23 - chemical energetics

enthalpy change (ΔH)

refers to the amount of heat energy transferred during a chemical reaction, at a constant pressure

lattice energy (ΔH_latt^ꝋ)

The enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions (under standard conditions)

is lattice energy endo or exo

• The ΔH_latt^ꝋ is always exothermic, as when ions are combined to form an ionic solid lattice there is an extremely large release of energy

  • The enthalpy change will always have a negative value

  • Because of the huge release in energy when the gaseous ions combine, the value will be a very large negative value

• The large negative value of ΔH_latt^ꝋ suggests that the ionic compound is much more stable than its gaseous ions

  • This is due to the strong electrostatic forces of attraction between the oppositely charged ions in the solid lattice

  • Gaseous ions are less stable than ions in an ionic lattice because they are not stabilized in a lattice by strong electrostatic attractions, so they have higher potential energy.

  • The more exothermic the value is, the stronger the ionic bonds within the lattice are

how to determine lattice energy (or how not to)

• The ΔH_latt^ꝋ of an ionic compound cannot be determined directly by one single experiment

• Multiple experimental values and an energy cycle are used to find the ΔH_latt^ꝋ of ionic compounds

• The lattice energy (ΔH_latt^ꝋ) of an ionic compound can be written as an equation

  • E.g. magnesium chloride is an ionic compound formed from magnesium (Mg²⁺) and chloride (Cl^-) ions:

Mg²⁺ (g) + 2Cl^- (g) → MgCl₂ (s)

standard enthalpy change of atomisation (ΔH_at^ꝋ)

The enthalpy change when 1 mole of gaseous atoms is formed from its element under standard conditions

is ΔH_at^ꝋ endo or exo

• The ΔH_at^ꝋ is always endothermic as energy is always required to break any bonds between the atoms in the element into its gaseous atoms

  • Since this is always an endothermic process, the enthalpy change will always have a positive value

• Equations can be written to show the standard enthalpy change of atomisation (ΔH_at^ꝋ) for elements

  • E.g. The standard enthalpy change of atomisation for sodium is the energy required to form 1 mole of gaseous sodium atoms:

Na (s) → Na (g) ΔH_at^ꝋ = +107 kJ mol ^-1

examiner tips and tricks (standard conditions)

Standard conditions in this syllabus are a temperature of 298 K and a pressure of 101 kPa

Make sure the correct state symbols are stated when writing these equations – it is crucial that you use these correctly throughout this entire topic

first electron affinity

• The first electron affinity (EA₁) is the enthalpy change when 1 mole of electrons is added to 1 mole of gaseous atoms, to form 1 mole of gaseous ions each with a single negative charge under standard conditions

X (g) + e^– → X^– (g)

• EA₁ is usually exothermic, as energy is released

• The value for EA₁ will usually be negative

• EA₁

  • X (g) + e^– → X^– (g)

  • Exothermic

second and third electron affinites

• An element can also accept more than one electron, in which case successive electron affinities are used

  • For example, the second electron affinity (EA₂) and third electron affinity (EA₃) of an element represent the formation of 1 mole of gaseous ions with 2- and 3- charges respectively

• The second and third electron affinities are endothermic, as energy is absorbed

  • This is because the incoming electron is added to an already negative ion

  • Energy is required to overcome the repulsive forces between the incoming electron and negative ion

  • The values will be positive

  • Only the first electron affinity is negative (exothermic) because it involves adding an electron to a neutral atom, which can attract that electron quite strongly.

• EA₂

  • X^– (g) + e^– → X^2– (g)

  • Endothermic

• EA₃

  • X^2– (g) + e^– → X^3– (g)

  • Endothermic

factors affecting electron affinity

The electron affinity of an element depends on how strongly the nucleus attracts an incoming electron.

