OCN 623: Thermodynamic Laws & Gibbs Free Energy. or how to predict chemical reactions without doing experiments

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OCN 623: Thermodynamic Laws & Gibbs Free Energy or how to predict chemical reactions without doing experiments

Definitions Extensive properties Depend on the amount of material e.g. # of moles, mass or volume of material examples in chemical thermodynamics: G -- Gibbs free energy H -- enthalpy

Intensive properties Do not depend on quantity or mass e.g. temperature pressure density refractive index

Reversible and irreversible processes Reversible process occurs under equilibrium conditions e.g. gas expanding against a piston P p p = P +!p reversible p = P + "p irreversible

No friction or other energy dissipation System can return to its original state Very few processes are really reversible in practice e.g. Daniell Cell Zn + CuSO 4 = ZnS0 4 + Cu with balancing external emf Compression of gases Vaporisation of liquids Is a concept used for comparison

Spontaneous processes Occur without external assistance e.g. expansion of a gas from region of high pressure to low pressure diffusion of a solute in a solvent Are irreversible by definition -- occur at a finite rate

1st law of thermodynamics Energy cannot be created or destroyed when one kind of energy disappears it is transformed into an equivalent quantity of another kind U = q-w du = dq - dw du = change in internal energy of the system dq = heat absorbed by the system + dw = work done by the system +

U is a thermodynamic function du depends only on the initial and final states of the system-not the path taken q and w are NOT thermodynamic functions Internal energy of system is increased by gaining heat (q) Internal energy of system is decreased when work is done by the system du = dq - dw

w rev. > w irrev. Work done by system during irreversible process is less than could be obtained if process was reversible e.g. if piston moved at finite rate, friction dissipates energy

Chemical thermodynamics HEAT Heat (q) is measured in terms of: amount of substance x molar heat capacity x temp rise For phase change q = moles of substance x molar latent heat q is the heat of reaction

WORK Work of expansion = P V where P = pressure and V is change in volume at constant volume V = 0, therefore w =0 no work is done

Electrical work Occurs in electrical cells e.g. Zn + Cu 2+ = Zn 2+ + Cu Electrical energy = I * E * t E is the emf (voltage), t = time, I = current flowing It=zF for 1 mol z= # of electrons transferred, F (Faraday)= 96,490 Coulombs/mol Electrical energy = zef

System at constant volume (all non-gas reactions are at constant volume) P V = w= 0 U = q v q v= heat at constant volume System at constant pressure (all reactions open to atmosphere) U = q p - P V Ideal gas law P V = nrt R is the gas constant = 8.314 Joules mol -1 K -1 therefore U = q p - nrt rearranging q p = U + nrt

q p is called the enthalpy H change in enthalpy H = U + P V in absolute terms H = U + PV H is a thermodynamic property, is defined in terms of thermodynamic functions: U, P and V For an infinitesimal change at constant pressure: dh = du + PdV

H f o 298 is the heat of formation of 1 mole of a compound from its elements at 298 o K H is -ve = exothermic reaction H is +ve = endothermic reaction (seen from system perspective) H is proportional to amount of material is an extensive property

H is equal in magnitude but opposite in sign for the reverse reaction, because it is a thermodynamic quantity H products - H reactants H for reaction is same regardless of number of steps between reactants and products H = U + nrt

Can use component reactions to calculate enthalpy change for unknown reaction Reaction Enthalpy H 2 + 0.5 O 2 = H 2 O H = -285.8 kj mol -1 (a) C + O 2 = CO 2 H = -393.3 kj mol -1 (b) C 2 H 6 + 3.5 O 2 = 2 CO 2 + 3H 2 O H = -1559.8 kj mol -1 (c) can use above to calculate H for ethane formation 2*b + 3*a - c 2C + 2O 2 + 3H 2 + 1.5 O 2 - C 2 H 6-3.5 O 2 = 2CO 2 + 3H 2 O - 2 CO 2-3H 2 O canceling yields 2C + 3 H 2 = C 2 H 6 H = 2* -393.3 + 3* -285.8 - (-1559.8) = -84.2 kj mol -1

But enthalpy change alone is insufficient to allow prediction of likelihood of reaction Entropy change is also needed 2nd law of thermodynamics ds = dq rev T change in entropy = amount of reversible heat absorbed by system divided by temperature

