change of the following reaction.

2SO2(g) + O2(g) → 2SO3(g)

Data: S(s) + O2(g) → SO2(g) ∆H = –296.8 kJ mol–1

2S(s) + 3O2(g) → 2SO3(g) ∆H = –791.4 kJ mol–1

Marks 2

Using ∆rH° = Σ m∆fH°(products) - Σ m∆fH°(reactants), the enthalpy change for the reaction can be written as

∆rH° = 2∆fH°(SO3(g)) - 2∆fH°(SO2(g)) As O2(g) is an element in its standard state, its ∆fH° = 0. The two reactions for which data are given correspond to the formation of SO2(g) and two moles SO3(g), respectively, from the elements in their standard states. Hence,

∆fH°(SO2(g)) = -296.8 kJ mol-1 2∆fH°(SO3(g)) = -791.4 kJ mol-1 or ∆fH°(SO3(g)) = -395.7

Hence,

∆rH° = (2 × -395.7) – (2 × -296.8) kJ mol-1 = -197.8 kJ mol-1

Answer: -197.8 kJ mol-1

CHEM1001 2007-J-7 June 2007 22/01(a)

• Calculate the standard heat of reaction for the following reaction.

Zn(s) + 2Cu+(aq) → 2Cu(s) + Zn2+(aq)

Data: ofH∆ = +51.9 kJ mol–1 for Cu+(aq)

ofH∆ = –152.4 kJ mol–1 for Zn2+(aq)

2

Using o o orxn f fH m H (products) n H (reactants)∆ = ∆ − ∆∆ = ∆ − ∆∆ = ∆ − ∆∆ = ∆ − ∆∑ ∑∑ ∑∑ ∑∑ ∑ , the heat of the

reaction as written is:

o o o 2 o orxn f f f f

1

H [2 H (Cu(s)) H (Zn (aq))] [ H (Zn(s)) 2 H (Cu (aq))

[(2 0) ( 152.4)] [(0) (2 51.9)] 256.2kJ mol

+ ++ ++ ++ +

−−−−

∆ = ∆ + ∆ − ∆ + ∆∆ = ∆ + ∆ − ∆ + ∆∆ = ∆ + ∆ − ∆ + ∆∆ = ∆ + ∆ − ∆ + ∆

= × + − − + × + = −= × + − − + × + = −= × + − − + × + = −= × + − − + × + = − o o

f fH (Cu(s)) H (Zn(s)) 0∆ = ∆ =∆ = ∆ =∆ = ∆ =∆ = ∆ = for elements already in their standard states.

Answer: -256.2 kJ mol-1

CHEM1001 2006-J-7 June 2006 22/01(a)

• A 1.00 g sample of ammonium nitrate, NH4NO3, is decomposed in a bomb calorimeter causing the temperature of the calorimeter to increase by 6.12 K. The heat capacity of the system is 1.23 kJ °C–1.

Marks 3

Describe this process as either endothermic or exothermic. exothermic

What is the molar heat of decomposition for ammonium nitrate?

The heat change is given by q = C × ∆T = 1.23 × 6.12 = 7.53 kJ. As the reaction is exothermic, the heat of decomposition for 1.00 g of NH4NO3 is ∆H = -7.53 kJ. The molar mass of NH4NO3 is:

molar mass = (2 × 14.01 (N)) + (4 × 1.008 (H)) + (3 × 16.00 (O)) = 80.052

1.000 g therefore corresponds to 1.00080.052

= 0.0125 mol. The molar heat of

decomposition is then: ∆H = -1-7.53= -602kJ mol

0.0125

Answer: -602 kJ mol-1

CHEM1001 2006-J-9 June 2006 22/01(a)

• Calculate the standard heat of reaction for the following reaction.

Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq)

Data: ∆Hof = +64.4 kJ mol–1 for Cu2+(aq)

∆Hof = –152.4 kJ mol–1 for Zn2+(aq)

Marks 2

The enthalpy of the reaction is given by:

o o orxn f f

o 2 o 2f f

1

H m H (products) n H (reactan ts)

[ H (Zn (aq))] [ H (Cu (aq))]

( 152.4) ( 64.4) 216.8 kJ mol

+ ++ ++ ++ +

−−−−

∆ = ∆ − ∆∆ = ∆ − ∆∆ = ∆ − ∆∆ = ∆ − ∆∑ ∑∑ ∑∑ ∑∑ ∑

= ∆ − ∆= ∆ − ∆= ∆ − ∆= ∆ − ∆

= − − + = −= − − + = −= − − + = −= − − + = −

(Note that of H (Zn(s))∆∆∆∆ and o

f H (Cu(s))∆∆∆∆ are both zero as these elements are in

the standard states).