• The stronger the attraction between the nucleus and the incoming electron, the more energy is released, making the electron affinity more exothermic (more negative)

• The factors that affect electron affinity are the same as those that influence ionisation energy:

• Nuclear charge

  • A higher nuclear charge means a stronger pull on the incoming electron, resulting in a more exothermic electron affinity

• Distance (atomic radius)

  • A larger distance between the nucleus and the outermost shell reduces the attractive force, making electron affinity less exothermic

• Shielding

  • More inner electron shells increase shielding, which weakens the nuclear attraction for the incoming electron, leading to a less exothermic electron affinity

trends in electron affinity of group 16 and group 17 elements

• Electron affinities of non-metals become more exothermic across a period, with a maximum at Group 17

• There is generally a downwards trend in the size of the electron affinities of the elements in Group 16 and 17

  • The electron affinities generally become less exothermic for each successive element going down both Groups, apart from the first member of each Group (oxygen and fluorine respectively)

Electron affinity table

Electron Affinity / kJ mol–1
Group 16	Group 17
O = –141	F = –328
S = –200	Cl = –345
Se = –195	Br = –325
Te = –190	I = –295

• An atom of chlorine has a higher nuclear charge than sulfur

  • This stronger nuclear charge results in a greater attraction between the nucleus and the incoming electron

  • Therefore, more energy is released when an electron is added to chlorine, making its first electron affinity (EA₁) more exothermic than that of sulfur

• As you move down Group 16 or Group 17:

  • The outermost electrons are farther from the nucleus, so the attractive force is weaker

  • There are more electron shells, increasing shielding and further reducing nuclear attraction

  • It becomes more difficult to add an electron to the outer shell

  • Less energy is released, so the electron affinity becomes less exothermic

• Fluorine as an exception

  • Fluorine has a very small atomic radius, resulting in:

  • High electron density around the nucleus.

  • Increased repulsion between the incoming electron and the existing electrons.

  • This repulsion weakens the overall attraction to the nucleus.

  • As a result, fluorine’s first electron affinity is less exothermic than expected, and it is actually lower than that of chlorine.

born-haber cycle

• A Born-Haber cycle is a specific application of Hess's Law for ionic compounds and enable us to calculate lattice enthalpy which cannot be found by experiment

• The basic principle of drawing the cycle is to construct a diagram in which energy increases going up the diagram

The basic principle of a Born-Haber cycle

The direction of the arrows in Born-Haber cycles indicates if a reaction is exothermic or endothermic

• The cycle shows all the steps needed to turn atoms into gaseous ions and from gaseous ions into the ionic lattice

• The alternative route to the ionic lattice begins from the enthalpy of formation of the elements in their standard states

calculations using born-haber cycles

• Once a Born-Haber cycle has been constructed, it is possible to calculate the lattice energy (ΔH_latt^ꝋ) by applying Hess’s law and rearranging:

ΔH_f^θ = ΔH_at^θ + ΔH_at^θ + IE + EA + ΔH_latt^θ

If we simplify this into three terms, this makes the equation easier to see:

• ΔH_latt^θ

• ΔH_f^θ

• ΔH₁^θ (the sum of all of the various enthalpy changes necessary to convert the elements in their standard states to gaseous ions)

• The simplified equation becomes

ΔH_f^θ = ΔH₁^θ + ΔH_latt^θ

So, if we rearrange to calculate the lattice energy, the equation becomes

ΔH_latt^θ = ΔH_f^θ - ΔH₁^θ

• When calculating the ΔH_latt^θ, all other necessary values will be given in the question

• A Born-Haber cycle could be used to calculate any stage in the cycle

  • For example, you could be given the lattice energy and asked to calculate the enthalpy change of formation of the ionic compound

  • The principle would be exactly the same

  • Work out the direct and indirect route of the cycle (the stage that you are being asked to calculate will always be the direct route)

  • Write out the equation in terms of enthalpy changes and rearrange if necessary to calculate the required value

• Remember: sometimes a value may need to be doubled or halved, depending on the ionic solid involved

  • For example, with MgCl₂ the value for the first electron affinity of chlorine would need to be doubled in the calculation, because there are two moles of chlorine atoms

  • Therefore, you are adding 2 moles of electrons to 2 moles of chlorine atoms, to form 2 moles of Cl^- ions

examiner tips and tricks

Make sure you use brackets when carrying out calculations using Born-Haber cycles as you may forget a +/- sign which will affect your final answer!

factors affecting lattice energy

charge and radius of the ions that make up the crystalline lattice

how does ionic radius affect lattice energy

• The lattice energy becomes less exothermic (numerically less negative) as the ionic radius increases

  • In larger ions, the charge is spread over a greater volume, so charge density is lower

  • The ions are also further apart in the lattice, increasing the distance between their centres

  • As a result, the electrostatic attraction between oppositely charged ions is weaker, and less energy is released when the lattice forms

• For example:

  • Both CsF and KF contain F⁻ ions

  • Cs⁺ is larger than K⁺, so the lattice energy of CsF is less exothermic than that of KF

Trends in the lattice energy of different metal halides

The lattice energies get less exothermic as the ionic radius of the ions increases

how does ionic charge affect lattice energy

The lattice energy becomes more exothermic (more negative) as the ionic charge increases.

