topic guide 5.1: chemical bonding, thermodynamics … 01, 2011 · thermodynamics and 5.1 ... with...

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Unit 5: Chemistry for Applied Biologists

The biochemical processes that take place in biological systems often involve large and structurally-complex molecules – for example, enzyme proteins, nucleic acids, steroid hormones and phospholipids. In this topic you will learn about the bonding within molecules such as these as well as the intermolecular bonding that is so crucial for many key biochemical processes.

You will also learn about the key ideas on rates of chemical reactions and energy changes in reactions. Both of these ideas will be developed further in later topic guides.

On successful completion of this topic you will: • understand how chemical bonding affects chemical and physical

properties of molecules (LO1).

To achieve a Pass in this unit you need to show that you can: • explain the physical properties of substances in terms of their chemical

bonding (1.1) • explain enthalpy changes in terms of bonding and interactions (1.2) • explain the effect of various factors on rates of chemical reactions (1.3).

Chemical bonding, thermodynamics and kinetics5.1

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5.1: Chemical bonding, thermodynamics and kinetics

1 Chemical bondingAll chemical bonding is due to electrostatic attractions between opposite charges. Chemists categorise substances as ionic or covalent depending on the nature of the bonding in the substance. The nature of the bonding in a substance determines its structure and, as we will see, the properties of a substance are determined by a combination of its structure and its bonding.

Before you startBefore starting this section of work you should ensure that you are confident with the following ideas:

• the nature of ionic and covalent bonding • the meaning of the words atom, molecule and ion • dot and cross diagrams to show the structure of atoms, molecules and ions • the different ways of representing the formulas of chemical substances, including molecular

formulae and structural formulae • how to write balanced chemical equations.

If you are not confident with these ideas you should consult a suitable level 2 textbook, such as BTEC First: Principles of Applied Science (Goodfellow, Hocking and Musa, 2012) and BTEC First: Application of Applied Science (Goodfellow, Hocking and Musa, 2012), or you could visit websites such as the BBC Bitesize Science revision site: http://www.bbc.co.uk/schools/gcsebitesize/science/ocr_gateway/chemical_concepts/fundamentalrev1.shtml.

Ionic and covalent bondingThese are the two strongest types of bonding present in compounds. Figure 5.1.1 summarises the key features of each type. A third type of bonding, metallic bonding, also occurs in metallic elements and alloys. Although this is less relevant in biological situations, it is included here for comparison.

ionic bonding

giant ionic lattice

Attraction between oppositely charged ions.

sodium chloride(NaCl)

A bond formed when two atoms share a pair of electrons. In covalent bonding, the positive nuclei of the atoms are attracted to the pair of electrons between the two nuclei. If two pairs of electrons are shared, a double covalent bond is formed; three shared pairs makes a triple covalent bond.

covalent bonding

Type ofbonding

Nature of bonding Type of structure resulting from bonding

Example of substance with this structure

simple molecule

carbon dioxide(CO2)

Attraction between the positively charged ions in a metallic lattice and a pool of mobile electrons that are spread out throughout the lattice.

metallic bonding

giant metallic lattice

iron(Fe)

macromolecule protein

giant covalent lattice

silicon dioxide(SiO2)

Figure 5.1.1: Key features of ionic, covalent and metallic bonding.

http://www.bbc.co.uk/schools/gcsebitesize/science/ocr_gateway/chemical_concepts/fundamentalrev1.shtml

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Covalent bonds and formulae

The structures of molecules can be shown as full structural formulae, shortened structural formulae or skeletal formulae:

• in a full structural formula, all of the atoms and bonds are shown • in a shortened structural formula, bonds to hydrogen atoms are not shown • in a skeletal formula, only the bonds that make up the skeleton of the

molecule are shown – carbon atoms and hydrogen atoms are not usually shown, although other atoms (such as oxygen and nitrogen) are.

The three different types of formula are shown in Figure 5.1.2.

