Enthalpy of Reaction and Calorimetry worksheet

1. Calcium carbonate decomposes at high temperature to form carbon dioxide and calcium oxide,calculate the enthalpy of reaction.

CaCO3 CO2 + CaO

Given that the enthalpy of formation of calcium carbonate is –1206.9 kJ/mol, the enthalpy offormation of carbon dioxide is –393.5 kJ/mol, and of calcium oxide is –635.5 kJ/mol, determinethe enthalpy of reaction.

ΔHrxn = ΔHf(products) - ΔHf(reactants)ΔHrxn = [1(-393.5 kJ/mol) + 1(-635.5 kJ/mol)] – [1(-1206.9 kJ/mol)]ΔHrxn = -1029.0 kJ/mol + 1206.9 kJ/molΔHrxn = +177.9 kJ/mol

1. Carbon tetrachloride can be formed by reacting chlorine with methane, calculate the enthalpy ofreaction.

CH4 + 2 Cl2 CCl4 + 2 H2

Given that the enthalpy of formation of methane is –74.4 kJ/mol and the enthalpy of formation ofcarbon tetrachloride is –135.4 kJ/mol, determine the enthalpy of reaction.

ΔHrxn = ΔHf(products) - ΔHf(reactants)ΔHrxn = [1(-135.4 kJ/mol)] – [1(-74.4 kJ/mol)]ΔHrxn = -135.4 kJ/mol + 74.4 kJ/molΔHrxn = -61.0 kJ/mol

1. When potassium chloride reacts with oxygen under the right conditions, potassium chlorate is formed:

2 KCl + 3 O2 2KClO3

Given that the enthalpy of formation of potassium chloride is –436.0 kJ/mol and the enthalpy offormation of potassium chlorate is –391.0 kJ/mol, determine the enthalpy of reaction.

ΔHrxn = ΔHf(products) - ΔHf(reactants)ΔHrxn = [2(-391.0 kJ/mol)] – [2(-436.0 kJ/mol)]ΔHrxn = -782.0 kJ/mol + 872.0 kJ/molΔHrxn = +90.0 kJ/mol

Calorimetry Practice

1. Compound A is burned in a bomb calorimeter that contains 2.50 liters of water. If the combustion of0.175 moles of this compound causes the temperature of the water to rise 45.00 C, what is the molarheat of combustion of compound A?

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ΔHrxn = - mcΔTn

= - (2.50 x 103 g)(4.18 J/g°C)(45.0°C) 0.175 mol

= - 470.3 kJ0.175 mol

= - 2 690 kJ/mol

1. Compound B is burned in a bomb calorimeter that contains 1.50 liters of water. When I burned 50.0grams of compound B in the calorimeter, the temperature rise of the water in the calorimeter was35.00 C. If the heat of combustion of compound B is 2,150 kJ/mol, what is the molar mass ofcompound B?

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ΔHrxn = - mcΔTn

= - msurrcΔTmsys

x Msys

-2 150 kJ/mol = - (1.50 x 103 g)(4.18 J/g°C)(35.0°C) 50.0 g

x Msys

Msys = 2 150 kJ • mol-1 x 50.0 g(1.50 x 103 g)(4.18 J/g°C)(35.0°C)

= 107 500 kJ • mol-1 •g219.45 kJ

= 490. g/mol

1. The molar heat of combustion of compound C is 1,250 kJ/mol. If I were to burn 0.115 moles of thiscompound in a bomb calorimeter with a reservoir that holds 2.50 L of water, what would the expectedtemperature increase be?

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ΔHrxn = - mcΔTn

-1 250 kJ/mol = - (2.50 x 103 g)(4.18 J/g°C)ΔT 0.115 mol

ΔT = 1 250 kJ • mol-1 x 0.115 mol(2.50 x 103 g)(4.18 J/g°C)

= 143.75 kJ10.45 kJ •°C-1 = 13.8 °C

INV. 5.1.1 MOLAR ENTHALPY of SOLUTION - MEDICAL COLD PACKS Part 21. Describe the difference between heat and temperature.

Heat energy is the energy due the position and random motion of particles. It isthermal energy being transferred. Temperature is a measure of average kineticenergy.

2. For the compounds used in this experiment (Nelson Chemistry 12 p. 347):(a) Calculate the number of moles present in a 10.00 g sample.

mol 0.134 KCl mol 0.0735 OH 3 OHNaCmol 0.125 NONH mol 0.0989 KNO

Similarly;

mol 0.187 g 53.50

ClNH mol 1 x g 10.0 Mm n

kJ/g 0.277 Cl)(NHH 53.50g/mol M g 10.0 m

2232

343

4ClNH

4sClNHClNH

4

44

==•

==

===

===

(b) Calculate the molar enthalpy

kJ/mol 17.2 KCl kJ/mol 19.6 OH 3 OHNaCkJ/mol 25.7 NONH kJ/mol 34.9 KNO

Similarly;

kJ/mol 14.8 ClNH mol 1

g 53.50 x g

kJ 0.277

M x Cl)(NHH Cl)(NHH

kJ/g 0.277 Cl)(NHH 53.50g/mol M g 10.0 m

2232

343

4

ClNH4s4sol

4sClNHClNH

4

44

==•

==

==

=Δ

===

(c) Write the thermochemical chemical equation for your compound's dissociation in water.

NH4Cl (s) + 14.8 → NH4+(aq) + Cl-

(aq)KNO3 (s) + 34.9 → K+

(aq) + NO3-(aq)

NH4NO3 + 25.7 → NH4+(aq) + NO3

-(aq)

NaC2H3O2•3H2O (s) + 19.6 → Na+(aq) + CH3CO2

-(aq) + 3 H2O (l)

KCl (s) + 17.2 → K+(aq) + Cl-

(aq)3. (a) When a substance dissolves its heat content, or enthalpy, will either increase or decrease. This

change in enthalpy is observed as a change in temperature of the water in the calorimeter.Sketch a graph showing the change in enthalpy for the dissolving of an ionic compound which is:

(i) exothermic (ii) endothermic

(b) For your compound, state whether the dissolving was an endothermic or exothermic process.

endo(c) Will the sign for _H in your case be positive or negative?

negative4. Describe two factors that affect the strength of an ionic bond in a compound.

Ion charge and ion size.5. (a) Define crystal lattice energy and hydration energy.