• Higher ionic charge leads to higher charge density

• This results in stronger electrostatic attractions between oppositely charged ions

• Consequently, more energy is released when the lattice is formed

• For example:

  • CaO is composed of Ca²⁺ and O²⁻ ions, while KCl contains K⁺ and Cl⁻ ions

  • The ions in CaO have higher charges, so the electrostatic forces are stronger

  • Therefore, the lattice energy of CaO is more exothermic than that of KCl

  • Additionally, Ca²⁺ and O²⁻ ions are smaller than K⁺ and Cl⁻, further increasing the exothermic value

standard enthalpy change of solution (ΔH_sol^θ

• the enthalpy change when 1 mole of an ionic substance dissolves in sufficient water to form a very dilute solution under standard conditions

• The symbol (aq) is used to show that the solid is dissolved in sufficient water

  • For example, the enthalpy changes of solution for potassium chloride are described by the following equations:

KCl (s) + aq → KCl (aq)

OR

KCl (s) + aq → K⁺ (aq) + Cl^– (aq)

• ΔH_sol^θ can be exothermic (negative) or endothermic (positive)

Enthalpy change of hydration

• The lattice energy (ΔH_latt^θ) of KCl is -711 kJ mol^-1

  • This means that 711 kJ mol^-1 is released when the KCl ionic lattice is formed

  • Therefore, to break the attractive forces between the K⁺ and Cl^- ions, +711 kJ mol^-1 is needed

• However, the ΔH_sol^θ of KCl is +26 kJ mol^-1

• This means that another +685 kJ mol^-1 (711 - 26) is required to break the KCl lattice

• This is compensated for by the standard enthalpy change of hydration (ΔH_hyd^θ)

  • The standard enthalpy change of hydration (ΔH_hyd^θ) is the enthalpy change when 1 mole of a specified gaseous ion dissolves in sufficient water to form a very dilute solution under standard conditions

Mg²⁺(g) + aq → Mg²⁺(aq)

• Hydration enthalpies are the measure of the energy that is released when there is an attraction formed between the ions and water molecules

  • Hydration enthalpies are exothermic

Ion–Dipole interactions during dissolution

• When an ionic solid dissolves in water, it breaks into positive and negative ions

• Water is a polar molecule, with a δ⁻ oxygen atom and δ⁺ hydrogen atoms

• The oxygen atom is attracted to positive ions (cations)

• The hydrogen atoms are attracted to negative ions (anions)

• These attractions form ion–dipole bonds between the water molecules and the ions

Interactions of polar water molecules and other ions in solution

The polar water molecules will form ion-dipole bonds with the ions in solution (a) causing the ions to become hydrated (b)

Energy Cycle Using Enthalpy Changes & Lattice Energy

• The standard enthalpy change of hydration (ΔH_hyd^θ) can be calculated by constructing energy cycles and applying Hess’s law

Example energy cycle

Energy cycle involving enthalpy change of solution, lattice energy, and enthalpy change of hydration

• The energy cycle shows that there are two routes to go from the ionic lattice to the hydrated ions in an aqueous solution:

  • Route 1: going from ionic solid → ions in aqueous solution (this is the direct route)

ΔH_sol^θ = enthalpy of solution

• Route 2: going from ionic lattice →  gaseous ions  →  ions in aqueous solution (this is the indirect route)

-ΔH_latt^θ + ΔH_hyd^θ = reverse lattice enthalpy + hydration enthalpies of each ion

• Lattice enthalpy usually means Lattice formation enthalpy, in other words bond forming

  • If we are breaking the lattice then this is reversing the enthalpy change so a negative sign is added in front of the term (alternatively it is called latticedissociation enthalpy)

• According to Hess’s law, the enthalpy change for both routes is the same, such that:

