Topic 13: Energetics II

Some revision cards for new spec 2015 Edexcel A Level Chemistry, energetics topics :) 

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  • Created on: 09-11-16 19:11

Lattice Enthalpy and definitions

Lattice enthalpy is like to ionic compounds what bond enthalpies are to covalent molecules: it's a measure of the strength of the bonds in the ionic lattice. In this course, the lattice enthalpy refers to -

The energy change when one mole of an ionic solid is made from its gaseous ions

This is sometimes indicated to be under standard conditions, i.e. 298oK and 100kPa....

But notice that the ions are gaseous - (like when you're calculating bond enthalpies) - i.e., not in the standard states of the elements that form those ions. 

So this is where lattice enthalpy varies from standard enthalpy of formation:

Na(s) + 1/2Cl2(g) ---> NaCl (s) - Formation (everything is standard states, elements used)

Na+ (g) + Cl- (g) ----> NaCl (s) - Lattice enthalpy of formation (gaseous ions to a solid)

As this is the lattice enthalpy of formation (as opposed to dissociation, which is the same thing but breaking down) the values are always negative - energy is released when new bonds are made. They are exothermic reactions.

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Factors affecting Lattice Enthalpy

Lattice enthalpy is affected by a few main factors:

  • The magnitude of the charge on the ions
  • The size of the ionic radii
  • The degree of covalent character (covered later).

In general, the smaller and more highly charged the ions are, the stronger the electrostatic forces of attraction and the more exothermic the lattice enthalpy.

(Note: always talk about lattice enthalpy in terms of more or less exothermic, as simply "more or less" is vague: technically, -890 kJmol-1 is less than -760 kJmol-1 ; but -890 kJmol-1 is releasing more energy, so describing the lattice enthalpy as "less" would be very confusing! You could also talk in terms of magnitude, which ignores the sign.) 

So, MgCl2 has a more exothermic lattice enthalpy than NaCl, as Mg has the 2+ charge and is smaller than Na+ . Furthermore, MgCl2 has more cation-anion interactions as there are two Cl- per cation as opposed to just one in NaCl. 

If the charges are all the same (like LiF and NaF) look at the ionic radii. Li+ is smaller than Na+ (one fewer quantum shells) so the lattice enthalpy is more exothermic. The ions are closer to each other.

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Atomisation and Electron Affinity

More definitions!

The standard enthalpy of atomisation is when one mole of gaseous atoms is made from an element in its standard state (i.e. under 298oK and 100kPa). For example:

Mg(s) --> Mg(g)

1/2Cl2(g) --> Cl(g)  (The diatomic molecule of chlorine gas is becoming one atom of gaseous Cl)

As the second example shows, you want just ONE mole of ATOMS, irrespective of whether the element is diatomic. You can probably tell then that only monoatomic gases - i.e. the noble gases - have a value of 0KJmol-1 for this.

Electron affinity is almost like the opposite of ionisation energy. The first electron affinity is the energy change when one mole of gaseous atoms gain an electron each to become one mole of uninegative  (-1) ions:

O(g) + e- --> O- (g) Second electron affinity tends to be endothermic (positive enthalpy change) as you are attracting an electron to something that is already negative and need an input of energy to overcome the forces of repulsion from the like charges.

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Born-Haber Cycles

Born-Haber cycles are like Hess' Law cycles but... more. You can think of them as the "process" used to make an ionic solid, starting from the elements in their standard states. The cycles use all the definitions mentioned already, as well as formation and ionisation (see AS), and can be used to find lattice enthalpy or any other missing enthalpy value.Imagine the 'process' you'd go through to produce MgCl2 (s) from Mg(s) and Cl2(g).

  • First you'd atomise them both to form Mg and 2Cl. 
  • Then you'd ionise them - the Mg would lose two electrons (first and second ionisation energy) and the 2Cl would gain one each (twice the 1st electron affinity of chlorine)
  • So now you have Mg2+ (g) and 2Cl- (g). So the next step would be the lattice enthalpy to form MgCl2 (s).

This can be presented as a cycle which includes the enthalpy of formation to produce a closed loop (remember energy can't be created or destroyed, so the total energy in the system is the same wherever you start and end... the enthalpy of formation is like the "net" energy difference between the original elements and the ionic salt produced.)