O

C CCH

H

H

O

O H

C

O

C OHH3C

OO

O

OH

(a) (b) (c)

Case study: Ionic bonding in receptor sitesMany enzymes and drug receptor sites contain charged groups of atoms such as –NH

3+ or –COO–.

The molecules that bind to these sites (known as ligands) may also have charged groups and these form strong ionic bonds to the receptor site.

Groups such as NH3

+ and COO− in receptor sites may become uncharged if the pH of their environment changes. How would this affect the strength of the bond between a receptor site and a charged ligand?

Case study: ChelationMolecules that contain groups such as amines (–NH

2 or –NH–) will be able to form dative covalent

bonds to certain metal ions. In this type of covalent bond, both electrons in the bond are provided by the nitrogen atom. This type of bonding is particularly significant when the arrangement of the amine groups allows several bonds to form to the same metal ion. This is known as chelation and it occurs in the structure of the haem group (a key component of haemoglobin).

The structure of haem is shown in Figure 5.1.3. • How many carbon–carbon double covalent bonds are there in the molecule shown in

Figure 5.1.3? • Identify one other type of double bond shown in this structure. • In what part of the molecule does chelation occur? • How many carbon atoms are there altogether in this molecule?

NN

OHO

N N

OHO

FeII

H3C

CH3

CH2

CH2

CH3H3C

Figure 5.1.3: Structure of haem.

Figure 5.1.2: Pyruvic acid is an important molecule in the metabolism of cells. Three ways of showing the structure

are shown here: (a) a full structural formula, (b) shortened structural formula and (c) skeletal formula.

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Intermolecular forcesMost biological substances have a molecular structure. The intermolecular forces (forces between molecules) play an important role in determining the properties of the substance.

Dipole–dipole interactions

In some cases, the electrons in a covalent bond are not equally shared between the two atoms. This occurs in bonds such as O–H, N–H, C=O and C–N and it leads to partial positive (δ+) and partial negative (δ-) charges on the atoms in the bond. Such a bond is described as a polar covalent bond and can cause molecules to have an overall dipole.

If two neighbouring molecules possess such a dipole then weak dipole–dipole interactions can occur between the molecules. Some examples of polar bonds are shown in Figure 5.1.4.

Hδ– δ+ δ– δ+ δ+ δ– δ+ δ–O HN OC NC

Take it furtherUnequal sharing of electrons in a bond occurs if the atoms have different electronegativities. The electronegativity of an atom is a measure of how strongly it attracts electrons in a bond. You can find out more about electronegativity and how it causes bond polarisation by visiting websites such as http://www.chemguide.co.uk/atoms/bonding/electroneg.html.

Hydrogen bonding

Hydrogen bonding occurs when there is a δ+ hydrogen in close proximity to a small δ- atom (usually oxygen or nitrogen atoms). These two atoms could be in neighbouring molecules (leading to an intermolecular hydrogen bond) or in different parts of the same molecule (leading to an intramolecular hydrogen bond).

Hydrogen bonds are stronger than dipole−dipole attractions because some electron sharing takes place between the two atoms. The electrons are provided by the non-bonded pair (‘lone pair‘) of electrons on the δ- atom. This is shown in Figure 5.1.5.

Hydrogen bonds are highly directional and are very important in maintaining the precise 3-D alignments of protein structures and in creating the complementary base pairing seen in the DNA molecule. Because these hydrogen bonds occur within a molecule, rather than between two separate molecules, they are described as intramolecular forces.

Van der Waals forces

The hydrocarbon chains and rings present in many biological molecules are non-polar. However, the electrons in the bonds between atoms are in constant random motion and may become temporarily arranged in an unsymmetrical way, leading to the formation of a temporary dipole.

Figure 5.1.4: Some common polar covalent bonds found

in biological molecules.

Key termsIntermolecular forces: Weak interactions between molecules such as hydrogen bonds, dipole−dipole interactions or Van der Waals forces.

