CHEM1001 2014-J-5 June 2014 22/01(a)

If the pressure remains constant at 1.00 atm, calculate the volume occupied by this mixture of gases after it was heated to 305 °C, before any reaction takes place.

Marks 6

The initial volume, V1, and temperature, T1, are related by: V1 / T1 = nR / P At the new volume, V2, and temperature, T2, they are similarly related by: V2 / T2 = nR / P As the number of moles and pressure are constant, these are equal: V1 / T1 = V2 / T2 At 298 K, the combined gases have a volume of 2.00 L so V1 = 2.00 L and T1 = 298 K. When T2 = 305 oC = 578 K: 2.00 L / 298 K = V2 / 578 K V2 = 3.88 L

Answer: 3.88 L

The molar heat capacity of N2H4 is 63 J K–1 mol–1 and that of N2O4 is 77 J K–1 mol–1. Calculate the heat capacity of this mixture.

From 2014-J-4, there are 0.0409 mol of N2H4 and N2O4 present in the mixture. Molar heat capacity are the heat capacity for one moles. The heat capacity of the mixture containing these amounts is therefore: heat capacity = (63 J K-1 mol-1 × 0.0409 mol) + (77 J K-1 mol-1 × 0.0409 mol) = 5.73 J K-1

Answer: 5.73 J K-1

Calculate the energy required to heat this mixture from 25 °C to 305 °C.

The temperature rise = ΔT = (305 – 25) = 280. K. Using q = CΔT, the heat needed is therefore: q = CΔT = (5.73 J K-1) × (280. K) = 1.60 kJ

Answer: 1.60 J or 1.60 × 103 kJ

ANSWER CONTINUES ON THE NEXT PAGE

CHEM1001 2014-J-5 June 2014 22/01(a)

Calculate the maximum mass of nitrogen gas that could be produced in this reaction.

From 2014-J-5, the balanced equation is: 2N2H4(g) + N2O4(g) → 3N2(g) + 4H2O(g) and there are 0.0409 mol of N2H4 and 0.0409 mol of N2O4 present. As each N2O4 requires 2N2H4, there is insufficient N2H4 for all of the N2O4 to react: N2H4 is the limiting reagent. From the balanced equation, 2 mol of N2H4 produces 3 mol of N2. Therefore: number of moles of N2 = 3/2 × 0.0409 mol = 0.0613 mol As the molar mass of N2 is 2 × 14.01 g mol-1 = 28.02 g mol-1. The mass of N2 in 0.0613 mol is therefore: mass = number of moles × molar mass = 0.0613 mol × 28.02 g mol-1 = 1.72 g

Answer: 1.72 g

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 2013-J-9 June 2013 22/01(a)

• A certain mixture of gases containing 0.24 mol of He, 0.53 mol of N2 and 0.05 mol of Ne is placed in a container with a piston that maintains it at a total pressure of 1.0 atm. This gas mixture is now heated from its initial temperature of 290 K to 370 K by passing 2.08 kJ of energy into it. Calculate the volume occupied by the gas at 370 K.

Marks 4

The total number of moles of gas = (0.24 + 0.53 + 0.05) mol = 0.82 mol. Using the ideal gas law, PV = nRT, the volume occupied is: V = nRT / P = (0.82 mol) × (0.08206 L atm K–1 mol–1) × (370 K) / (1.0 atm) = 25 L

Answer: 25 L

Calculate the heat capacity of the gas mixture (in J K–1 mol–1).

The temperature change = ΔT = (370 – 290) = 80 K. This is produced using a heat change of 2.08 kJ. Using q = nCΔT, the molar heat capacity is: C = q / nΔT = (2.08 × 103 J) / (0.82 mol × 80 K) = 30 J K-1 mol-1

Answer: 30 J K-1 mol-1

THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.

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 2009-J-10 June 2009 22/01(a)

• How much heat is evolved, in kJ, when 5.00 g of Al reacts with a stoichiometric amount of Fe2O3 according to the following equation?

2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s) ∆Ho = –852 kJ mol–1

1

The atomic mass of Al is 26.98 g mol-1 so 5.00 g corresponds to:

number of moles of Al = ����

������ ����

.��

��.�� ����� = 0.185 mol

From the chemical equation, 852 kJ mol-1 is evolved when 2 mol of Al reacts so 426 kJ is evolved when 1 mol of Al reacts. Therefore, when 0.185 mol of Al reacts, the heat evolved is (0.185 mol × 426 kJ mol-1) = 78.9 kJ.

Answer: 78.9 kJ

CHEM1001 2009-J-11 June 2009 22/01(a)

• The specific heat of Si is 0.71 J g–1 K–1. How much heat is required to heat a Si wafer weighing 0.45 g from 20.0 oC to 26.0 oC?

Marks 1

The heat, q, required to increase a solid of mass m by ∆T is given by q = mCv∆T where Cv is the specific heat capacity. Hence:

q = (0.45 g) × (0.71 J g-1 K -1) × ((26.0 – 20.0) K) = 1.9 J

Answer: 1.9 J

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

• If 78.2 J is required to raise the temperature of 45.6 g of lead by 13.3 °C, what is the specific heat of lead in J g–1 K–1?