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Born-Haber Cycle II

Image result for born haber cycle mgo (http://www.docbrown.info/page03/3_51energy/BHcycleMgCl2.gif)This is a graphical representation of that whole cycle (we "started" in the bottom left, with the elements).  You can see that:

  • Arrows point up for endothermic (the system gains energy)
  • Arrows point down for exothermic (losing energy to surroundings).
  • The most stable state (i.e. lowest energy) is the ionic salt.
  • (Note this has the Lattice Enthalpy of dissosciation, so is endothermic)
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Calculations from Born-Haber Cycles

Born-Haber cycles can be used to find any missing value.

  • Like with Hess' Law Cycles, direction of the arrow is important. To find a magnitude of an unknown arrow, you'd start at the top of the arrow and move around the cycle to the bottom of the arrow (i.e. the point). 
  • For example, to find blue arrow 4, you'd move "against"  the red arrows 4,3,2,1, "with" the grey arrow and "against" blue arrow 5. Remember when you move against an arrow you reverse the sign of the enthalpy change. (E.g., if it's negative it becomes positive).
  • You can see also that the enthalpy of formation, which is the GREY arrow, is equal to all the other arrows (i.e. enthalpy values) added up (taking into account the charge), i.e. 

 Enthalpy of Formation = 1+2+3+4+5 

This is ALWAYS true, so to find any missing value, just rearrange this equation.

 Image result for born haber cycle mgcl2 (http://wps.prenhall.com/wps/media/objects/602/616516/Media_Assets/Chapter06/Text_Images/FG06_09.JPG)

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Covalent Character

Experimental lattice enthalpy is found from calculations using Born-Haber cycles and values found from experiments. Theoretical lattice enthalpy is found when X-Rays are used to find the ionic spacing and size, data which can be used to calculate the strength of ionic bonding in the lattice. The calculations assume that all the ions are perfectly spherical and evenly charge distributed.

However, experimental data (Born-Haber) sometimes varies quite a lot from theoretical calculations. This means there must be another interaction occurring - the bonding is not completely 100% ionic. These ionic lattices are said to have a degree of covalent character.

Image result for polarisation of an anion

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Polarisation

Covalent character arises when the cation in the ionic lattice is a good polariser, and the anion is polarisable. This occurs when:

  • The ions have large charges
  • The cation is small
  • The anion is large

A small, highly charged cation - such as Mg2+ - is very charge dense. It's able to distort the electron cloud of the large anion (its outer electrons are susceptible to distortion as they have a lot of energy). This ability to distort is known as the cations polarising power. 

For example, NaCl has a very close theoretical value to its experimental. Na+ has only a 1+ charge and Cl- is fairly small and not easily distorted. There isn't much covalent character.

MgI2 would be a lot more covalent, as the I- anion is larger and Mg2+ more charge dense. Its experimental value is more exothermic than the theoretical predicts.

So if you see a large % difference between the prediction and the observed enthalpy - there's some covalent character.

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Enthalpy of Solution and Hydration

MORE definitions!

Enthalpy of solution is a factor that determines if an ionic solid is soluble or not. It can be either positive or negative. It is:

The enthalpy change when one mole of an ionic solid dissolves in water to produce its aqueous ionsat infinite dilution.

(Remember: Solution Salt) MgO (s) ---> Mg2+(aq) + O2- (aq)

Enthalpy of hydration is when one mole of an ion in the gaseous state is hydrated completely to form one mole of aqueous ions:

Mg2+ (g) ---> Mg2+ (aq) It's always negative, as bonds are being formed between the ions and water. Smaller, more highly charged ions have more exothermic enthalpies of hydration.Image result for enthalpy of solution symbol (http://figures.boundless-cdn.com/32461/full/83cteivdthi6ycdkaigi.png)- A hydrated Na+ ion, surrounded by the d- oxygen of water.

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Hess' Law and Lattice Enthalpy

But how does this relate to lattice enthalpy? Enthalpy of solution and hydration have something in common with each other - (aq) ions. They also have the different states of lattice enthalpy - the gaseous ions (enthalpy of hydration) and the ionic solid (enthalpy of solution). So, a Hess' law triangle can be used to find one of these unknowns when you have the other two...