Lattice Energy – the energy required to separate the ions of a crystal to an infinite distance from each other.Hydration Energy – the energy released when ions associate with water to become aqueous.

(b) Which of the properties in part (a) reduces the solubility of an ionic compound?

As lattice energy increases, solubility decreases(c) Which property, when increased, increases the solubility?

As hydration energy increases, solubility increases.6. For an ionic solid to melt, external energy must be applied to overcome the attractive forces holding

its ions together. For example, sodium hydroxide must be heated to temperatures exceeding 318°C forit to melt. Despite such a high melting point, sodium hydroxide readily and exothermally dissolvesand dissociates into its ions in water at room temperature.

(a) What is the source of the energy required to separate the ions from the solid sodium hydroxide?

Hydration energy of ions.(b) Why does the resulting sodium hydroxide solution feel hot?

Hydration energy >> Lattice energytherefore exothermic dissolving process.

7. Ammonium chloride is also soluble despite having a melting point of 320°C. Why does an ammoniumchloride solution feel cold to the touch?

Hydration energy << Lattice energytherefore endothermic dissolving process.

8. Magnesium oxide, MgO, and magnesium chloride, MgCl2, are very similar, white, ionic solids withthe following properties:

Compound Melting Point SolubilityMgO 2800°C insolubleMgCl2 1412°C very soluble

(a) Give the formula of the ions of each compound.

Mg+2 and O-2 ; Mg+2 and Cl-

(b) Account for the drastic difference in physical properties.

MgO has a smaller and more negatively charged ion therefore the force of attractionand the crystal lattice energy is higher.

9. Consider the following data from the Group 1 chlorides. These compounds are similar in structure,each being formed from a + cation and the -1 chloride ion. However, they are quite different withrespect to lattice energy, hydration energy, and enthalpy of solution. Explain these differences.

Substance Lattice Energy (kJ•mol-1)

Hydration Energy(kJ•mol-1)

Enthalpy of Solution(kJ•mol-1)

LiCl 833 -883 -50NaCl 766 -770 -4KCI 630 -626 +4

As you move down the Group, the lattice energy of the chlorides decreases due togreater ion separation, hydration energy of the ions decreases since the larger metalion is further from the hydrating water and therefore less Potential Energy is lost andthe bond forms at a greater distance. The enthalpy of solution decreases because asmetal size increases the decreasing hydration energy is more significant than thedecreasing lattice energy.

SUPPLEMENTARY ENTHALPY PROBLEMS

1. Construct an enthalpy diagram showing the enthalpy changes for a one step conversion ofgermanium, Ge(s), into GeO2 (s) and a two step conversion - first to the monoxide, GeO(s) followed bythe oxidation to GeO2 (s) . The relevant thermochemical equations are as follows;

Ge(s) + 1/2 O2 (g) GeO(s) ∆H° = -255 kJGe(s) + O2(g) GeO2 (s) ∆H° = -534.7 kJ

Using this diagram, determine the value of ∆H° for the reaction

GeO(s) + 1/2 O2 (g) GeO2 (s)

2. Nitrogen monoxide oxidizes to the toxic red brown gas nitrogen dioxide in air. Construct anaccurate enthalpy diagram for this reaction. Use the diagram to explain which of the oxides of nitrogenhas stronger covalent bonds.

2 NO(g) + O2 (g) 2 NO2 (g)

∆H° = 2 ∆Hf°(NO2 (g) ) - 2 ∆Hf° (NO(g) )∆H° = 2(33.81 kJ) - 2(90.29 kJ)∆H° = -112.98 kJ

PE

(kJ)

0Ge + O2

GeO-255

-534.7GeO2

ΔH = -534.7 kJ - (-255 kJ)

= -279.7 kJ

PE (kJ)

0

- 112.98

2 NO + O2

2 NO2

3. One of the “building blocks” for proteins such as those in muscles and in sinews, is an amino acidcalled glycine, C2H5NO2. The equation for its combustion is

4 C2H5NO2 (s) + 9 O2 (g) 8 CO2 (g) + 10 H2O(l) + 2 N2 (g)

The value of its ∆H°combustion is -973.49 kJ/mol. Calculate ∆H

°f for glycine.

4(-973.49 kJ) = 10(-285.9 kJ) + 8(-393.5 kJ) - 4∆Hf°(glycine)-3893.96 kJ = -2859 kJ -3148 kJ - 4∆Hf°(glycine)4∆Hf°(glycine) = -2113.04 kJ∆Hf°(glycine) = 528.26 kJ

4. On Easter Sunday, April 3, 1983, nitric acid spilled from a tank car near Denver, Colorado. The spillwas neutralized with sodium carbonate. The reaction is

2 HNO3 (aq) + Na2CO3 (s) 2 NaNO3 (aq) + H2O(l) + CO2 (g)

(a)Calculate ∆H° for this reaction. [∆H°f for NaNO3 (aq) is -467 kJ/mol, ∆H

°f for HNO3 (aq) is -

207.16 kJ/mol, ∆H°f for Na2CO3 (aq) is -1129.60 kJ/mol]

∆H° = ∆Hf° CO2 + ∆Hf° H2O + 2 ∆Hf° NaNO3 - 2 ∆Hf° HNO3 - ∆Hf° Na2CO3∆H° = -393.5 kJ - 285.9 kJ + 2(-467 kJ) - 2(-207.16 kJ) - (-1129.60 kJ)∆H° = -69.48 kJ

(b) Approximately 9.1 x 104 L of nitric acid was spilled. Assuming a molar concentration of 15.4M,how much sodium carbonate was required for complete neutralization and how much heat wasevolved.

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2 HNO3 + Na2CO3 → 2 NaNO3 + H2O + CO2V = 9.1 x 104 L m = ?C = 15.4 M M = 105.99g/mol

m Na2CO3 = CHNO3 VHNO3 x # mol Na2CO3# mol HNO3

x M Na2CO3

= 15.4 mol HNO3L

x 9.1 x 104 L x 1 mol Na2CO32 mol HNO3

x 105.99g1 mol Na2CO3

= 7.4 x 107 gHeat evolved = q = ΔHrxn(HNO3) x nHNO3

= - 69.48 kJ x 12 mol HNO3

x 15.4 mol HNO3L

x 9.1 x 104 L

= - 4.87 x 107 kJ(c) According to the Denver Post for April 4, 1983, authorities feared a volatile reaction might occurduring the neutralization. Considering the magnitude of ∆H°, what was their major concern?