ΔH_sol^θ = -ΔH_latt^θ + ΔH_hyd^θ

ΔH_hyd^θ = ΔH_sol^θ + ΔH_latt^θ 

• Each ion will have its own enthalpy change of hydration, ΔH_hyd^θ, which will need to be taken into account during calculations

  • The total ΔH_hyd^θ is found by adding the  ΔH_hyd^θ values of both anions and cations together

Energy Cycle Calculations

• The energy cycle involving the enthalpy change of solution (ΔH_sol^θ), lattice energy (ΔH_latt^θ), and enthalpy change of hydration (ΔH_hyd^θ) can be used to calculate the different enthalpy values

• According to Hess’s law, the enthalpy change of the direct and of the indirect route will be the same, such that:

ΔH_hyd^θ = ΔH_latt^θ + ΔH_sol^θ

• This equation can be rearranged depending on which enthalpy value needs to be calculated

• For example, ΔH_latt^θ can be calculated using:

 ΔH_latt^θ = ΔH_hyd^θ - ΔH_sol^θ

• Remember: the total ΔH_hyd^θ is found by adding the ΔH_hyd^θ values of both anions and cations together

• Remember: take into account the number of each ion when completing calculations

  • For example, MgCl₂ has two chloride ions, so when completing calculations this will need to be accounted for

  • In this case, you would need to double the value of the hydration enthalpy, since you are hydrating 2 moles of chloride ions instead of 1

The standard enthalpy change of hydration (ΔH_hyd^θ) is affected by

• the amount that the ions are attracted to the water molecules

• The factors which affect this attraction are the ionic charge and radius

how does ionic radius affect (ΔH_hyd^θ)

• The standard enthalpy change of hydration ( ΔH_hyd^θ) becomes more exothermic(more negative) as the ionic radius decreases

• Smaller ions have a higher charge density, resulting in stronger ion–dipole attractions with water molecules

• Therefore, more energy is released when these ions are hydrated

• For example:

  • Mg²⁺ in MgSO₄ is smaller than Ba²⁺ in BaSO₄

  • As a result, ΔH_hyd^θ of MgSO₄ is more exothermic than that of BaSO₄

how does ionic charge affect (ΔH_hyd^θ)

• The enthalpy of hydration becomes more exothermic as the ionic charge increases

• Higher charge leads to a greater charge density, strengthening ion–dipole attractions with water molecules

• This means more energy is released during hydration

• For example:

  • Ca²⁺ and O²⁻ in CaO have higher charges than K⁺ and Cl⁻ in KCl

  • Therefore, ΔH_hyd^θ of CaO is more exothermic than that of KCl

Comparing enthalpies of hydration

entropy

• The entropy (S) of a given system is the number of possible arrangements of the particles and their energy in a given system

  • In other words, it is a measure of how disordered a system is

• When a system becomes more disordered, its entropy will increase

• An increase in entropy means that the system becomes energetically more stable

entropy examples

Examples

• Thermal decomposition of calcium carbonate

CaCO₃ (s) → CaO (s) + CO₂ (g)

• The entropy of the system increases

  • In this decomposition reaction, a gas molecule (CO₂) is formed

  • The CO₂ gas molecule is more disordered than the solid reactant (CaCO₃), as it is constantly moving around

  • The system has become more disordered and there is an increase in entropy

• Melting ice to form liquid water:

H₂O (s) → H₂O (l)

• The water molecules in ice are in fixed positions and can only vibrate about those positions

• In the liquid state, the particles are still quite close together but are arranged more randomly, in that they can move around each other

• Water molecules in the liquid state are therefore more disordered

  • Thus, for a given substance, the entropy increases when its solid form melts into a liquid

• In both examples, the system with the higher entropy will be energetically themost stable (as the energy of the system is more spread out when it is in a disordered state)

Entropy between physical states

Melting a solid will cause the particles to become more disordered resulting in a more energetically stable system

examiner tips and tricks

Make sure you don’t confuse the system with your surroundings!