Answer: -216.8 kJ mol-1

CHEM1001 2005-J-8 June 2005 22/01(a)

• A sealed 1.000 L flask at 30 °C contains air at a pressure of 1.000 atm. A 5.00 g sample of liquid water is injected into the flask and the flask heated to a temperature of 150 °C, causing the water to vaporise. What is the final pressure in the flask?

Marks 3

The pressure inside the flask will increase due to both the air and the vaporised water. H2O has a molar mass of 2×1.008 (H) + 16.00 (O) = 180.016. The number of moles of water is therefore:

moles of water = mass of water 5.00

= = 0.278 molmolar mass of water 18.016

For a perfect gas, PV = nRT. The pressure due to the water is therefore:

partial pressure of water = nRT /V = (0.278 mol)(0.08206 atm L mol-1 K -1)((150 + 273) K) / (1.000 L) = 9.64 atm

Note the use of R = 0.08206 L atm K-1 mol-1 allows the use of V = 1.000 L and gives the answers in atmospheres.

The pressure due to the air already in the flask increases due to the increase in temperature. As the number of moles of air and the volume stays the same:

1 2

1 2

P P=

T T or

212

1

P T (1.000)× (150 + 273)P = = =1.40atm

T (30 + 273)

(Alternatively, PV = nRT can be used to work out the number of moles of air present and then used again to find the pressure exerted at the higher temperature).

The total pressure is therefore:

P = (9.64 + 1.40) atm = 11.0 atm

Answer: 11.0 atm

CHEM1001 2005-J-9 June 2005

Consider the following reaction.

2C8H18(l) + 25O2(g) 16CO2(g) + 18H2O(l) E = –10909 kJ mol–1

A mixture of C8H18 (10.00 g) and O2 (30.00 g) is allowed to react. Assuming that the

reaction goes to completion, how much energy will be produced?

Marks

4

The molar mass of C8H18 is (8 × 12.01 (C) + 18×1.008 (H)) g mol-1

= 114.224 g

mol-1

. The number of moles of C8H18 = 𝟏𝟎.𝟎𝟎 𝐠

𝟏𝟏𝟒.𝟐𝟐𝟒 𝐠 𝐦𝐨𝐥−𝟏 = 0.08755 mol

The molar mass of O2 is (2 × 16.00) g mol-1

= 32.00 g mol-1

. The number of moles

of O2 is therefore 𝟑𝟎.𝟎𝟎 𝐠

𝟑𝟐.𝟎𝟎 𝐠 𝐦𝐨𝐥−𝟏 = 0.9375 mol.

25 moles of O2 is required for every 2 moles of C8H18 so 12.5 mol of O2 is

required for every 1 mol of C8H18.

The ratio of O2 : C8H18 is actually 0.9375

10.710.08755

so O2 is the limiting reagent.

10909 kJ is produced for every 25 mol of O2 that reacts. As 0.9375 mol of O2 are

present, the energy produced is:

ΔE = 𝟎.𝟗𝟑𝟕𝟓

𝟐𝟓 × 10909 kJ = 409.1 kJ

Answer: 409.1 kJ

CHEM1001 2004-J-5 June 2004

• Water solutions of NaOH (100 mL, 2.0 M) and HCl (100 mL, 2.0 M), both at 24.6 °C, were mixed together in a coffee cup calorimeter. The temperature of the solution rose to 38.0 °C during the reaction process. Write a balanced chemical equation to describe the reaction in the calorimeter.

Marks5

H+(aq) + OH–(aq) H2O(l)

Is the process an endothermic or exothermic reaction? Temperature increases so exothermic

Assuming a perfect calorimeter, determine the standard enthalpy change for the neutralisation reaction. Assume the density of water is 1.00 g mL–1. The heat capacity of water is 4.18 J K–1 g–1.

The total volume = 100.0 + 100.0 = 200.0 mL. If the density = 1.00 g mL-1, this has a mass of 200.0 g. The heat change is given by:

q = c × m × ΔT = (4.18 J K-1 g-1) × (200.0 g) × ((38.0 – 24.6) K) = 11202 J or 11.2 kJ

Both solutions have the same concentration. 200 mL of a 2.0 M solution contains (0.200 L) × (2.0 mol L-1) = 0.40 mol. The reaction is a 1:1 reaction between NaOH and HCl and equal amounts of each are present. The reaction of 0.20 mol of NaOH with 0.20 mol of HCl generates 11.2 kJ so the heat change for the reaction of 1 mol with 1 mol is:

heat change for 1 mol =

. = 56 kJ mol-1

The reaction is exothermic as the temperature rises. The enthalpy change is therefore:

ΔrH = −56 kJ mol-1

CHEM1001 2004-J-7 June 2004

Aluminium metal is a very effective agent for reducing oxides to their elements. For

example, it is used as a component of the solid fuel in the space shuttle, and in the

thermite reaction shown in lectures:

Fe2O3(s) + 2Al(s) Al2O3(s) + 2Fe(s)

Write a balanced equation for the reduction of CuO(s) to the base metal by Al(s).