The large amount of heat energy released in this neutralization reaction could cause the decompositionof the nitrate compounds into toxic oxides of nitrogen.

Heats of Reaction

Using the Standard values of Η°f , calculate the heats of reaction, ΔΗ°rxn, or heats of formation, ΔΗ°fAll answers must include the intermediate steps with the formation reaction from the elements and theirmanipulation to form the overall reaction.

For #1-6, the equation: CANNOT BE USED.

eg. SO2(g) + H2O(l) H2SO4(aq) ∆H° kJ

(1) 1/8 S8 + O2 SO2 -296.8(2) H2 + 1/2 O2 H2O -285.8(3) H2 + 1/2 O2 + 1/8 S8 H2SO4 -909.3Then multiply or reverse, whatever is needed so that they add to the original reaction.

1. C6H6 (l) + 15/2 O2 (g) —> 6 CO2 (g) + 3 H2O (g)∆H° kJ

(1) 6 C (s) + 3 H2 (g) —> C6H6 (l) + 49.0

(2) C (s) + O2 (g) —> CO2 (g) - 393.5

(3) H2 (g) + 1/2 O2 (g) —> H2O (g) - 241.8

-1 x (1) C6H6 (l) —> 6 C (s) + 3 H2 (g) - 49.0

6 x (2) 6 C (s) + 6 O2 (g) —> 6 CO2 (g) - 2361.0

3 x (3) 3 H2 (g) + 3/2 O2 (g) —> 3 H2O (g) - 725.4_____________________________________________________________________________________

C6H6 (l) + 15/2 O2 (g) —> 6 CO2 (g) + 3 H2O (g) - 3135.4

2. 2 HNO3 (aq) + NO (g) —> 3 NO2 (g) + H2O (l)∆H° kJ

(1) 1/2 H2 (g) + 1/2 N2 (g) + 3/2 O2 (g) —> HNO3 (aq) - 207.0

(2) 1/2 N2 (g) + 1/2 O2 (g) —> NO (g) + 90.2

(3) 1/2 N2 (g) + O2 (g) —> NO2 (g) + 33.2

(4) H2 (g) + 1/2 O2 (g) —> H2O (l) - 285.8

-2 x (1) 2 HNO3 (aq) —> H2 (g) + N2 (g) + 3 O2 (g) + 414.0

-1 x (2) NO (g) —> 1/2 N2 (g) + 1/2 O2 (g) - 90.2

3 x (3) 3/2 N2 (g) + 3 O2 (g) —> 3 NO2 (g) + 99.6

(4) H2 (g) + 1/2 O2 (g) —> H2O (l) - 285.82 HNO3 (aq) + NO (g) —> 3 NO2 (g) + H2O (l) + 137.6

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ΔHrxno = ΣnΔHproductso − ΣnΔHreactants

o

3. C2H2 (g) + C2H6 (g) —> 2 C2H4 (g)∆H° kJ

(1) 2 C (s) + H2 (g) —> C2H2 (g) + 228.2

(2) 2 C (s) + 3 H2 (g) —> C2H6 (g) - 83.8

(3) 2 C (s) + 2 H2 (g) —> C2H4 (g) + 52.5

-1 x (1) C2H2 (g) —> 2 C (s) + H2 (g) - 228.2

-1 x (2) C2H6 (g) —> 2 C (s) + 3 H2 (g) + 83.8

2 x (3) 4 C (s) + 4 H2 (g) —> 2 C2H4 (g) + 105.0_____________________________________________________________________________________

C2H2 (g) + C2H6 (g) —> 2 C2H4 (g) - 39.4

4. PbO2 (s) + CO (g) —> PbO (s) + CO2 (g)∆H° kJ

(1) Pb (s) + O2 (g) —> PbO2 (s) - 277.4(2) C (s) + 1/2 O2 (g) —> CO (g)- 110.5

(3) Pb (s) + 1/2 O2 (g) —> PbO (s) - 219.0(4) C (s) + O2 (g) —> CO2 (g) - 393.5

-1 x (1) PbO2 (s) —> Pb (s) + O2 (g) + 277.4-1 x (2) CO (g) —> C (s) + 1/2 O2 (g) + 110.5

(3) Pb (s) + 1/2 O2 (g) —> PbO (s) - 219.0(4) C (s) + O2 (g) —> CO2 (g) - 393.5

_____________________________________________________________________________________PbO2 (s) + CO (g) —> PbO (s) + CO2 (g) - 224.6

5. Al2(SO4)3 (s) + 3 H2O (l) —> Al2O3 (s) + 3 H2SO4 (aq)∆H° kJ

(1) 2 Al (s) + 3/8 S8 (s) + 6 O2 (g) —> Al2(SO4)3 (s) - 3405.5(2) H2 (g) + 1/2 O2 (g) —> H2O (l) - 285.8(3) 2 Al (s) + 3/2 O2 (g) —> Al2O3 (s) - 1675.7(4) H2 (g) + 1/8 S8 (s) + 2 O2 (g) —> H2SO4 (aq) - 909.3

-1 x (1) Al2(SO4)3 (s) —> 2 Al (s) + 3/8 S8 (s) + 6 O2 (g) +3405.5-3 x (2) 3 H2O (l) —> 3 H2 (g) + 3/2 O2 (g) + 857.4

(3) 2 Al (s) + 3/2 O2 (g) —> Al2O3 (s) - 1675.7 3 x (4) 3 H2 (g) + 3/8 S8 (s) + 6 O2 (g) —> 3 H2SO4 (aq) - 2727.9_____________________________________________________________________________________

Al2(SO4)3 (s) + 3 H2O (l) —> Al2O3 (s) + 3 H2SO4 (aq) - 140.7

6. 3 SO2 (s) + 2 HNO3 (aq) + 2 H2O (l) —> 3 H2SO4 (aq) + 2 NO (g)∆H° kJ

(1) 1/8 S8 (s) + O2 (g) —> SO2 (s) - 296.8(2) 1/2 H2 (g) + 1/2 N2 (g) + 3/2 O2 (g) —> HNO3 (aq) - 207.0(3) H2 (g) + 1/2 O2 (g) —> H2O (l) - 285.8(4) H2 (g) + 1/8 S8 (s) + 2 O2 (g) —> H2SO4 (aq) - 804.3(5) 1/2 N2 (g) + 1/2 O2 (g) —> NO (g) + 90.2