The system consists of the molecules that are reacting in a chemical reaction

The surroundings are everything else such as the solvent, the air around the reaction, test-tube, etc

states and their entropies

• All elements have positive standard molar entropy values

• The order of entropy for the different states of matter are as follows:

gas > liquid > solid

• There are some exceptions such as calcium carbonate (solid) which has a higher entropy than mercury (liquid)

• Simpler substances with fewer atoms have lower entropy values than complex substances with more atoms

  • For example, calcium oxide (CaO) has a smaller entropy than calcium carbonate (CaCO₃)

• Harder substances have lower entropy than softer substances of the same type

  • For example, diamond has a smaller entropy than graphite

how does entropy of a substance change during a change in state - solid to liquid and liquid to gas

Solid to liquid

• The entropy increases when a substance melts

  • Increasing the temperature of a solid causes the particles to vibrate more

  • The regularly arranged lattice of particles changes into an irregulararrangement of particles

  • These particles are still close to each other but can now rotate and slide over each other in the liquid

  • As a result, there is an increase in disorder

Liquid to gas

• The entropy increases when a substance boils

  • The particles in a gas can now freely move around and are far apart from each other

  • The entropy increases significantly as the particles become very disordered

how does entropy of a substance change during a change in state - gas to liquid and liquid to solid

• Similarly, the entropy decreases when a substance condenses (change from gas to liquid) or freezes (change from liquid to solid)

  • The particles are brought together and get arranged in a more regular arrangement

  • The ability of the particles to move decreases as the particles become more ordered

  • There are fewer ways of arranging the energy so the entropy decreases

Graph of entropy against temperature

The entropy of a substance increases when the tempe

how does entropy of a substance change during a change in state - dissolving and crystallisation

Dissolving

• The entropy also increases when a solid is dissolved in a solvent

• The solid particles are more ordered in the solid lattice as they can only slightly vibrate

• When dissolved to form a dilute solution, the entropy increases as:

  • The particles are more spread out

  • There is an increase in the number of ways of arranging the energy

Crystallisation

• The crystallisation of a salt from a solution is associated with a decrease in entropy

  • The particles are spread out in solution but become more ordered in the solid

Entropy changes during the dissolving of a solid

When a solid is dissolved in a solvent to form a dilute solution, the entropy increases as the particles become more disordered

entropy changes in reactions

• Gases have higher entropy values than solids

• So, if the number of gaseous molecules in a reaction changes, there will also be a change in entropy

• The greater the number of gas molecules, the greater the number of ways of arranging them, and thus the greater the entropy

Examples

• The decomposition of calcium carbonate (CaCO₃)

 CaCO₃ (s) → CaO (s) + CO₂ (g)

• The CO₂ gas molecule is more disordered than the solid reactant (CaCO₃) as it can freely move around whereas the particles in CaCO₃ are in fixed positions in which they can only slightly vibrate

  • The system has therefore become more disordered and there is an increase in entropy

• The formation of ammonia in the Haber process

N₂ (g) + 3H₂ (g) ⇋ 2NH₃ (g)

• In this case, all of the reactants and products are gases

  • Before the reaction occurs, there are four gas molecules (1 nitrogen and 3 hydrogen molecules) in the reactants

  • After the reaction has taken place, there are now only two gas molecules (2 ammonia molecules) in the products

  • So there are fewer ways of arranging the energy of the system over the products

  • The system has become more ordered causing a decrease in entropy

  • The reactants (N₂ and H₂) are energetically more stable than the product (NH₃)

Entropy Change Calculations

• The standard entropy change (ΔS_system^θ) for a given reaction can be calculated using the standard entropies (S^θ) of the reactants and products

• The equation to calculate the standard entropy change of a system is:

ΔS_system^θ = ∑S_products^θ - ∑S_reactants^θ

(where ∑ = sum of)

• For example, the standard entropy change for the formation of ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) can be calculated using this equation

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

ΔS_system^θ = (2 x ΔS^θ(NH₃)) - (ΔS^θ(N₂) + 3 x ΔS^θ(H₂))

examiner tips and tricks

Use the stoichiometry  of the equation and the correct state of the compounds when calculating the entropy change of a reaction

Gibbs free energy

• The feasibility of a reaction does not only depend on the entropy change of the reaction but can also be affected by the enthalpy change

• Therefore, using the entropy change of a reaction only to determine the feasibility of a reaction is inaccurate

• The Gibbs free energy (G) is the energy change that takes into account both the entropy change of a reaction and the enthalpy change

• The Gibbs equation is:

ΔG^θ = ΔH_reaction^θ - TΔS_system^θ

• The units of ΔG^θ are in kJ mol^-1

  • The units of ΔH_reaction^θ are in kJ mol^-1

  • The units of T are in K

  • The units of ΔS_system^θ are in J K^-1 mol^-1

examiner tips and tricks

Examiner Tips and Tricks

Careful: When calculating ΔG^θ the value for ΔS_system^θ must be divided by 1000 

The Gibbs Equation: Calculations

• The Gibbs equation can be used to calculate the Gibbs free energy change of a reaction

ΔG^θ = ΔH_reaction^θ - TΔS_system^θ

• The equation can also be rearranged to find values of ΔH_reaction^ꝋ, ΔS_system^ꝋ or the temperature, T

• For example, if for a given reaction, the values of ΔG^ꝋ, ΔH_reaction^ꝋ and ΔS_system^ꝋare given, the temperature can be found by rearranging the Gibbs equation as follows:

Reaction Feasibility

• The Gibbs equation can be used to calculate whether a reaction is feasible or not

ΔG^θ = ΔH_reaction^θ - TΔS_system^θ

• When ΔG^θ is negative, the reaction is feasible and likely to occur

• When ΔG^θ is positive, the reaction is not feasible and unlikely to occur

Reaction Feasibility & Temperature Changes - overview

• The feasibility of a reaction can be affected by the temperature

• The Gibbs equation will be used to explain what will affect the feasibility of a reaction for exothermic and endothermic reactions

exothermic reactions and Gibbs free energy

• In exothermic reactions, ΔH_reaction^θ is negative

• If the ΔS_system^θ is positive:

  • Both the first and second term will be negative

  • Resulting in a negative ΔG^θ so the reaction is feasible

  • Therefore, regardless of the temperature, an exothermic reaction with a positive ΔS_system^θ will always be feasible

• If the ΔS_system^θ is negative:

  • The first term is negative and the second term is positive

  • At high temperatures, the -TΔS_system^θ will be very large and positive and will overcome ΔH_reaction^θ

  • Therefore, at high temperatures ΔG^θ is positive and the reaction is not feasible

  • The reaction is more feasible at low temperatures, as the second term will not be large enough to overcome ΔH_reaction^θ resulting in a negative ΔG^θ

• This corresponds to Le Chatelier’s principle which states that for exothermic reactions an increase in temperature will cause the equilibrium to shift position in favour of the reactants, i.e. in the endothermic direction

  • In other words, for exothermic reactions, the products will not be formed at high temperatures

  • The reaction is not feasible at high temperatures

Summary of factors affecting Gibbs free energy for exothermic reactions

If ΔH ....	And if ΔS ....	Then ΔG is	Spontaneous?	Because
is negative < 0 exothermic	is positive > 0 more disorder	always negative < 0	Always	Forward reaction spontaneous at any T
is negative < 0 exothermic	is negative < 0 more order	negative at low T positive high T	Dependent on T	Spontaneous only at low T TΔS < H

endothermic reactions and Gibbs free energy

• In endothermic reactions, ΔH_reaction^θ is positive

• If the ΔS_system^θ is negative:

  • Both the first and second term will be positive

  • Resulting in a positive ΔG^θ so the reaction is not feasible

  • Therefore, regardless of the temperature, endothermic with a negative ΔS_system^θ will never be feasible

• If the ΔS_system^ꝋ is positive:

  • The first term is positive and the second term is negative

  • At low temperatures, the -TΔS_system^θ will be small and negative and will not overcome the larger ΔH_reaction^θ

  • Therefore, at low temperatures ΔG^θ is positive and the reaction is less feasible

  • The reaction is more feasible at high temperatures as the second term will become negative enough to overcome the ΔH_reaction^θ resulting in a negative ΔG^θ

• This again corresponds to Le Chatelier’s principle which states that for endothermic reactions an increase in temperature will cause the equilibrium to shift position in favour of the products

  • In other words, for endothermic reactions, the products will be formed at high temperatures

  • The reaction is therefore feasible

Summary of factors affecting Gibbs free energy for endothermic reactions

If ΔH ....	And if ΔS ....	Then ΔG is	Spontaneous?	Because
is positive > 0 endothermic	is negative < 0 more order	always positive > 0	Never	Reverse reaction spontaneous at any T
is positive > 0 endothermic	is positive > 0 more disorder	negative at high T positive low T	Dependent on T	Spontaneous only at high T TΔS > H