Marks

5

3CuO(s) + 2Al(s) Al2O3(s) + 3Cu(s)

Given the following thermochemical data, evaluate the enthalpy change per gram of

reactant for the CuO and Fe2O3 reactions above.

Compound Hf (kJ mol–1

)

Fe2O3 -821

Al2O3 -1668

CuO -157

Using ΔrxnH° = ΣmΔfH°(products) - ΣnΔfH°(reactants), for the reaction of CuO

and Al, ΔrxnH ° = ([-1668] – [3 × -157]) kJ mol-1

= -1197 kJ mol-1

.

The formula mass of CuO is (63.55 (Cu) + 16.00 (O)) g mol-1

= 79.55 g mol-1

. The

atomic mass of Al is 26.98 g mol-1

. The enthalpy change is for the reaction of 3

moles of CuO and 2 moles of Al, corresponding to a total mass of (3 × 79.55 + 2 ×

26.98) g = 292.61 g.

The change per gram of reactants is therefore ΔH° = −𝟏𝟏𝟗𝟕 𝐤𝐉

𝟐𝟗𝟐.𝟔𝟏 𝐠 = -4.09 kJ g

-1

For the reaction of Fe2O3 and Al, ΔrxnH ° = ([-1668] – [-821]) = -847 kJ mol-1

.

The formula mass of Fe2O3 is ((2 × 55.85) + (3 × 16.00)) g mol-1

= 159.7 g mol-1

.

The enthalpy change is for the reaction of 1 mole of Fe2O3 and 2 moles of Al,

corresponding to a total mass of (159.7 + 2 × 26.98) g = 213.66 g.

The change per gram of reactants is therefore = −𝟖𝟒𝟕 𝐤𝐉

𝟐𝟏𝟑.𝟔𝟔 𝐠 = -3.96 kJ g

-1

Answer: CuO/Al: 4.09 kJ g-1

, Fe2O3/Al: 3.96 kJ g-1

ANSWER CONTINUES ON THE NEXT PAGE

Which would make the best rocket fuel on the basis of most energy provided per mass

of fuel (i.e. biggest “bounce per ounce”)?

The CuO / Al system is a better fuel on the basis of its “bounce per once”.

CHEM1001/CHEM1101 combined 2003-N-9 November 2003

Diborane (B2H6) is a highly reactive compound, which was once considered as a

possible rocket fuel for the US space program. Calculate the heat of formation of

diborane at 298 K from the following reactions.

Marks

2

Reaction Hr (kJ mol–1

)

1 2B(s) + 3/2O2(g) B2O3(s) –1273

2 B2H6(g) + 3O2(g) B2O3(s) + 3H2O(g) –2035

3 H2(g) + 1/2O2(g) H2O(l) –286

4 H2O(l) H2O(g) +44

The heat of formation of a compound refers to its formation from the elements in

their standard states.

In the reactions above, (1) and (3) correspond to the formation of B2O3(s) and

H2O(l) from their elements so:

ΔfH°(B2O3(s)) =–1273 kJ mol–1

and ΔfH°(H2O(l)) = –286 kJ mol–1

Formation of H2O(g) from its elements corresponds to formation of H2O(l) [-286

kJ mol-1

] followed by vaporization of H2O(l) [+44 kJ mol-1

(reaction (4))]. The heat

of formation of H2O(g) is therefore:

ΔfH°(H2O (g)) = (–286 + 44) kJ mol-1

= -242 kJ mol–1

Reaction (2) corresponds to the combustion of B2H6(g). The heat of combustion is

given by:

ΔrxnH° = ΣmΔfH°(products) - ΣnΔfH°(reactants)

= [ΔfH°(B2O3(s) + 3ΔfH°(H2O(g))] – [ΔfH°(B2H6(g) + 3ΔfH°(O2(g))]

= -2035 kJ mol-1

All of the ΔfH° values except that for B2H6(g) are known from above – the

ΔfH°(O2(g)) is zero because it is an element in its standard state. Substituting these

values in gives:

([-1273 + 3 × -242] – [ΔfH°(B2H6(g)) + 0] kJ mol-1

= -2035 kJ mol-1

or

ΔfH°(B2H6(g)) = [-1273 + 3 × -242 + 2035] kJ mol-1

= +36 kJ mol-1

Answer: +36 kJ mol

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