-3 x (1) 3 SO2 (s) —> 1/8 S8 (s) + O2 (g) + 890.4-2 x (2) 2 HNO3 (aq) —> H2 (g) + N2 (g) + 3 O2 (g) + 414.0-2 x (3) 2 H2O (l) —> 2 H2 (g) + O2 (g) + 571.6 3 x (4) 3 H2 (g) + 3/8 S8 (s) + 6 O2 (g) —> 3 H2SO4 (aq) - 2412.9 2 x (5) N2 (g) + O2 (g) —> 2 NO (g) + 180.4_____________________________________________________________________________________3 SO2 (s) + 2 HNO3 (aq) + 2 H2O (l)—> 3 H2SO4 (aq) + 2 NO (g) - 356.5

7. Given that ∆H°rxn = - 1196.0 kJ for the following, calculate the ∆H°f for ClF3 (g).

2 ClF3 (g) + 2 NH3 (g) —> N2 (g) + 6 HF (g) + Cl2 (g)

∆H°rxn = Σ n ∆H°f products - Σ n ∆H°f reactants

-1196.0 kJ = (1(0 kJ) + 6(-271.1 kJ) + 1(0 kJ)) - (2 mol ∆H°f (ClF3) + 2 (-45.9 kJ))

-1196.0 kJ = - 1626.6 kJ - (2 mol ∆H°f (ClF3) -91.8 kJ)

2 mol ∆H°f (ClF3) = - 338.8 kJ

∆H°f (ClF3) = - 169.4 kJ/mol

8. Given that ∆H°rxn = - 52.3 kJ for the following, calculate the ∆H°f for HNO2 (g).

HNO2 (g) + 1/2 O2 (g) —> HNO3 (aq)

∆H°rxn = Σ n ∆H°f products - Σ n ∆H°f reactants

-52.3 kJ = (1(-207.0 kJ)) - (1 mol ∆H°f (HNO2) + 1/2 (0 kJ))

-52.3 kJ = -207.0 kJ - 1 mol ∆H°f (HNO2)

∆H°f (HNO2) = - 154.7 kJ/mol

9. Given that ∆H°rxn = - 386.2 kJ for the following, calculate the ∆H°f for Fe(CO)5 (g).

Fe2O3 (s) + 13 CO (g) —> 2 Fe(CO)5 (g) + 3 CO2 (g)

∆H°rxn = Σ n ∆H°f products - Σ n ∆H°f reactants

-386.2 kJ = (2 mol ∆H°f (Fe(CO)5) + 3(-393.5 kJ/mol)) - (1(-824.2 kJ) + 13(-110.5 kJ))

-386.2 kJ = 2mol ∆H°f (Fe(CO)5) - 1180.5 kJ + 824.2 kJ + 1436.5 kJ

∆H°f (Fe(CO)5) = - 733.2 kJ/mol

10. Given that ∆H°rxn = + 47.2 kJ for the following, calculate the ∆H°f for Fe3O4 (s).

2 Fe3O4 (s) + CO2 (g) —> 3 Fe2O3 (s) + CO (g)

+47.5 kJ = (3(-824.2 kJ) + 1(-110.5 kJ)) - (2 mol ∆H°f (Fe3O4) + 1(-393.5 kJ))

+47.5 kJ = - 2472.6 kJ - 110.5 kJ - 2mol ∆H°f (Fe3O4) + 393.5 kJ

∆H°f (Fe3O4) = - 1118.6 kJ/mol

Practice Thermochemistry Questions1. a) Urea (from urine) hydrolyzes slowly in the presence of water to produce ammonia and carbon

dioxide. What is the standard ∆H°, ∆S° and ∆G° for this reaction when 1 mole of urea reacts withwater?

CO(NH2)2(aq) + H2O(l) _ CO2(g) + 2 NH3(g)

∆H° = ∑ ∆H°products - ∑ ∆H°reactants= [CO2(g) + (2) NH3(g)] - [CO(NH2)2(aq) + H2O(l)]= [-393.5 kJ + (2) -45.9 kJ] - [-335.5 kJ + -285.8 kJ]= (-485.3 kJ) - (-621.3 kJ)= +136.0 kJ

∆S° = ∑ ∆S°products - ∑ ∆S°reactants= [CO2(g) + (2) NH3(g)] - [CO(NH2)2(aq) + H2O(l)]= [213.7 J/K + (2) 192.8 J/K] - [104.6 J/K + 69.9 J/K]= (599.3 J/K) - (174.5 J/K)= +424.8 J/K

∆G° = ∆H° - T ∆S°= +136.0 kJ – 298.15 K x +0.4248 kJ/K= +9.3 kJ

b) What is the spontaneity of this reaction?

As _G° is positive, the reaction is not spontaneous.

2. What is _G° for the combustion of liquid ethyl alcohol (C2H5OH) to give CO2(g) and H2O(g)? Isthe reaction spontaneous?

C2H5OH (l) + 3 O2(g) _ 2 CO2(g) + 3 H2O(g)

∆G° = ∑ ∆G°products - ∑ ∆G°reactants= [(2) CO2(g) + (3) H2O(g)] - [C2H5OH(l) + (3) O2(g)]= [(2) -394.4 kJ + (3) -228.6 kJ] - [-174.8 kJ + (3) 0 kJ]= (-1 474.6 kJ) - (-174.8 kJ)= -1 299.8 kJ

3. Calculate _H° and _S° for the following reaction and decide in which direction each of thesefactors will drive the reaction.

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

∆H° = ∑ ∆H°products - ∑ ∆H°reactants= [(2) NH3(g)] - [N2(g) + (3) H2(g)]= [(2) -45.9 kJ] - [0 kJ + (3) 0 kJ]= -91.8 kJ ∴ spontaneous

∆S° = ∑ ∆S°products - ∑ ∆S°reactants= [(2) NH3(g)] - [N2(g) + (3) H2(g)]= [(2) +192.8 J/K] - [+191.6 J/K + (3) 130.7 J/K]= (+385.6 J/K) - (583.7 J/K)= -198.1 J/K ∴ non spontaneous

Supplementary Thermochemistry Problems

1. a) Calculate the enthalpy change, ΔHrxn°, for the following reaction using equations 1, 2 and 3.