1

The heat, q, required to change the temperature by an amount ∆T is given by:

q = cm∆T where m is the mass and c is the specific heat capacity. Hence as the temperature change is 13.3 °C or 13.3 K,

78.2 J = c × (45.6 g) × (13.3 K) or c = 0.129 J g-1 K -1

Answer:

CHEM1001 2007-J-6 June 2007 22/01(a)

A 60.0 g piece of Ag metal is heated to 90.0 °C and dropped into 120.0 g of water at

25.0 °C in a well insulated container. The final temperature of the Ag-H2O mixture

is 26.7 °C. Calculate the specific heat of silver.

Data: The specific heat of water is 4.18 J g–1

K–1

.

Marks

3

The temperature of the Ag metal decreases from 90.0 °C to 26.7 °C. The heat

change is given by:

heat change of Ag = qAg = cAg mAg ΔT = cAg × (60.0) × (26.7 – 90.0)

The temperature of the water changes from 25.0 °C to 26.7 °C. The heat change

is given by:

heat change of water = qwater = cwater mwater ΔT

= (4.18) × (120.0) × (26.7 – 25.0)

When the Ag metal is dropped into the water, the heat lost by the silver is gained

by the water until both reach the same temperature: -qAg = qwater:

-cAg × (60.0) × (26.7–90.0) = (4.18) × (120.0) × (26.7 – 25.0)

cAg = (4.18) (120.0) (26.7 25.0)

(60.0) (90.0 26.7)

0.22 J g

–1 K

–1

Answer: 0.22 J g

–1 K

–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-6 June 2005

A 0.50 g sample of ammonium nitrate, NH4NO3(s), was dissolved in 35.0 g of water

in a coffee cup calorimeter. The temperature of the solution dropped from 22.7 to

21.6 oC. Write a balanced equation to describe the reaction in the calorimeter.

Marks

5

NH4NO3(s) NH4+(aq) + NO3

–(aq)

Describe this process as either endothermic or

exothermic. Temperature decreases so

endothermic

Assuming a perfect calorimeter what is the heat of solution of ammonium nitrate,

expressed in kJ mol–1

? Assume the density of the solution is 1.00 g mL–1

and that the

heat capacity of the solution is 4.18 J K–1

g–1

.

The molar mass of NH4NO3 is (14.01 (N) + 4 1.008 (H) + 14.01 (N) + 3 16.00

(O)) g mol-1

= 80.052 g mol-1

. The sample of 0.50 g therefore corresponds to:

number of moles = 𝟎.𝟓𝟎 𝐠

𝟖𝟎.𝟎𝟓𝟐 𝐠 𝐦𝐨𝐥−𝟏 = 0.0062 mol

The total mass of NH4NO3 and water is (0.50 + 35.0) g = 35.5 g. The heat change

is given by:

q = c m ΔT

= (4.18 J K-1

g-1

) (35.5 g) ((22.7 – 21.6) K) = 163 J or 0.163 kJ

This is the heat change produced by 0.0062 mol. The heat change produced by 1

mol is therefore 𝟏𝟔𝟑 𝐉

𝟎.𝟎𝟎𝟔𝟐 𝐦𝐨𝐥 = 26 kJ mol

-1

The reaction is endothermic as the temperature drops. The enthalpy change is

therefore:

ΔrH = 26 kJ mol-1

Heat radiating fins are used to dissipate heat and prevent damage to electronic

components. Is it better to make the fins out of aluminium or iron? Give reasons

for your answer.

Data: Specific heat of Al = 0.900 J K–1

g–1

Specific heat of Fe = 0.444 J K–1

g–1

2

The fins are required to remove heat from the electronic components.

Aluminium has a higher heat capacity so it can absorb more heat per gram.

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/CHEM1101 combined 2003-N-7 November 2003

• A 50.0 mL solution contained 10.00 g of NaOH in water at 25.00 °C. When it was added to a 250.0 mL solution of 0.200 M HCl at 25.00 °C in a “coffee cup” calorimeter, the temperature of the solution rose to 33.95 °C. Assuming the specific heat of the solution is 4.18 J K–1 g–1, that the calorimeter absorbs a negligible amount of heat, and that the density of the solution is 1.00 g mL–1, calculate ΔHr (in kJ mol–1) for the following reaction. H+(aq) + OH–(aq) → H2O(l)

Marks4

The total volume = (50.0 + 250.0) mL = 300.0 mL. If the density = 1.00 g mL-1, this has a mass of 300.0 g. The heat change is given by:

q = c × m × Δ = (4.18 J K-1 g-1) × (300.0 g) × ((33.95 – 25.00) K) = 11223 J or 11.22 kJ

NaOH has a molar mass of (22.99 (Na) + 16.00 (O) + 1.008 (H)) g mol-1 = 40.00 g mol-1 so 10.00 g corresponds to 0.2500 mol. 250.0 mL of a 0.200 M solution of HCl contains (0.2500 mL) × (0.200 mol L1) = 0.0500 mol. As the reaction is a 1:1 reaction between NaOH and HCl, only 0.0500 mol of the NaOH can react. The heat change is therefore associated with 1 mol is therefore:

heat change for 1 mol =

. = 225 kJ mol-1

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

ΔrH = −225 kJ mol-1

When the experiment was repeated using 12.00 g of NaOH in water, the temperature increase was the same. Explain.

As noted above, 10.00 g of NaOH contains more moles than 250.0 mL of 0.200 M HCl. 12.00 g of NaOH therefore contains even more moles and so is still in excess. This excess cannot react and so there is no change in the amount of heat produced.

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

-1