title: lesson 5 bond enthalpies learning objectives: – understand bond enthalpies and what they...

Title: Lesson 5 Bond Enthalpies

Learning Objectives:– Understand bond enthalpies and what they mean

– Use bond enthalpies to calculate enthalpy changes

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Using the equations below:

C(s) + O2(g) → CO2(g) ∆Ho = –394 kJ mol-1

Mn(s) + O2(g) → MnO2(s) ∆Ho = –520 kJ mol-1

What is ∆H, in kJ, for the following reaction?

MnO2(s) + C(s) → Mn(s) + CO2(g)

A. 914B. 126C. –126D. –914

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Playing with Magnets

Take two magnets (preferably round ones)

Place them near each other and allow them to attract each other. Is this an endothermic or an exothermic process? How do

you know?

Pull your two magnets apart. Is this an endothermic or an exothermic process? How do

you know?

Experiment with magnets of different strengths. How do your energy changes relate to the strength of the

magnets?

What has all this got to do with chemical bonds? Are there any limits to this analogy (places where it

breaks down)?

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Why is bond making exothermic?

When two atoms are a long way apart, the electrical potential energy of the two together is at a maximum.

If two atoms are bonded together, in order to separate them, work must be done against the force holding them together – electrical potential energy of the atoms increases.

As atoms come together, because of the attractive force between them, the potential energy decreases and this energy is released as heat.

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Average Bond Enthalpies Bond enthalpies are defined in terms of

breaking bonds.

Try to come up with a suitable definition for bond enthalpies Include all relevant detail: amounts, states,

conditions and so on What sign would bond enthalpies have?

ΔHE,298

6 of 36 © Boardworks Ltd 2009

Breaking and forming bonds

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The bond breaking reaction: For O=O: O2(g) 2O(g) H = 498 kJ mol-1

For H-F: HF(g) H(g) + F(g) H = 568 kJ mol-1

From gaseous covalent bonds to gaseous atoms

It gets complicated when there is more than 2 atoms in a molecule... E.g. There are more than one O-H bonds in Water...

1st bond enthalpy2nd bond enthalpy

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Average Bond Enthalpy

The average energy required to break one mole of a chemical bond to produce two moles of gaseous atoms at 298K.

Check Table 11 of the Data Booklet

Bond enthalpies are always positive

These are average values, they do not take into account: Variations in strengths of the same bond in different compounds

For example: the strength of a C-H bond will vary slightly depending on whether it is in methane, ethene, glucose and so on

Energy changes due to state changes (from intermolecular forces) Energy changes due to dissolution in water

This means enthalpy changes calculated from bond enthalpies will generally be less accurate than those calculated by other means

HR + ΣE [ bonds BROKEN ]- ΣE [ bonds FORMED ]

NB approximation to H because:

2. Gaseous state may not apply

1. Average values used

“+” because bond breaking is

ENDOthermic

“-” because bond formation is EXOthermic

Reactants Products

Gaseous Atoms

BONDS BROKEN - ENDOTHERMIC

BONDS FORMED - EXOTHERMIC

HR

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H H

Bond Enthalpy and Enthalpy Change As the enthalpy increases, bonds are being

broken, as it decreases, bonds are being made.

∆H is negative Weaker bonds broken Stronger bonds made

∆H is positive Stronger bonds broken Weaker bonds made

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Bond Enthalpies

All bond enthalpies refer to reactions in the gaseous state.

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Making bonds is an exothermic process

The same amount of energy is absorbed when a bond is broken as is given out when a bond is made

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Solutions

15 of 36 © Boardworks Ltd 2009

Calculating enthalpy change

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Bond Enthalpies in Calculations – Method 1 Determine the enthalpy change for the following reaction

H2(g) + Cl2(g) 2 HCl(g)

H2(g) + Cl2(g) 2 HCl(g)

2 H(g) + 2 Cl(g)

∆H = 679 – 864 = -185 kJ mol-1

Note: the experimentally measured value is also -185 kJ mol-1

1 x H-H + 1 x Cl-Cl = 436 + 243 = 679

2 x H-Cl = 2 x 432 = 864

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Bond Enthalpies in Calculations – Method 2 Determine the enthalpy change for the following

reaction

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

∆H = ∑ E(bonds broken) - ∑ E(bonds formed)= (4 x C-H + 2 x O=O) - (2 x C=O + 4 x O-H)

= (4 x 413 + 2 x 498) - (2 x 746 + 4 x 464)= 2648 – 3348= -700 kJ mol-1

Note: the experimentally measured value is -890 kJ mol-1

Why the big difference?

e.g.1Using the table of bond energies, calculate the energy

required to "atomise" (break all the bonds in the molecule) one mole of propan-1-ol, CH3CH2CH2OH.