FeO + CO → Fe + CO2Given: ΔHrxn° (kJ)

1) Fe2O3 + 3 CO → 2 Fe + 3 CO2 -252) 3 Fe2O3 + CO → 2 Fe3O4 + CO2 -473) Fe3O4 + CO → 3 FeO + CO2 +38

3 x 1) 3 Fe2O3 + 9 CO → 6 Fe + 9 CO2 (3) -25 = -75-1 x 2) 2 Fe3O4 + CO2 → 3 Fe2O3 + CO (-1)-47 = +47-2 x 3) 6 FeO + 2 CO2 → 2 Fe3O4 + 2 CO (-2)+38 = -76

Then: 6 FeO + 6 CO → 6 Fe + 6 CO2 -104FeO + CO → Fe + CO2 -17.3

2. Calculate the enthalpy change, ΔHrxn°, for the following reaction using equations 1, 2 and 3.

3 N2H4 + 4 ClF3 → 3 N2 + 12 HF + 2 Cl2Given: ΔHrxn° (kJ)

1) 2 ClF3 + 2 NH3 → N2 + 6 HF + Cl2 -11962) N2H4 + O2 → N2 + 2 H2O -6223) 4 NH3 + 3 O2 → 2 N2 + 6 H2O -1530

2 x 1) 4 ClF3 + 4 NH3 → 2 N2 + 12 HF + 2 Cl2 (2) -1196 = -2392 3 x 2) 3 N2H4 + 3 O2 → 3 N2 + 6 H2O (3)-622 = -1866-1 x 3) 2 N2 + 6 H2O → 4 NH3 + 3 O2 (-1)-1530 = 1530

Then 3 N2H4 + 4 ClF3 → 3 N2 + 12 HF + 2 Cl2 -2728

3. Calculate the enthalpy change, ΔHrxn°, for the following reaction using equations 1, 2 and 3.

2 H3BO3 → B2O3 + 3 H2OGiven: ΔHrxn° (kJ)

1) H3BO3 → HBO2 + H2O -0.0200

2) 2 B2O3 + H2O → H2B4O7 -17.5

3) H2B4O7 + H2O → 4 HBO2 -11.3

4 x 1) 4 H3BO3 → 4 HBO2 + 4 H2O (4)-0.0200 = -0.0800

-1 x 2) H2B4O7 → 2 B2O3 + H2O (-1)-17.5 = +17.5

-1 x 3) 4 HBO2 → H2B4O7 + H2O (-1)-11.3 = +11.3

Then 4 H3BO3 → 2 B2O3 + 6 H2O +28.72 H3BO3 → B2O3 + 3 H2O +14.4

4. Using the average bond energies, N≡N (941 kJ/mol), N=N (418 kJ/mol), N-N (160 kJ/mol), N-H (391kJ/mol) and H-H (432 kJ/mol) estimate the ΔHf° for hydrazine, N2H4.

N2 + 2 H2 → N2H4∆Hf° = ∆BE reactants – ∆BE products

= [(1) N≡N + (2) H-H] – [(1) N-N + (4) N-H]= [941 kJ + (2)432 kJ] – [160 kJ + (4)391 kJ]= [1805 kJ] – [1724 kJ]= +81 kJ/mol

5. The ΔHrxn° of the gaseous reaction:

XeF2 + H2 → 2 HF + Xe

is found to be –430 kJ. Use this value and average bond energies [H-F (565 kJ/mol)] to determinethe average bond energy of the Xe-F bond.

∆Hf° = ∆BE reactants – ∆BE products -430 kJ = [(2) Xe-F + (1) H-H] – [(2) H-F] -430 kJ = [(2) Xe-F + 432 kJ] – [(2)565 kJ] -430 kJ = [(2) Xe-F +432 kJ] – [1130 kJ] -430 kJ = (2) Xe-F -698 kJ(2) Xe-F = 268 kJ Xe-F = 134 kJ/mol

KINETICS – Rate laws

1. Using the data provided below, draw a Born-Haber cycle for the formation of lithium fluoride andcalculate the enthalpy of formation of the compound

Li (s) + 1/2 F2 (g) → LiF (s)

Given: ΔH vap (Li metal) = 155 kJ/mol BE of F-F = 150 kJ/mol1st IE of Li atom = 520 kJ/mol E.A. of F atom = 333 kJ/molLattice energy = 1012 kJ/mol

Li (s) + 1/2 F2 (g)

Li (g) + 1/2 F2 (g)

Li (g) + F (g)

Li +(g) + F (g) + e-

ΔHvap(Li) = +155 kJ

1/2 BE (F-F) = +75 kJ

1st I.E. Li = + 520 kJLi + (g) + F - (g)

E.A. of F = - 333 kJ

LiF (s)

ΔHf = -595 kJ

lattice energy = -1012 kJ

2. Draw a Bohr-Haber cycle for the following reaction:

H2O (l) → H2 (g) + 1/2 O2 (g)

Given: BE for H-H = 431. kJ/mol BE for O=O = 485. kJ/molBE for H-O = 458.9 kJ/mol ΔHvap for water = 40.65 kJ/mol

H2O (l)

2 H (g) + O (g)

ΔHvap(H2O) = +40.65 kJ

2 BE (H-O) = +917.8 kJ

ΔHf = -285 kJ

H2O (g)

H2 (g) + O (g)

BE (H-H) = -431 kJ

BE (O=O) = -485 kJ

3. For the reaction: 2 NO (g) + H2 (g) → N2O (g) + H2O (g)the following data was obtained:

Trial [NO] i (mol / L) [H2] i (mol / L) Initial Rate (mol / L•s)

1 0.150 0.800 0.5002 0.0750 0.800 0.1253 0.150 0.400 0.250

a) What is the rate law for the reaction?

€

Rate = k [NO]n[H2]m

Rate 1Rate 2

= k [NO]1n[H2]1m

k [NO]2n[H2]2m 0.5000.125

= k (0.150)n(0.800)m

k (0.0750)n(0.800)m

so 4 = 2n then 22 = 2n ∴ n = 2and Rate ∝ [NO]2[H2]m

Rate 1Rate 3

= k [NO]1n[H2]1m

k [NO]3n[H2]3m 0.5000.250

= k (0.150)n(0.800)m

k (0.150)n(0.400)m

so 2 = 2m then 21 = 2n ∴ m = 1∴ Rate ∝ [NO]2[H2]1 and Rate = k [NO]2[H2]1

b) What is the value of the rate constant?