Polar: A bond or molecule that has an uneven distribution of charge. If charge is distributed evenly, the bond or molecule is described as non-polar.

Dipole: Two oppositely-charged regions separated by a short distance.

O

O

Hδ–

δ–H δ+

δ+ Lone pair of electrons

Hydrogen bond

Figure 5.1.5: An intermolecular hydrogen bond between two OH groups.

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If a second molecule comes close to this temporary dipole then a dipole may be induced on this second molecule, as shown in Figure 5.1.6.

Non-polar molecule

δ+ δ–δ+

δ–

δ+δ–

Temporary dipole

Induced dipole

Intermolecular force

If you look at the charge distribution on the two molecules shown in Figure 5.1.6 you will see that, at this particular instant, there will be an attraction between the two dipoles. However, further random motion of the electrons means that the dipoles soon disappear, and therefore so does the force of attraction.

Large molecules or non-polar groups within molecules tend to produce stronger Van der Waals forces of attraction because they contain more electrons.

Physical properties Knowledge of the bonding in different substances (including the intermolecular forces discussed above) allows chemists to explain the physical properties of many substances. Table 5.1.1 shows some examples where an important physical property of each substance is explained in terms of its chemical bonding.

Substance Physical property Explanation in terms of structure and bonding

amino acids high melting point (often above 200 oC)

At neutral pH amino acids form structures called zwitterions, containing –NH3

+ and COO– ions. Crystals of amino acids are held together by ionic bonds between these groups.

solutions of sodium and potassium salts (e.g. NaCl)

high electrical conductivity

The charged ions are free to move when the salts are dissolved in water. If a voltage is applied, for example across the plasma membrane of a neuron, then Na+ or K+ will move towards the side of the membrane with the negative electrical potential.

water high surface tension Hydrogen bonds between δ+ hydrogen atoms and δ- oxygen atoms form an extensive 2-dimensional network on the surface of the water. The difference in surface tensions between water and lipids plays a part in the formation of the lipid bilayer in the plasma membrane.

methane low boiling point Methane does not contain any polar bonds and hence the molecule itself is non-polar. The only intermolecular forces are Van der Waals forces (which are weak) and so little energy is required to separate the molecules.

glucose high solubility in water

Glucose contains several hydroxyl (–OH) groups. These hydrogen bond with atoms in water molecules, providing enough energy to separate the glucose molecules.

Figure 5.1.6: The formation of Van der Waals forces between

two non-polar molecules.

Table 5.1.1: Table showing some explanations of physical

properties of some substances in terms of their chemical bonding.

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Activity1 The structures of four molecules found in biological systems are shown in Figure 5.1.7. Use

these structures and ideas about intermolecular forces to explain:a why water has a higher boiling point than methaneb why propane-1,2,3-triol is more soluble in water than propan-1-ol.

2 Many drug molecules are made inactive by converting them into polar or ionic forms. This reduces their lipophilicity – the ability to dissolve in non-polar lipid solvents. Use ideas about structure and bonding (with particular reference to intermolecular bonding) to explain why.

OH OHHO

(a) (b) (c) (d)

OH H CH H

H

H

OH

Figure 5.1.7: Structures of: (a) water, (b) methane, (c) propan-1-ol, (d) propane-1,2,3-triol.

2 Enthalpy changes in chemical reactionsBefore you startBefore starting this section of work you should ensure that you are confident with the following ideas:

• the meaning of the terms exothermic and endothermic • the energy changes which occur when bonds are broken and formed • how to explain whether a reaction is exothermic or endothermic using ideas about bond

breaking and forming.

If you are not confident with these ideas, consult a suitable level 2 textbook, such as BTEC First: Principles of Applied Science (Goodfellow, Hocking and Musa, 2012) and BTEC First: Applications of Applied Science (Goodfellow, Hocking and Musa, 2012), or you could visit websites such as the BBC Bitesize Science revision site.