  • the +2258 is the lattice enthalpy of dissociation of Image result for hess law enthalpy of solution, hydration, lattice enthalpy (http://www.chemguide.co.uk/physical/energetics/dhsolcacl2.gif)CaCl2 (opposite to our lattice enthalpy but the same value, you could reverse the arrow and make the sign negative if you wanted to)
  • the -1650 and 2(-364) are the enthalpies of hydration of Ca2+ and 2Cl- respectively.
  • Do Hess' law as usual to find the missing value, enthalpy of solution.
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Entropy

Entropy is the measure of disorder in a system. For example, a gas is more disordered than a solid as the particles have a random arrangement and movement rather than vibrating in fixed, ordered positions.Image result for disorder solids liquids gas (http://www.brainfuse.com/quizUpload/c_83128/ComparisonBetweenTheStates.GIF)

Entropy change has the units JK-1mol-1 and the symbol Related image (http://vt-s3-files.s3.amazonaws.com/uploads/formula_image/image/24475/gif.latex). Note that it is measured in Joules, not kilojoules, as these are SI units - so when you use molar entropies with enthalpy data you will need to convert one of the values (x or / by 1000).

Entropy is an important consideration when you are deciding whether or not a reaction is spontaneous...

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Spontaneous Reactions

Spontaneous reactions happen without any continuous external intervention such as continually heating the reactants strongly. For example, ice will not melt spontaneously at -20oC.

The reverse reaction of a spontaneous process is not spontaneous (e.g., ice freezes at -20oC but will not melt; gases will diffuse together but not separate out into discreet gases again...)

You may think that spontaneous reactions have to be exothermic, as they release heat to form a more stable (lower energy) state ; whilst endothermic reactions need that external input of heat.

However, spontaneous reactions can be endothermic.

This means that enthalpy change alone does not dictate the spontaneity of a reaction. 
Entropy change also does, and the natural progression is for a system's entropy change to be positive - entropy increases.

However, the entropy of a system doesn't have to increase for a reaction to be spontaneous - think of water freezing at -20oC (A solid is more "ordered", with a lower entropy than the liquid.) What is true is that the total entropy must increase for a reaction to be spontaneous.

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Total Entropy

So what is "total entropy"? Total entropy doesn't just take into account the entropy change of your reactants to products (delta S System) but also the entropy change of the surroundings (delta S surroundings) - that is, maybe the beaker the reaction takes place in and the air around it.

Total entropy change = Entropy change of surroundings + entropy change of system

Where Entropy change of system = Entropy of Products - entropy of reactants (remember that a +ve value indicated that entropy increases, i.e. the products are more disordered than reactants)

and Entropy change of surroundings = -Enthalpy change of reaction / Temperature in K (remember to convert enthalpy change into joules per mole, and don't forget the - sign)

For a reaction to be spontaneous, total entropy change must be positive.

Related image (http://wilsonjoneswilson.org/joneswilson/chem350/chem350notes/thermodynamics/second_law.png)

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Entropy change of surroundings

Looking at the entropy of surroundings equation in more detail, we can see that:

  • EXOTHERMIC reactions will have a positive value, as enthalpy change will be negative and the two negatives make a positive. This makes sense: if you imagine the system transferring heat outwards, it's warming up its surroundings and giving them more energy - making the air particles, for example, move around faster and thus be more disordered (higher entropy).
  • ENDOTHERMIC reactions always have a decrease in entropy of the surroundings for the reverse reasons.
  • It also tells you that the entropy change depends on the temperature. This also makes common sense: imagine your surroundings are already very hot (T is very large), the heat released by your reaction taking place won't have much of a relative impact on the surroundings' particles' entropy! A large value of T makes the change in entropy of surroundings more insignificant.Image result for entropy change of surroundings equation (http://apolympics.weebly.com/uploads/8/8/4/9/8849208/9256421_orig.png)
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Summary of spontaneous reactions

Total entropy change will be positive (and therefore could be spontaneous) if:

  • Both surroundings and system are positive
  • Surroundings is positive and larger than negative system
  • Surroundings is negative and smaller than positive system.

Why does water freeze below 0oC?