1-22-2

2

2

2

2

12

2

sL mol 27.8 smol

L 27.8

mol 0.800

L x mol 0.0225

L x sLmol 0.500 k

Lmol 0.800

Lmol 0.150k

sLmol 0.500 1; Trial Using

][H[NO]k Rate

••=•

=

•=

=•

=

c) What is the rate when the [NO] is 0.300 mol/L and [H2] is 0.400 mol/L?

sLmol/ 1.00 L

mol 0.400 x L

mol 0.300 x smol

L 27.8 Rate

][H[NO]k Rate2

2

2

12

2

•=

•=

=

4. For the reaction: O3 (g) + NO (g) → O2 (g) + NO2 (g)the following data was obtained:Trial [O3] i (mol / L) [NO] i (mol / L) Initial Rate of Formation of NO2 (mol / L•s)

1 0.0010 0.0010 x2 0.0010 0.0020 2x3 0.0020 0.0010 2x4 0.0020 0.0020 4x

What is the rate law for the reaction?

€

Rate = k [O3]n[NO]m

Rate 2Rate 1

= k [O3]2n[NO]2m

k [O3]1n[NO]1m 2 xx

= k (0.0010)n(0.0020)m

k (0.0010)n(0.0010)m

so 2 = 2m then 21 = 2m ∴ m = 1and Rate ∝ [O3]n[NO]1

Rate 3Rate 1

= k [O3]3n[NO]3m

k [O3]1n[NO]1m 2 xx

= k (0.0020)n(0.0010)m

k (0.0010)n(0.0010)m

so 2 = 2n then 21 = 2n ∴ n = 1∴ Rate ∝ [O3]1[NO]1 and Rate = k [O3]1[NO]1

5. Write the following for the reactionN2 + 3H2 → 2NH3

a) The rate expression for the reaction (assume this reaction is elementary and only occurs in 1 step.)3

21

2 ][H][Nk Rate =

b) The order of the reaction for each reactant1st order for N2 and 3rd order for H2

c) The overall order of the reactionoverall order = 1 + 3 = 4th order

6. The rate constant for the reactionHNO3 + NH3 → NH4NO3

is 14.5 L/(mol.sec). If the concentration of the nitric acid is 0.050 M and the concentration of theammonia is 0.10 M, what will the rate of this elementary reaction be?

sLmol 0.073 Rate

Lmol 0.10 x

Lmol 0.050 x

s molL 14.5 Rate ][NH][HNOk Rate 1

31

3

•=

•==

7. For the reaction: 5 Br- (aq) + BrO3- (aq) + 6 H+ (aq) → 3 Br2 (l) + 3 H2O (l)

the following data was obtained at 298 K:

Trial [Br-]i (mol/L) [BrO3-]i (mol /L) [H+]i (mol/L) Initial Rate of Disappearance

of BrO3- (mol/L•s)

1 0.00100 0.00500 0.100 2.50 x 10-4

2 0.00200 0.00500 0.100 5.00 x 10–4

3 0.00100 0.00750 0.100 3.75 x 10–4

4 0.00100 0.01500 0.200 3.00 x 10–3

a) Determine the order of the reaction for each reactant.

€

Rate = k [Br- ]n[BrO3-]m[H+ ]p

Rate 2Rate 1

= k [Br- ]2n[BrO3-]2m[H+ ]2p

k [Br- ]1n[BrO3-]1m[H+ ]1p 5.00 x 10-4

2.50 x 10-4 = k (0.00200)n(0.00500)m(0.100)p

k (0.00100)n(0.00500)m(0.100)p

so 2 = 2n then 21 = 2n ∴ n = 1and Rate ∝ [Br- ]1[BrO3-]m[H+ ]p

Rate 3Rate 1

= k [Br- ]31[BrO3-]3m[H+ ]3p

k [Br- ]11[BrO3-]1m[H+ ]1p 3.75 x 10-4

2.50 x 10-4 = k (0.00100)1(0.00750)m(0.100)p

k (0.00100)1(0.00500)m(0.100)p

so 12

= 12

m then ∴ m = 1

∴ Rate ∝ [Br- ]1[BrO3-]1[H+ ]p

Rate 1Rate 1

= k [Br- ]41[BrO3-]41[H+ ]4p

k [Br- ]11[BrO3-]11[H+ ]1p 3.00 x 10-3

2.50 x 10-4 = k (0.00100)1(0.01500)1(0.200)p

k (0.00100)1(0.00500)1(0.100)p

so 12 = 3 x 2p then 4 = 22 = 2p ∴ p = 2∴ Rate ∝ [Br- ]1[BrO3-]1[H+ ]2

b) Write the rate law for the overall reaction.

21-3

1- ]H[][BrO]k[Br Rate +=

c) Determine the value of the specific rate constant for the reaction at 298 K.

smolL 500.

)Lmol 100.0)(Lmol 00500.0)(Lmol 00100.0(sLmol10 x 2.50 k

]H[][BrO][BrRate k ]H[][BrO]k[Br Rate

3

3

21-1-1-

1-1-4-

21-3

1-21-

31-

•=

•••

••=

==+

+

8. When two compounds, A and B, are mixed together, they form the compound C, by a reaction that isnot well understood. Fortunately, the following rate information was experimentally determined, asshown below:

A (mol/L) B (mol/L) Rate (mol/(L sec))0.050 0.050 4.0 x 10-30.100 0.050 8.0 x 10-30.050 0.100 1.6 x 10-2

a) Determine the rate expression for this reaction.