Enthalpy changesChemical reactions are frequently accompanied by energy transfers, due to the breaking or formation of bonds. For isolated chemical reactions, the easiest form of energy transfer to measure is the amount of heat absorbed or released by a chemical reaction, which (under conditions of constant pressure) is equal to the enthalpy change (ΔH) of a reaction.

In biochemical reactions, a related term, the Gibbs energy change of a reaction, is a more useful quantity to measure.

Enthalpy profiles

Although the absolute enthalpy of a system is impossible to measure directly, it is often helpful to create an enthalpy profile diagram to show how the enthalpy of a system changes during a reaction.

Portfolio activity (1.1)Choose a substance with some biological importance and use ideas about structure and bonding to explain some of its physical properties. Suitable examples could include pyruvic acid, a fatty acid such as oleic acid and ammonia.

LinkYou will learn more about the Gibbs energy change of a reaction in Topic guide 5.2.

Key termsEnthalpy: A measure of the total energy of a system (the sum of the chemical energy and the thermal energy).

System: The specific portion of the universe that is being studied, for instance the chemicals involved in a chemical reaction. Everything outside the system is considered the surroundings or environment.

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Progress of reaction

Enth

alpy

Eact (catalysed)

ATP + H2O

ADP + H+ + Pi

ΔH (hydrolysis) = –16 kJ mol−1

Eact (uncatalysed)

ΔH (phosphorylation) = +16 kJ mol−1

Consider the hydrolysis of ATP, and the reverse reaction (phosphorylation of ADP). These reactions have critical roles in the transfer and release of energy in cells.

The reactions can be represented by the following equation (although not technically a balanced chemical equation, this is the usual way in which this process is represented):

ATP + H2O ⇌ ADP + P

i + H+

where ATP is adenosine triphosphate, ADP is adenosine diphosphate and Pi is

’inorganic phosphate‘ – a mixture of H2PO

4– and HPO

42– ions.

The enthalpy profile for these reactions is shown in Figure 5.1.8. It is a visual way of describing what happens to the enthalpy of the system during the reaction and will help you to understand several important ideas:

1 The forward reaction (hydrolysis), in which the enthalpy of the system decreases, is an exothermic process.

2 The backward reaction (phosphorylation), in which the enthalpy of the system increases, is an endothermic process.

3 ΔH is defined as Hproducts

− H

reactants. It therefore has a negative value for an

exothermic reaction and a positive value for an endothermic process. If the reaction is carried out under standard conditions (1 bar pressure, 1 mol dm–3 solution, all substances in their usual states at these conditions) then the enthalpy change is known as the standard enthalpy change ΔHƟ

298. The Ɵ

superscript used in the symbol for this enthalpy change is used to indicate that the quantity is measured under these standard conditions; it is also necessary to state the temperature separately. You will see this symbol used in several other thermodynamic quantities in this unit.

4 The activation energy (Eact

) can be seen as the minimum energy that must be possessed by the reactants before a reaction will proceed to form products. It can be thought of as an ’energy barrier‘ for a reaction.

5 In the presence of a catalyst, the activation energy is lowered.

Figure 5.1.8: Enthalpy profile diagram of the hydrolysis of ATP

and the phosphorylation of ADP.

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Defined enthalpy changesSeveral specific types of enthalpy changes are to be found in tables of thermodynamic data and are frequently used in thermodynamic calculations (see Table 5.1.2).

Enthalpy change Description Example

enthalpy of combustion enthalpy change when 1 mole of a substance undergoes complete combustion

C2H

5OH(l) + 3.5O

2(g) ➝ 2CO

2(g) + 3H

2O(l)

ΔHcƟ

298 = −1367.3 kJ mol−1

enthalpy of formation enthalpy change when 1 mole of a substance is formed from its constituent elements

6C(s) + 6H2(g) + 3O

2(g) ➝ C

6H

12O

6(s)

ΔHfƟ

298 = −1273.3 kJ mol−1

enthalpy of reaction enthalpy change when the number of moles of reactants stated in the equation react together

2H2(g) + O

2(g) ➝ 2H

2O(l)

ΔHƟ298

= −286 kJ mol−1

Activity • All the reactions shown in Table 5.1.2 are exothermic. What can you conclude about the

relative size of the energy required to break bonds in these reactions compared to the energy released by forming bonds?