  • System=negative, as Ice is more ordered than water ( (g) --> (l) is a negative change)
  • Surroundings=positive: the process is exothermic. If it's more than system, it's spontaneous

If we calculate total entropy at 1oC / 274oK:

(Enthalpy of freezing water is -6010 Jmol-1; molar entropy of water is 69.9 JK-1mol-1 and ice is 47.9JK-1mol-1) 

System: 47.9 - 69.9 = -22.0JK-1mol-1 Surroundings: +6010/274 = 21.93 Total: -0.066JK-1mol-1 : won't happen.

Repeating at -1oC / 272oK, you get 22.096 - 22 = +0.096Jmol-1K-1, So it WILL happen!

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Predicting entropy changes

We know that Solid < liquid << gas in terms of entropy. So you can predict if a reaction will have a positive entropy or not: for example,

Ba(OH)2.8H2O(s) + 2NH4Cl(s) -> BaCl2 (s) + 10H2O(l) + 2NH3 (g)  

Although this is endothermic (and so Surroundings will be negative) the entropy of  the system is sufficiently increased and so overall it is a positive entropy change - you can maybe guess here, as gases and aqueous solutions have been produced from solids. The number of moles also gives you a clue: mole moles=more particles=more possible arrangements in space=more disordered=higher entropy. Here, there are 3 moles on the LHS and 13 on the RHS, so a large increase! 

To find out for sure which factor is greater in determining the feasibility, you need to have data:

  • The enthalpy change of the reaction (remember to convert to Joules!)
  • The molar entropies of the substances involved (check what the values quoted are for, e.g. if you're given the entropy of O2 or O for example; so you can multiply or divide as needed)
  • The temperature in K that the reaction is happening at.
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Predicting whether an ionic solid will dissolve

When an ionic substance dissolves, two changes happen. The lattice breaks down, and the ions are hydrated. You may remember that this is the enthalpy of solution, which is equal to the lattice enthalpy of dissociation (the opposite of lattice e. of formation) plus the enthalpy of hydration (look back at Hess' cycle on card 10).

  • Breaking down the lattice is endothermic. It's also got a positive entropy as the number of moles of particles increases. (NaCl (s) --> Na+ (g) + Cl- (g) +787 kJmol-1)
  • The hydration of ions is exothermic. It's got a negative entropy, however, as the water molecules become more ordered (clustered together). (Na+ (g) + Cl-(g)--> Na+(aq)+Cl-(aq) -784kJmol-1)

These factors are differently weighted for ionic compounds, meaning some will dissolve readily at certain temperatures whilst others will not.

You can adapt the total entropy equation to be

Total entropy = Entropy of system + (-enthalpy of solution / Temperature) 

where the last part is the entropy of the surroundings: the enthalpy of solution represents the overall enthalpy change during the reaction.

If this value is positive overall, the salt is soluble at that temperature.

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Gibbs Energy

The Gibbs Energy is another value that can be used to find out if a reaction is feasible. It is equal to Gibbs free energy = enthalpy change - Temperature x entropy change of the system 

(You can derive this from total entropy change equation and show that the Gibbs free energy is equal to -Tx Total entropy change) 

This equation is useful because it means you don't have to calculate the entropy change of the surroundings separately; it's all in one equation. It is important to note that:

  • If the Gibbs Energy is positive, the reaction is not thermodynamically feasible
  • If the Gibbs Energy is negative, the reaction is thermodynamically feasible 

So, for example, if the enthalpy change is very negative (i.e. exothermic) and the -TdeltaS also negative (e.g. a positive entropy change and temperature) then the reaction will happen. Image result for gibbs free energy

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Applying Gibbs Free Energy - Feasible Temperatures

The Gibbs Free Energy equation can be used to 

  • Show a reaction is or isn't feasible (remember -ve value = feasible)
  • Calculate the minimum temperature at which a reaction will become feasible

As a reaction is possible when Related image (http://engineering.dartmouth.edu/maps/symbolRef_files/DeltaG.gif) is negative and not when it's positive, it's a logical conclusion that the reaction is on the point of becoming feasible when Related image (http://engineering.dartmouth.edu/maps/symbolRef_files/DeltaG.gif)= 0. When the equation is made to equal 0, you get:

Related image (http://www.newworldencyclopedia.org/math/6/0/2/60238408112cb59c20361e2926cafe20.png)= 0 therefore 

 H/Image result for delta symbolSsystem = T

if Related image (http://www.newworldencyclopedia.org/math/6/0/2/60238408112cb59c20361e2926cafe20.png) is compared to y=mx + c, you can plot a linear graph with y-axis Related image (http://engineering.dartmouth.edu/maps/symbolRef_files/DeltaG.gif) and x-axis T. The gradient will be -Image result for delta symbolS  (which we assume remains effectively constant at different temperatures, unless told otherwise) and the y-intercept will be the enthalpy change of the reaction. The x-intercept (i.e. where Gibbs energy is 0) will be the temperature at which the reaction becomes feasible - the part of the line that lies below the x-axis is the feasible part.

