4.2 Entropy

Entropy = a measure of how much disorder there is in a substance, how many different ways particles can be arranged.

Systems are MORE energetically stable when disorder/entropy is HIGH.

EXAMPLE: A gas will want to escape its bottle because the room it’s in is much bigger and the particles can be arranged in lots of different ways.

SOLID

LIQUID

GAS

No randomness, therefore lowest entropy.

e.g. S^ѳ (H₂O_(s)) = 7.4 JK^-1mol^-1

(see below)

Some randomness, some entropy.

e.g. S^ѳ (H₂O_(l)) = 70 JK^-1mol^-1

(see below)

Most randomness, highest entropy.

e.g. S^ѳ (H₂O_(g)) = 189 JK^-1mol^-1

(see below)

*Note that zero entropy will only occur in a perfectly ordered crystal

Affecting Factors:

More quanta (packets of energy) = More ways to arrange particles = More entropy
More particles = More arrangements = More entropy.

E.G. X -> 2Y 2 moles of Y produced from 1 mole of X therefore entropy has increased

Increase in temperature = Increase in energy = More entropy

E.G. – from solid to liquid entropy has increased a bit

– from liquid to gas entropy has increased a lot

Complicated/complex molecules = more entropy

DEFINITIONS:

Standard entropy of a substance, S^ѳ, is the entropy of one mole of a substance under standard conditions of 298K and 1atm. The units are JK^-1mol^-1.

We expect exothermic reactions to be the spontaneous ones; however some endothermic reactions are spontaneous too. This is to do with entropy. If entropy is high enough, the reaction will be spontaneous, whether the reaction is exo/endothermic.

EXAMPLE:

NaHCO₃(s) + H⁺(aq) —> Na⁺(aq) + CO₂(g) + H₂O(l)

1 mole 1 mole 1 mole 1 mole 1 mole

Solid aqueous ions aqueous ions gas liquid

Here, the products have high entropy states (e.g. gas) and there are more moles (e.g. reactants to products = 2:3) And so, overall entropy has increased = SPONTANEOUS (also depends on ΔH – see below)

LEARN THESE:

ΔS_sys = S_products– S_reactants

ΔS_total = ΔS_sys+ ΔS_surr

Where;

ΔS_sys = Entropy change of a system, the entropy change between the reactants and the products

ΔS_surr = Entropy change of a surrounding

ΔS_total = Total entropy change, the sum of the entropy changes of the system and the surroundings

EXAMPLE:

NH₃(g) + HCl(g) —> NH₄Cl(s)

Info: ΔH = -315kJmol^-1

S^ѳ (NH_3(g)) = 192.3 JK^-1mol^-1S^ѳ (HCl_(g)) = 186.8 JK^-1mol^-1S^ѳ (NH₄Cl_(s)) = 94.6 JK^-1mol^-1

Find entropy of the system

ΔS_sys = S_products– S_reactants

= 94.6 – (192.3 + 186.8)

= – 284.5 JK^-1mol^-1

Find entropy of surroundings

= – (-315000)/298 [Note: ΔH = -315kJmol^-1 is in KILOJOULES, therefore x1000]

= + 1057 JK^-1mol^-1

Find total entropy

ΔS_total = ΔS_sys+ ΔS_surr

= -284.5 + 1057

= + 772.5 JK^-1mol^-1 [Note: must include sign (and units) with final answer]

When will a reaction be spontaneous?

Total entropy must increase
+ ΔS_total = kinetically favourable (wants to react; spontaneous)
— ΔS_total = kinetically stable (will not react on its own; not spontaneous)

* You can predict ionic compound solubility using the same idea; if ΔS_total is positive √, if negative X

ENDOTHERMIC experiments that are spontaneous:

Ba(OH)₂(s) and NH₂Cl(s)

Ba(OH)₂.8H₂O(s) + 2NH₂Cl(s) —> BaCl₂(s) + 10H₂O(l) + 2NH₃(g)

When you add barium hydroxide to ammonium chloride:

Smell of ammonia gas
Temperature drops below 0˚C

Cold pack – NH₄NO₃(s) and H₂O(l)

NH₄NO₃(s) —H₂O(l)—> NH₄⁺(aq) + NO₃^–(aq)

When you dissolve ammonium nitrate crystals in water:

Looking at the states in both these experiments, we have an INCREASE in entropy (from solids to liquids/aqueous). These reactions are spontaneous EVEN THOUGH the ΔS_surr is negative (because if ΔH is positive for endothermic reactions the equation of ΔS_surr means the overall ΔS_surr will be negative – see above equation) the ΔS_sys is GREAT ENOUGH to overcome it, meaning ΔS_total will be positive still.

DEFINITIONS:

Thermodynamic stability – where the ΔS_total is negative, at RTP, the reaction will simply not occur. E.G. limestone –> CaO + CO₂

Kinetic inertness – when the ΔS_total of a reaction is positive, a reaction can happen spontaneously, however the rate of reaction at RTP is so slow because the activation energy needed for it to start is so high. E.G. diamond –> graphite

The enthalpy change of hydration, ΔH_hyd – the enthalpy change when 1 mole of aqueous ions is formed from gaseous ions. E.G. Na⁺(g) —> Na⁺(aq)

The standard lattice enthalpy, ΔH^ѳ_latt – the enthalpy change when 1 mole of a solid ionic compound is formed from gaseous ions under standard conditions (298K and 1atm). E.G. Na⁺(g) + Cl^–(g)—> NaCl(s)

The enthalpy change of solution, ΔH_sol – the enthalpy change when 1 mole of solute is dissolved in sufficient solvent, so no further enthalpy change occurs on further dilution. E.G. NaCl(s) —> NaCl(aq)

Factors affecting ΔH^ѳ_latt AND ΔH_hyd include;

Ionic charge; = larger charge

= more exothermic lattice energy

= MORE NEGATIVE LATTICE ENTHALPY/ENTHALPY OF HYDRATION

E.G. NaCl has ΔH^ѳ_latt = -780kJmol^-1 whereas MgCl₂ has ΔH^ѳ_latt = -2526kJmol^-1 because magnesium has a charge of 2+ which is greater than sodium’s 1+

Ionic radii; = smaller ionic radii

= more exothermic lattice enthalpy

= higher charge density

= MORE NEGATIVE LATTICE ENTHALPY/ENTHALPY OF HYDRATION

E.G. Sodium’s ionic radius is bigger than magnesium’s (because Mg has one more proton which has a stronger positive nuclear attraction to its electrons – see unit 1/2) therefore magnesium will have a more negative lattice enthalpy/hydration enthalpy.

Finding the enthalpy of solution

where we use a similar principle to Hess’ Law;

ΔH₁ = ΔH₂+ ΔH₃

REMEMBER (for ΔH_sol ): GASEOUS IONS DOWN, AQUEOUS IONS UP

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