• Use the information in the table to suggest the value of the enthalpy change for the reactions represented by the following equations:a H

2(g) + ½O

2(g) ➝ H

2O(l)

b 2H2O(l) ➝ 2H

2(g) + O

2(g)

Mean bond enthalpy

This is the enthalpy change when 1 mole of bonds is broken (assuming the reaction is done in the gaseous state).

Bond breaking, which requires energy, is endothermic (hence the + sign in the table); conversely bond formation is exothermic.

Values quoted in tables, such as Table 5.1.3, are usually the mean of values from a range of molecules.

Bond Mean bond enthalpy / kJ mol–1

C–C +347

C–H +413

C–O +336

O–H (in water) +464

O=O +498

C=O +805

ActivityResearch the definitions of the following enthalpy changes found in thermodynamic tables: enthalpy of dissociation, enthalpy of solution, lattice enthalpy, enthalpy of hydration.

Table 5.1.2: Some common specific types of enthalpy changes.

Key term(Chemical) thermodynamics: The study of energy changes in chemical reactions.

Table 5.1.3: Some common bonds and their mean bond enthalpies

(Source: Nuffield Advanced Chemistry Book of Data, 1984).

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Hess’s Law and enthalpy calculations

Hess’s Law states that ’the total enthalpy change for a process is independent of route‘.

This idea can be used to predict approximate values for the enthalpy change of reactions if the structures of reactants and products are known. Even if a reaction actually takes place via a multi-step mechanism, the enthalpy change for this process will be identical to one in which the reaction proceeds simply by the breaking of all the bonds in the reactants followed by the forming of bonds in the products.

Using this idea:

enthalpy change of reaction = Σ enthalpy of bonds broken − Σ enthalpy of bonds formed

In reality this is only an approximate prediction as the mean values may vary from the true ones and it does not take into account intermolecular forces, which are responsible for the enthalpy changes that occur when substances condense or vaporise.

ActivityThe equation for the complete combustion of methane is:

CH4(g) + 2O

2(g) ➝ CO

2(g) + 2H

2O(l)

The displayed formulae of these molecules are shown in Figure 5.1.9. 1 Write out a list of the number and type of bonds broken. Do the same for the bonds formed. 2 Use the mean bond energies from Table 5.1.3 to calculate a value for the enthalpy change of

the reaction.3 Find the equation for the complete combustion of glucose (the reaction which occurs in

aerobic respiration) and the molecular structures of the reactants and products. Use mean bond energies (from Table 5.1.3) and the molecular structures you have found to calculate a value for enthalpy change for this reaction.

+O

H H

OH H

CH H

H

H

OO

OOCO O

Figure 5.1.9: Equation showing the displayed formulae for the complete combustion of methane.

Using enthalpy of formation data

This is another application of Hess’s Law, and makes use of the enthalpy of formation data found in tables of thermodynamic quantities:

ΔHƟ = Σ ΔHƟformation

(products) – Σ ΔHƟformation

(reactants)

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Take it furtherThermodynamic data, such as enthalpy of formation values, can be found in online databases such as the NIST databook http://webbook.nist.gov/chemistry/, or by searching for compounds individually in a search engine.

Activity • Find values for the standard enthalpy of formation of products and reactants in the equation

for the complete combustion of methane and use these values to calculate a value for the standard enthalpy change of this reaction. Compare your answer with the one obtained in the previous activity. Comment on any difference.

• Carry out the same kind of calculation to find a value for the standard enthalpy change of combustion of glucose. Again, comment on the difference between this value and the one obtained in the previous activity.