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Gibbs Free Energy and K

Delta G (http://engineering.dartmouth.edu/maps/symbolRef_files/DeltaG.gif) = -RT lnK      and K= e^(-Delta G (http://engineering.dartmouth.edu/maps/symbolRef_files/DeltaG.gif)/ RT )

Delta G (http://engineering.dartmouth.edu/maps/symbolRef_files/DeltaG.gif) = Gibbs Free Energy R = The gas constant, 8.31 Jmol-1K-1 T = The temperature in Kelvin

Gibbs Free Energy can be related to another value, K. This is the thermodynamic equilibrium constant (it differs from Kc and Kp, which are experimental) and has no units. 

It is important to remember that all reactions are technically in equilibrium, however, in some the position is so extreme we say they have "gone to completion" (eq. lies to the right, favours products, K is very large) or "doesn't happen" (eq. to the left, reactants favoured, K very small.)

The last equation above is a rearrangement of the first equation and can be used to show where the position of equilibrium lies. If Gibbs Energy is negative, the exponent will be positive and K will be greater than one. This will favour the products (eq lies to the right). If it is positive, K will be <1 and the reactants are favoured. 

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Gibbs Energy and Temperature

So, this shows a reaction with a positive Gibbs energy doesn't not happen, it's just that the equilibrium position favours the reactants (sometimes by quite a lot, so nothing seems to happen).

You can use two factors to deduce the equilibrium position, the sign and magnitude of the Gibbs Energy. When it is more negative that -40 kJmol-1, K will be in the order around 10^9 (the reaction is basically complete) and if it is more positive than +40 kJmol-1, K will be very small.

But the position of equilibrium is also affected by the temperature of the reaction. Combining the equations Delta G (http://engineering.dartmouth.edu/maps/symbolRef_files/DeltaG.gif) = H - TS and Delta G (http://engineering.dartmouth.edu/maps/symbolRef_files/DeltaG.gif)= -RTlnK gives -RTlnK = HTS. Rearranged to give lnK = - H/RT + S/R , which shows that:

  • In an exothermic reaction, H is negative, so the expression -H/RT becomes positive. As T increases, this value decreases and so lnK and K decrease. A decreased K = favours the reactants more = favours the reverse reaction = favours the exothermic reaction.
  • This can be applied to an endothermic reaction too.

This is a qualitative explanation of the "oppose the change" you met in Year 1

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Understanding Feasibility

However, this thermodynamic data (i.e. entropy and enthalpy change) does not tell you whether a reaction will definitely happen - only if it is feasible.

You also need to consider kinetic stability - the activation energy. For example, the reaction between hydrogen and water has a very negative Gibbs Energy., so you'd assume that it's spontaneous. However, it doesn't happen unless the hydrogen is lit - this is as the activation energy is very large.

(This may be a 1 mark exam question at the end of a longer question; if you have reached a final answer which has a negative Gibbs or a positive Total Entropy but the question asks why doesn't it happen, then it will probably be because the activation energy is too large. This is also used in multiple choice where the question may be "which of these is the thermodynamic/ kinetic explanation for why this reaction does not happen.... remember that kinetics refers to rates and activation energy, and thermodynamics is all about enthalpy changes and entropy)

 As with all questions in energetics and enthalpy, often the reactions in practice are different to the equations on paper as standard conditions can't or are not be maintained. For example, a standard (aq) solution is at 298 oK and 100kPa, and 1.0moldm-3. This may result in a reaction with a positive (and thus unfeasible) value of Gibbs. However, if a higher concentration is used in the lab, then it may become feasible (Gibbs will be negative, the position of equilibrium could be changed.).

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