CLICK HERE FOR BOND

ENERGY DATA

Bonds to break:

2 × (+347) kJ mole-1

7 × (+413) kJ mole-1

1 × (+358) kJ mole-1

1 × (+464) kJ mole-1

Total = + 4407 kJ mole-1

2 C-C

7 C-H

1 C-O

1 O-H

2 Using the table of bond energies, calculate the energy (enthalpy) change for :

  H2(g) + F2(g) 2HF(g)

H =

- 2 (+ 568)

= - 542 kJ mole-1

BONDS BROKEN BONDS FORMED

H - H

F - F

2 H - F+ 436

+ 158

+ 594 - 1136

+ 594 - 1136

CLICK HERE FOR BOND

ENERGY DATA

3 Using the table of bond energies, calculate the energy (enthalpy) change for :

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

H =

= - 818 kJ mole-1

- 2 (+ 805)

BONDS BROKEN BONDS FORMED

4 C - H

2 O = O

2 C = O4 (+ 413)

2 (+ 498)

+ 2648 - 3466

4 O - H - 4 (+ 464)

+ 2648 - 3466

CLICK HERE FOR BOND

ENERGY DATA

Draw out the structure so you can see the bonds!

22 of 36 © Boardworks Ltd 2009

Calculating enthalpy change problems

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Solutions

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2. Use the bond energy data table to calculate heat of reaction for:

CH4(g) + 2Br2(g) CH2Br2(g) + 2HBr(g)

By Hess’s Law: H =

CH4(g) + 2Br2(g) CH2Br2(g) + 2HBr(g)

= - 100 kJ mole-1

H

+4(413) + 2(193) -2(413) - 2(290) -2(366)

Click here for E data

Alternative route

BONDS BROKEN - ENDOTHERMIC

BONDS FORMED - EXOTHERMIC

C(g) + 4H(g) + 4Br(g)

+4E[C-H]+2E[Br-Br]

-2E[C-H]-2E[C-Br]-2E[H-Br]

Atoms here because bond energies given

5 Using the table of bond energies, calculate the energy (enthalpy) change for :

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

H CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

H =

4E[C-H] -2E[C=O]

4(+413) + 2(498) -2(+695) - 4(+464)

= - 598 kJ mole-1

E data

C(g) + 4H(g) + 4O(g)

Atoms here because bond energies given

+ 2E[O=O] -4E[O-H]

1 Construct appropriate energy cycles and use the bond energy data table to calculate heats of reaction for:

(a) CH4(g) + Br2(g) CH3Br(g) + HBr(g)

(b) 2CO(g) + O2(g) 2CO2(g)

(c) CH3CH2OH(g) + HCl(g) CH3CH2Cl(g) + H2O(g)

2 Given that : P(g) + 3Cl(g) PCl3(g) ; -983 kJ mole-1

2P(s) + 3Cl2(g) 2PCl3(g) ; -610 kJ mole-1

P(s) P(g) ; +314 kJ mole-1

 Connect these reactions using an appropriate energy cycle and then

use Hess's Law to calculate H for Cl-Cl(g) 2Cl(g) ie E[Cl-Cl]

Answer (a) Answer (b) Answer (c)

Answer

CLICK HERE FOR BOND ENERGY DATA

- 116 kJ mole-1 - 568 kJ mole-1 - 20 kJ mole-1

+243 kJ mole-1

H C N O S F Cl Br I

H 436 413 391 464 364 568 432 366 298

C   347 286 358 272 467 346 290 228

N     160 201   490 319 264 184

O       144        

S         266 326 255 213  

F           158 255 238  

Cl             243 217 209

Br               193 180

I               151

C=C 612 CC 838

O=O 498 NN 945

N=N 410 CO 1077

C=O 805 CN 887

SOME BOND ENERGY VALUES (all in kJ mole-1) For SINGLE covalent bonds :

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Key Points

Bond enthalpies tell you the amount of energy required to break a bond They are always positive (endothermic)

Bond enthalpies are always positive

Enthalpies of reaction calculated from bond enthalpies are less accurate due to their being averages.

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Ozone Depletion Oxygen is present in two forms – O2 and O3 – both play a

role in protecting Earth from harmful UV radiation.

They form a protective screen which ensures that radiation that reaches the surface of the Earth is different from that emitted by the Sun.

The double bond in the O2 is stronger that the 1.5 bond in ozone so is broken by a radiation of higher energy and shorter wavelengths.

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The natural formation and depletion of ozone Temperature of the atmosphere generally decreases with height

but at 12km above the Earth’s surface the temperature starts to rise because UV radiation is absorbed in a number of photochemical reactions.

The oxygen atoms have unpaired electrons. They are reactive free radicals so react with another oxygen to form ozone:

The second step is exothermic; bonds are formed and the energy released raises the temperature of the stratosphere.

Bonds in ozone are weaker than in O2, lower energy UV light is needed to break them:

Oxygen free radicals then react with another ozone molecule to form two oxygen molecules

This is another exothermic reaction (bond formation) – this maintains the relatively high temperature of the stratosphere.

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Steady State The level of ozone in the atmosphere stays constant

(<10ppm) if the formation of ozone is balanced by the rate of its removal.

The processes on the previous slide can be outlined by the Chapman Cycle:

These processes are significant because dangerous UV light has been absorbed and the stratosphere can be maintained at a warmer temperature.

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Solutions