Portfolio activity (1.2)Choose a biochemical reaction involving fairly simple molecules. A suitable example could be the oxidation of ethanol to ethanoic acid by reaction with oxygen. Construct an enthalpy profile for this reaction and explain why the reaction is exothermic (or endothermic, if appropriate) using ideas about bond enthalpies.

In your answer you should: • write a balanced chemical equation for the reaction • calculate a value for the enthalpy change for this reaction, obtained by using enthalpy of

formation data • include an enthalpy profile showing the relative enthalpies of products and reactants • carry out a calculation using bond energies to explain the size and sign of the enthalpy change.

3 Rates of chemical reactionsThe rate of a chemical reaction is defined as the rate at which reactants are converted into products.

For a simple reaction, such as the decomposition of hydrogen peroxide, the rate of the reaction can be measured by calculating the rate at which the concentration of hydrogen peroxide falls:

2H2O

2 (aq) ➝ 2H

2O (l) + O

2(g)

Rate = change in concentration of hydrogen peroxide

time taken for this change

Mathematically this can be written as:

Rate = Δ [H

2O

2]

Δt

where [H2O

2] = the concentration of hydrogen peroxide.

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The graphs shown in Figure 5.1.10 show how the concentration of reactants and products changes during a reaction, and how the rate of reaction at a particular time can be estimated by using a tangent method to find the gradient of the line on the graph.

[rea

ctan

ts]

∆t

Time

[pro

duct

s]

Time

∆ concentration

This reaction can be carried out as a simple test tube reaction, usually using a catalyst of powdered manganese(IV) oxide, but it is also a common cellular process because hydrogen peroxide is a by-product of the electron transport chain, which is the final stage of aerobic respiration. In the cellular process, the reaction is catalysed by the enzyme catalase (also known as peroxidase).

Factors affecting rates of reactionFor the decomposition of hydrogen peroxide described above, similar factors affect the rate of both the test tube reaction and the cellular process:

1 concentration of the reactant (hydrogen peroxide)

2 temperature at which the reaction is carried out

3 presence of a catalyst (manganese oxide or the enzyme catalase)

4 size (and hence surface area) of any solid particles (including the catalyst) involved in the reaction.

The rate of reactions that involve gases as reactants will also be affected by the pressure of the gases.

Collision theory

The effect of these factors can be explained by using the ideas of collision theory:

‘the rate of a chemical reaction depends on the number of successful collisions per second between the reacting molecules.’

A successful collision is one in which the reacting molecules collide with sufficient energy (the activation energy) to overcome the energy barrier to a reaction, and in the correct orientation.

Figure 5.1.10: These graphs show how the concentration of reactants

and products might vary during a reaction (for example, during the

decomposition of hydrogen peroxide).

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Effect of factors on rate

Collision theory can be used to explain how the various factors listed above affect the rate of simple reactions:

1 When the concentration of dissolved reactant molecules increases, there will be more frequent collisions between the reactant particles – the rate will increase.

2 When the temperature is higher, a greater fraction of collisions will be between molecules possessing combined energy greater than the activation energy so a greater fraction of collisions will be successful and the rate will increase. Figure 5.1.11 shows how the distribution of molecular energies changes with temperature and how this will affect the fraction of successful collisions and therefore the activation energy.

N

E

Eact

N

E

Eact

N

E

Eact

(a) (b) (c)

3 When solid particles are broken up into smaller particles, the surface area on which a collision can occur increases so there will be more frequent collisions between reactant molecules and this surface area.

4 When a catalyst is present, the activation energy is lowered. So as you can see in Figure 5.1.11, a greater fraction of collisions are successful and the rate will increase.

Enzymes as catalysts

Enzymes, like all catalysts, provide an alternative pathway for a reaction with a reduced activation energy. The active site of the enzyme binds to a specific substrate molecule (the hydrogen peroxide in the case of catalase) to form an enzyme–substrate complex. This facilitates the conversion of substrate into product molecule(s), which are subsequently released from the enzyme.

