Entropy and Free Energy. Why do reactions take place? Feasible reactions take place spontaneously,...
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Transcript of Entropy and Free Energy. Why do reactions take place? Feasible reactions take place spontaneously,...
![Page 1: Entropy and Free Energy. Why do reactions take place? Feasible reactions take place spontaneously, although the rate may be slow. Generally the more negative.](https://reader033.fdocuments.us/reader033/viewer/2022051516/56649f455503460f94c665ab/html5/thumbnails/1.jpg)
Entropy and Free Energy
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Why do reactions take place?
• Feasible reactions take place spontaneously, although the rate may be slow.
• Generally the more negative ΔH the more likely the reaction.
• So why are endothermic reactions possible?
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Randomness
When a gas evaporates it spreads out.
This increases its degree of randomness.
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The same is true on mixing gases;
+
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And dissolving solids;
+
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Entropy
• Endothermic reactions are feasible if they increase disorder.
• IE Product particles are more randomly arranged than reactant particles.
• Randomness is expressed mathematically as entropy (S).
• A feasible endothermic reaction will have a positive entropy change (ΔS).
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• All entropies are +ve.• Entropies of elements in
their standard states are not zero.
• Entropies increase with temperature as the particles spread out.
• So entropies are quoted at 298K and 101kPa.
• NB Entropies of solids < liquids < gases
Substance S (JK-1mol-1)
Iron 27
Iron Oxide 88
Calcium carbonate
93
Ice 48
Water 70
Steam 189
Carbon dioxide
214
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Calculating entropy changes
• 1) Add the entropies of the products.
• 2) Add the entropies of the reactants.
• 3) Subtract the entropies of the reactants from that of the products.
• 4) Then if ΔS is positive the reaction will be feasible.
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• Eg; Calcium carbonate decomposes when heated to form calcium oxide.
• CaCO3 → CaO + CO2 ΔH = +178kjmol-1
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CaCO3 → CaO + CO2
Entropies of products;
• CaO = 40• CO2 = 214• 40 + 214 = 254 • Entropy of reactants;
• CaCO3 = 93• Entropy change• ΔS = 254 - 93• = +161 JK-1mol-1
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Gibbs Free Energy
• The feasibility of a reaction is determined by;
• 1) ΔH• 2) ΔS• These two factors are combined to
calculate Gibbs Free Energy (G).• ΔG = ΔH – TΔS• If ΔG is negative a reaction is feasible.
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Influence of temperature on reactions
• ΔG is temperature dependent.
• This means that reactions can become feasible as temperature is raised.
• Eg; CaCO3 → CaO + CO2
• At 298K ΔH = 178 kjmol-1
• ΔS = 161 jK-1mol-1 = 0.161 kjK-1mol-1
• T ΔS = 298 x 0.161 = 47.98kjmol-1
• ΔG = 178 – 47.98 = 130 kjmol-1
• So the reaction is not feasible.
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• Instead the reverse reaction occurs;
CaO + CO2 → CaCO3 ΔG = -130 kjmol-1
• But at 1500K • ΔH = 178 kjmol-1
• ΔS = 161 jK-1mol-1 = 0.161 kjK-1mol-1
• T ΔS = 1500 x 0.161 = 241.5kjmol-1
• ΔG = 178 – 241.5 = -63.5 kjmol-1
• The reaction now has a negative ΔG so it has become feasible.
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Zero values of ΔG
• When ΔG is 0 the reaction is just feasible.
• The temperature at which this occurs can be calculated.
• Eg; CaCO3 → CaO + CO2
• ΔG = ΔH – TΔS• 0 = 178 – T0.161• T = 178 / 0.161 = 1106K
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Kinetic factors
• Neither ΔH nor ΔS gives any indication of the rate of a reaction.
• Some reactions are predicted to be feasible on the basis of ΔH and ΔS but in practice are so slow that they unfeasible.
• This is because of a kinetic barrier.
• IE There is a high activation energy.
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reactants
products
energy
exergonic reaction
Reaction profile
activation energy, Eaa
transition state
(or activated complex)
bonds breaking
bonds forming
Course of reaction Replay Close window
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Eg; Stability of graphite
• C + O2 → CO2
• ΔH = -394kjmol-1
• S CO2 = 213.6 jK-1mol-1
• S O2 = 205 jK-1mol-1
• S C = 5.7 jK-1mol-1
• ΔS = +2.91 jK-1mol-1
• ΔG = -394 – (298 x 0.00291)
• = -394.86 kjmol-1
• So the reaction is feasible at 298K.
• But in practise it is too slow.