€

Rate = k [A]n[B]m

Rate 2Rate 1

= k [A]2n[B]2m

k [A]1n[B]1m 8.0 x 10-3

4.0 x 10-3 = k (0.100)n(0.050)m

k (0.050)n(0.050)m

so 2 = 2n then 21 = 2n ∴ n = 1and Rate ∝ [O3]n[NO]1

Rate 3Rate 1

= k [A]3n[B]3m

k [A]1n[B]1m 1.6 x 10-2

4.0 x 10-3 = k (0.050)n(0.100)m

k (0.050)n(0.050)m

so 4 = 2m then 22 = 2m ∴ m = 2∴ Rate ∝ [A]1[B]2 and Rate = k [A]1[B]2

b) Determine the rate constant for this reaction with units.

smolL 32

)Lmol 050.0)(Lmol 050.0(sLmol10 x 4.0 k

[B][A]Rate k [B]k[A] Rate

2

2

21-1-

1-1-3-

2121

•=

••

••=

==

KINETICS – Rate Mechanisms and the RDS

1. For the following reaction,

NO (g) + O3 (g) → NO2 (g) + O2 (g)

Given: Ea(forward) = 10.0 kJ/mol and ΔH (forward) = -200.0 kJ/mola) Draw the energy level diagram for this reversible reaction.

b) Calculate the Ea(reverse) and the ΔH (reverse).

kJ/mol 210.0 kJ/mol 200.0 kJ/mol 10.0

H(forward) E E a(forward)a(reverse)

+=+=

Δ+=

For the reaction the following data was obtained:

Trial [O3]i (mol / L) [NO]i (mol / L) Initial Rate of Formation of NO2 (mol / L•s)

1 0.0010 0.0010 x2 0.0010 0.0020 2x3 0.0020 0.0020 4x

c) What is the rate law for the reaction?

€

Rate = k [O3]n[NO]m

Rate 2Rate 1

= k [O3]2n[NO]2m

k [O3]1n[NO]1m 2 xx

= k (0.0010)n(0.0020)m

k (0.0010)n(0.0010)m

so 2 = 2m then 21 = 2m ∴ m = 1and Rate ∝ [O3]n[NO]1

Rate 3Rate 1

= k [O3]3n[NO]3m

k [O3]1n[NO]1m 4 xx

= k (0.0020)n(0.0020)m

k (0.0010)n(0.0010)m

so 4 = 2n x 2 then 21 = 2n ∴ n = 1∴ Rate ∝ [O3]1[NO]1 and Rate = k [O3]1[NO]1

d) The proposed mechanism is 3-step. The 1st is the RDS producing O• and NO3, which areboth rxn intermediates. Derive the other 2 elementary steps for the mechanism.

NO (g) + O3 (g) → NO3 (g) + O• (g) RDS (slow) O• (g) + O3 (g) → O2 (g) + O2 (g)NO3 (g) + NO (g) → NO2 (g) + NO2 (g)

Then: 2 NO (g) + 2 O3 (g) → 2 NO2 (g) + 2 O2 (g) NO (g) + O3 (g) → NO2 (g) + O2 (g) overall rxn

2. Propose a mechanism for the chain reaction:

H2 (g) + I2 (g) → 2 HI (g)

Where, I•, is a catalyst. Label each step as either chain initiation, propagation or termination.

HI 2 2I 2H Overall 2I I In Terminatio

I HI 2I H H HI 2H In Propagatio

I I UV 2I Initiation

→+→+

•+→+

•+→+

+ →

••

•

•

••

3. For the reaction: 2 NO (g) + Br2 (g) → 2 NOBr (g) the following data wasobtained:

Trial [NO]i (mol / L) [Br2]i (mol / L) Initial Rate of Formation of NOBr (mol / L•s)

1 0.0160 0.0120 3.24 x 10-4

2 0.0160 0.0240 6.48 x 10-4

3 0.0320 0.0060 6.48 x 10-4

a) Write the rate law equation that is consistent with the data.

€

Rate = k [NO]n[Br2]m

Rate 2Rate 1

= k [NO]2n[Br2]2m

k [NO]1n[Br2]1m 6.48 x 10-4

3.24 x 10-4 = k (0.0160)n(0.0240)m

k (0.0160)n(0.0120)m

so 2 = 2m then 21 = 2m ∴ m = 1and Rate ∝ [NO]n[Br2]1

Rate 3Rate 1

= k [NO]3n[Br2]3m

k [NO]1n[Br2]1m 6.48 x 10-4

3.24 x 10-4 = k (0.0320)n(0.0060)m

k (0.0160)n(0.0120)m

so 2 = 2n x 0.5 then 4 = 22 = 2n ∴ n = 2∴ Rate ∝ [NO]2[Br2]1 and Rate = k [NO]2[Br2]1

b) Calculate the value of the specific rate constant, k, and specify units.

1-s2L2-mol 105 s2mol

2L 105

)1-Lmol (0.01202)1-Lmol (0.0160

1-s1-Lmol 4-10 x 3.24 1]2[Br2

1[NO]1Rate k

]2[Br2[NO]k Rate

••=•

=

••

••==

=

c) The following mechanism was proposed for the reaction:

Br2 (g) + NO (g) → NOBr2 (g) slow

NOBr2 (g) + NO (g) → 2 NOBr (g) fast

Is this mechanism consistent with the given experimental observations? Justify your answer.

The slow equation is the RDS giving a rate law of Rate = k[Br2][NO], but the empirical rate law

is Rate = k[Br2]2[NO]. The proposed mechanism is NOT consistent with the experimental data.

4. Propose a mechanism for the chain reaction:

H2 (g) + I2 (g) → 2 HI (g)

Where, I•, is a reaction intermediate. Label each step as either chain initiation, propagation ortermination.

HI 2 2I 2H Overall

2H H Hn Terminatio

H HI 2H I

I HI 2I Hn Propagatio

H H UV 2H Initiation

→+

→+

•+→+

•+→+

+ →

••

•

•

••

5. The reaction:

NO2 (g) + CO (g) → NO (g) + CO2 (g)

is believed to occur by the following reaction mechanism:

NO2 (g) + NO2 (g) → NO3 (g) + NO (g) ( slow )

NO3 (g) + CO (g) → NO2 (g) + CO2 (g) ( fast )

a) What is the rate law according to this mechanism?

The RDS gives Rate = k[NO2]2

b) What would be the rate law if the reaction occurred directly in a single step?

Rate = k[NO2][CO]

6. Draw an energy level diagram for:

X + Y → Z

using the following information:

a) The reaction is reversibleb) The reaction consists of three elementary stepsc) The rate determining step is the second stepd) The forward reaction is endothermice) The Ea is twice the ∆H for the forward reaction.

KINETICS - REVIEW

1. At high temperatures, the following reaction occurs:

4 NH3 (g) + 5 O2 (g) _ 4 NO (g) + 6 H2O (g)

One experiment showed a concentration of ammonia of 0.120 mol/L at 10 s and of 0.100 mol/L at55 s. Calculate the average rate, over this time interval, for each of the following:

a) the rate of decomposition of NH3.

€

Δ[NH3]Δt

= - 0.100 - 0.120( ) mol/L55−10( ) s

= - (-0.0200) mol/L45 s

= 4.4 x 10-4 mol/L•s

b) the rate of production of water.