Portfolio activity (1.3)Sucrose can be broken down (‘hydrolysed’) to produce glucose and fructose, in the presence of an acid catalyst or the enzyme sucrase. This reaction occurs during the digestion of sucrose and is also widely used in the food industry. Discuss and explain the factors that might affect the rate of this reaction.

In your answer you should: • identify the factors which will affect the rate of the reaction • describe how each factor is likely to affect the rate • use collision theory to explain the patterns you describe.

Figure 5.1.11: Graphs showing the distribution of molecular energies at

(a) low temperature (b) high temperature (c) at low temperature but in the

presence of a catalyst. N represents the relative number of molecules

possessing kinetic energy E. The area shaded yellow represents collisions of

these molecules that will be successful.

ActivityMany enzymes are used in industrial processes. Conditions are chosen for these processes to ensure maximum rate for the enzyme-catalysed reactions. However, the temperature used in these processes is rarely much above 40 °C.

Research how high temperatures affect the activity of an enzyme and explain why higher temperatures cannot be used.

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Case study: Enzymes in industryEnzymes are commonly used in the food industry either in solution or immobilised (attached to or entrapped in a solid support). Examples of reactions that make use of enzymes are:

• the production of high fructose syrup from glucose (a desirable reaction in the confectionary industry) using immobilised glucose isomerase enzyme

• the use of solutions of microbial rennet to replace rennet from animal sources in the cheese industry. The chymosin enzyme in the rennet breaks down the casein proteins in the milk. Casein is an emulsifier – it helps stabilise the colloid structure of milk, so without casein, solid curds precipitate out (from which cheese is made).

1 Suggest the advantages of using immobilised enzymes rather than a solution of free enzymes. Are there any possible disadvantages?

2 What is a colloid? Find out why colloids will often break down unless some stabilising substance (such as an emulsifier) is present.

ChecklistAt the end of this topic you should be familiar with the following ideas:

ionic and covalent bonds are the main types of bonding holding chemical substances together

polar covalent bonds can lead to the formation of intermolecular forces such as hydrogen bonds

Van der Waals forces can form between non-polar molecules or groups

the nature and strength of the bonding (including intermolecular forces) in a chemical substance can be used to explain its physical properties

ideas about bond strengths can be used to explain the sign of the enthalpy change accompanying a chemical reaction

data about enthalpy change can be used to deduce important information about reactions

collision theory can be used to explain the effect of changing various factors on the rate of reaction, including those using enzymes as catalysts.

Further readingBTEC First: Principles of Applied Science (Goodfellow, Hocking and Musa, 2012), Chapters 1 and 2 contain a very basic introduction to ideas about chemical reactions, bonding and rates of reaction.

BTEC First: Application of Applied Science (Goodfellow, Hocking and Musa, 2012), Chapter 5 contains a very basic introduction to ideas about energy changes.

BBC Bitesize revision is useful for review of level 2 ideas if any of the material in this topic guide is unfamiliar. Go to www.bbc.co.uk/schools/gcsebitesize/science and navigate to the Chemistry section of any of the GCSE courses listed, where you will find a menu of available topics.

Chemguide (www.chemguide.co.uk) covers key ideas of Chemistry at level 3. In some cases this will provide extra input beyond what you will find in these topic guides. The section on Structure and Bonding links well to this topic guide and the section on Physical chemistry includes ideas about enthalpy changes and rates of reactions.

The Chemistry WebBook (http://webbook.nist.gov/chemistry) produced by the National Institute of Standards and Technology, gives access to databases containing thermochemical data.

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AcknowledgementsThe publisher would like to thank the following for their kind permission to reproduce their photographs:

Shutterstock.com: Anton Prado Photo.

All other images © Pearson Education

In some instances we have been unable to trace the owners of copyright material, and we would appreciate any information that would enable us to do so.

topic guide 5.1: chemical bonding, thermodynamics … 01, 2011 · thermodynamics and 5.1 ... with...

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