Entropy is the degree of disorder of the system chance of finding something in a fixed volume- higher pressure, greater chance, liquid lower entropy than gas, solid lower than liquid Change in entropy measures capacity for spontaneous change diffusion of a solute from high concentration to low concentration, or a gas from high pressure to low pressure A system undergoing spontaneous change is moving to a greater degree of disorder

Entropy is an extensive property Units heat/temperature = J(oules) K -1 mol -1 for solids dq = moles x molar heat capacity x temperature rise dq = C m dt C m = molar heat capacity and dt = temperature change S m = from T1 to T2 C m dt T S m = C m ln T 2 T 1 where S m is molar entropy change

For phase change S = L T where L is the latent heat of fusion e.g. for water L = 6 kj mol -1 S = 6000 = 22.0 J K -1 mol -1 for melting ice 273.1 reversible reaction the net change in entropy is zero irreversible reaction there is a net increase in entropy

Since nearly all reactions are irreversible, entropy is increasing -- called times arrow always increasing --cannot run time backwards Third law of thermodynamics The entropy of a perfectly crystalline material at the absolute zero (-273.1 C) is zero entropy allows prediction of the feasibility of reactions

Gibb s free energy Gibbs function G = H - TS equals the enthalpy - temperature x entropy At constant volume q = U Helmholtz function A = U - TS Gibbs function used most (constant pressure) for reactions of liquids or solids U = H no expansion so, A = G

Tendency of reaction to occur is combination of energy and entropy change Reactions proceed in direction that minimises the internal energy, i.e. H is lowered Reactions proceed to maximise degrees of freedom i.e. entropy increases G is negative for a spontaneous reaction

G = H - TS differentiating : dg = dh - TdS -SdT at constant temperature: dg = dh - TdS G is a thermodynamic function G for a reaction = G products - G reactants

Spontaneity of reactions sign of!g!h!s Comments always - - + Spontaneous at all T; reverse never spontaneous always + + - Never spontaneous at any T reverse reaction occurs High T - spontaneous at High T Low T + + + only low T - - - Spontaneous at low T high T + reverse rx at high T

For an electrical system G = -zef where E = the emf of the cell (or half cell) F = Faraday (a positive number) z = number of electrons involved in the reaction Effect of pressure on G G = G + nrtlnp can calculate Gibbs free energy for any material as long as know the standard state (G )value

Chemical potential Free energy per mole chemical potential µ = G/n µ is an intensive property Material flows from region of high chemical potential to region of low chemical potential Effect of pressure on µ µ = µ o + RT ln P

For an ideal solution µ = µ o + RT ln C for non ideal solution µ = µ o + RT ln a where a is activity and a i = γ i x i γ i is activity coefficient x i is mole fraction ln<1 is -ve µ < µ o chemical potential of a mixture< pure substance

Diffusion down a concentration gradient from a pure substance to a mixture is thus a result of thermodynamics Gibbs free energy and equilibrium Reaction aa + bb = cc + dd Equilibrium constant K = [C] c [D] d [A] a [B] b K is ratio of products to reactants G = G o + RT ln K

van t Hoff isotherm!ga,b!g!g c,d A,B 50% Eq pos C,D

At equilibrium G = 0 G o = -RT ln K Knowing G o for reaction, can calculate equilibrium constant e.g 0.5 N 2 + 0.5 O 2 = NO H = 89.96 kj mol -1 S NO O N 2 210.7 205.1 192.3 S for rx = 210.7- (0.5* 205.1) - (0.5*192.3) = 12.0 J K -1 mol -1

G o = H - T S G o = 89960-298(12) J mol -1 = 86.38 kj mol -1 now as G = -RT lnk lnk = - 86380 8.314 x 298 log K = -15.20 K = 6.31 x 10-16 i.e.predict the reaction is well to the left hand side (not surprising as G is large and positive)

Effect of temperature on equilibrium ln K 1 = - H 1-1 K2 R T1 T2

For exothermic reactions lnk increases as 1/T increases, (T decreases), favouring product formation For endothermic reactions lnk increases as 1/T decreases (T increases) favouring product formation lnk Exothermic Endothermic 1/T