Thermodynamics Flashcards
∆H
Heat energy exchange, enthalpy
Exothermic
-ve ∆H. Heat is released into surroundings
Therefore +ve∆T
Endothermic
+ve ∆H
First law
Total energy remains constant
Standard Conditions
100kPa pressure
298K temp
1 mol dm-3 concentration
∆Hr
For reactions in molar quantities in a chemical equation
∆Hc
Change when 1 mol reacts completely with O2
∆Hf
Change when 1 mol of compound is formed from its constituents
∆Hneut
Change when acid us neutralised by base to form 1 mol H2O
∆Hat
To form 1mol gaseous atoms
∆Hi.e
To remove 1e- per atom in 1mol gas
∆He.a
To add 1e- per atom to 1mol gaseous atoms
∆Hsolution
To dissolve 1mole of solute to form solution
∆Hhyd
To dissolve 1mole gaseous ions in water to form 1mole hydrated ions in solutions
Lattice enthalpy, ∆Hl.e
For formation of 1mol ionic compound.
Exothermic
ID indirectly by Born-Haber Cycle, and applying Hess’ Law
∆H Equation
Q=mc∆T
m= mass of surroundings
c = specific heat capacity of surroundings
∆T = temp change (final-initial temp)
Hess’ Law
If a reaction can go via +1 route, final and initial conditions are same; total enthalpy change is the same
∆Hc equation
∆H=∑∆Hc (reactants) - ∑∆Hc (products)
∑∆Hf
∆H = ∑∆Hf (products) - (reactants)
For elements, ∑∆Hf=0
Bond enthalpy
Change to break 1mol of bonds
Bond breaking vs bond making
Breaking = endo ∑(bond enthalpies in reactants) Making = exo -∑(bond enthalpies in products)
Bond enthalpy equation
∆H = ∑(BEs in reactants) - ∑(BEs in products)
Small BEs
Break first, therefore have quick reactions
Entropy, S
Measures no. of ways to arrange molecules
Increased with disorder (0K= perfect crystals and 0 entropy
Lattice Enthalpy factors
Ionic size and charge increase
Size: decreases charge density
Decreases between ion attraction
Lattice energy becomes less negative
Ionic Charge and LE
Cation:
↑charge = more attraction
↓size =more attraction
Anion:
↑charge = more attraction
↑size = less attraction
Hydration enthalpy factors
Charge:
↑ionic charge
↑H2O attraction
Hydration enthalpy factors for size
Cations:
↓size
↑H2O attraction
Anions:
↑size
↓H2O attraction
BE equation
∆H = ∑(BEs of reactants) - ∑(product BEs)
Entropy equation
∆S = ∑S(products) - ∑S(reactants)
Free energy ∆G equation
∆G = ∆H-T∆S
Spontaneous (feasible) process criteria
-ve ∆G
-ve∆H + +ve∆S
-ve∆G
+ve∆H + -ve∆S
+∆G
-ve∆H + -ve∆S
-ve ∆G at low temps (∆H > T∆S)
+ve∆H + +ve∆S
-∆G at high temps
T∆S > ∆H
-ve ∆G and equil constant
Equil constant has large value (products predominate)
+ve ∆G and equil
Equil constant is small value
Enthalpy and entropy comparison
Lower enthalpy system = more stable one
More disorderly system = more stable one
-more moles also = more entropy
∆Ssystem
∆Ssystem = ∑S(products) - ∑S(reactants)
Positive v negative entropy
Positive = favour for a process
∆Ssurroundings
∆Ssurroundings = -(∆H/T)
∆G≤0 implications
∆G≤0 = feasible/spontaneous reaction
Therefore ∆G MUST be negative
∆Stotal
∆Ssystem + ∆Ssurroundings
∆Stotal >0 = feasible process
When ∆Stotal increases, so does magnitude of equil constant:
∆S = RInK
Temperature factors
K, not C
∆G = 0, temp is at point where reaction can become spontaneous
(T = ∆H/∆S)