€

Δ[H2O]Δt

= Δ[NH3]Δt

x 6 mol H2O4 mol NH3

= 4.4 x 10-4 mol/L•s x 32

= 6.6 x 10-4 mol/L•s

c) the rate of production of NO.

€

Δ[NO]Δt

= Δ[NH3]Δt

x 4 mol NO4 mol NH3

= 4.4 x 10-4 mol/L•s x 11

= 4.4 x 10-4 mol/L•s

d) the rate of consumption of oxygen.

€

Δ[O2]Δt

= Δ[NH3]Δt

x 5 mol O24 mol NH3

= 4.4 x 10-4 mol/L•s x 54

= 5.5 x 10-4 mol/L•s

2. For the following reactions,

For the following reaction and its reverse activation energies were determined as follows:

C2H4(g) + H2 (g) → C2H6(g) Ea = 180 kJ/molC2H6(g) → C2H4(g) + H2(g) Ea = 317 kJ/mol

Calculate ∆H for the reaction of C2H4 with hydrogen

€

ΔHrxn = (180 - 317 ) kJ/mol = -137 kJ/mol

3. For the reaction:

2 A + 2 B C + D

The following data about the reaction above were obtained from three experiments:

Experiment [A] mol/L [B] mol/L Rate of Formation of C (mol/L •min)1 0.60 0.15 6.3x10-32 0.20 0.60 2.8x10-33 0.20 0.15 7.0x10-4

a) What is the rate equation for the reaction?

€

Rate = k [A]n[B]m

Rate 2Rate 3

= k [A]2n[B]2m

k [A]3n[B]3m 2.8 x 10-3

7.0 x 10-4 = k (0.20)n(0.60)m

k (0.20)n(0.15)m

so 4 = 4m then 22 = 2m ∴ m = 1and Rate ∝ [A]n[B]1

Rate 1Rate 3

= k [A]1n[B]11

k [A]3n[B]31 6.3 x 10-3

7.0 x 10-4 = k (0.60)n(0.15)1k (0.20)n(0.15)1

so 9 = 3n then 32 = 3n ∴ n = 2∴ Rate ∝ [A]2[B]1 and Rate = k [A]2[B]1

b) What is the numerical value of the rate constant k? What are its dimensions?

1-min2L 2-mol 27.8 min2mol

2L 0.12

mol 0.15

L x 2mol 0.36

2L x minL

mol 3-10 x 6.3 k

Lmol 0.152

Lmol 0.60k

minLmol 3-10 x 6.3 1; Trial Using

1[B]2[A]k Rate

••=•

=

•=

=•

=

c) Propose a reaction mechanism for this reaction.

A + A + B A2B

A2B + B A2 + B2Then 2 A + 2 B A2 (or C) + B2 (or D)

4. The following results were obtained when the reaction represented below was studied at 25˚C.

2 ClO2(g) + F2(g) → 2 ClO2F(g)

Experiment [ClO2]i (mol/L) [F2]i (mol/L) Rate of Increase of [ClO2F]i (mol/L •s)1 0.010 0.10 2.4×10-32 0.010 0.40 9.6×10-33 0.020 0.20 9.6×10-3

a) Write the rate law expression for the reaction above.

€

Rate = k [ClO2]n[F2]m

Rate 2Rate 1

= k [ClO2]2n[F2]2m

k [ClO2]1n[F2]1m 9.6 x 10-3

2.4 x 10-3 = k (0.010)n(0.40)m

k (0.010)n(0.10)m

so 4 = 4m then ∴ m = 1and Rate ∝ [ClO2]n[F2]1

Rate 3Rate 1

= k [ClO2]3n[F2]31

k [ClO2]1n[F2]11 9.6 x 10-3

2.4 x 10-3 = k (0.020)n(0.20)1k (0.010)n(0.10)1

so 4 = 2n x 2 then 21 = 2n ∴ n = 1∴ Rate ∝ [ClO2]1[F2]1 and Rate = k [ClO2]1[F2]1b) Calculate the numerical value of the rate constant and specify the units.

1-s 1L 1-mol 2.4 smol

L 2.4

mol 0.10

L x mol 0.010

L x sL

mol 3-10 x 2.4 k

Lmol 0.10

Lmol 0.010k

sLmol 3-10 x 2.4 1; Trial Using

1]2[F1]2[ClOk Rate

••=•

=

•=

=•

=

c) In experiment 2, what is the initial rate of decrease of [F2]?

€

2 ClO2 + F2 → 2 ClO2FΔ[F2]Δt

= Δ[ClO2F]Δt

x 1 mol F22 mol ClO2F

= 9.6 x 10-3 mol/L•s x 12

= 4.8 x 10-3 mol/L•s

d) Which of the following reaction mechanisms is consistent with the rate law developed in (a).Justify your choice.

I. ClO2 + F2 → ClO2F2 (fast)ClO2F2 → ClO2F + F (slow)ClO2 + F → ClO2F (fast)

II. F2 → 2 F (slow)2 (ClO2 + F → ClO2F) (fast)

I. mechanism with consistent is which ]2][F2k[ClO Rate overall

]2][F2[ClOrkfk

2k Rate : 1 Step from Subing

]2F2[ClO2k Rate :(RDS) 2 Step From

]2][F2[ClOrkfk ]2F2[ClO

]2F2[ClOrk ]2][F2[ClOfk : 1 Step Fromreverse rate forward rate

=∴

=

=

=

=

=

5. Freon gases, e.g. CCl2F2 (g) , from leaking air conditioners and refrigeration units also make theirway into the upper atmosphere.. These gases catalyze the destruction of ozone.

CCl2F2 (g) + O3 (g) _ CClF2• (g) + ClO• (g) + O2 (g)

Where, Cl•, is a reaction intermediate. Propose a mechanism for this reaction.

•++→+

•+→+

+ →

•

•

••

ClO 2O CClF 3O FCCl

ClO 2O 3O Cl

Cl 2CClF UV 2F2CCl

222

6. Draw a potential energy diagram for the reaction

X + Y → Z

showing the following information.

(a) The mechanism consists of two elementary steps.(b) The second elementary step is the rate-determining step.(c) The overall reaction is exothermic.

(d) What would happen to the shape of the graph if a catalyst were added?

The Activation energy would decrease.