free energy / chemical potential Flashcards
what is the equation for Helmholtz free energy (A)
differentiate this
A = U - TS
constant T:
dA = dU - TdS
as dbarWrev = dU - TdS
dA = dbarWrev
condition must be satisfied by a reversible process so dw = dA is condition for system to be at equilibrium
what are equilibrium conditions in a system at constant temperature and volume which does no work
dA = 0
A is a minimum no further change can occur
how is dw shown under constant pressure
how is this shown at equilibrium
dwrev = -PdV + dwadd dwadditional = work other than PV
dwrev = dU - dq rev
dqrev = TdS
dwadd - PdV = dU - TdS
what is gibbs free energy equation for a system that does additional work
differntiate this at constant P and T
find condition for equilibrium
G = U + PV - TS = H -TS
dG = dU + PdV - TdS
as dwadd = dU - TdS + PdV
dwadd = dG for reversible change
is condition for equilibrium at cont T and P.
what are conditions for spontaneous process of dwadd
dwadd will be less negative than for corresponding reversible change
show gibbs free energy for system that only does PV work
G = U + PV - TS dG = dU + PdV + VdP - SdT - TdS
as only PV work
dU = dqrev + dwrev
dU = TdS - PdV
dG = VdP - SdT
show the pressure dependence of Gibbs free energy
constant T dT = 0 dG = VdP as PV = nRT dG = (nRT)(dP/P) so deltaG = nRTln(Pb/Pa)
can relate to std free energy which is free energy of one mole of gas at 1atm
G = stdG + RTln(P)
show temp dependence of Gibbs free energy
const P dP = 0 dG = -SdT -S = dG/dT as G = H - TS G = H - T(dG/dT) [d(deltaG/T)/dT] = -deltaH/T2 which is the Gibbs-helmholtz equation
what is the clapeyron equation
how is It found
(dP/dT) = deltaH / TdeltaV
consider two phases eg liquid and vapour in equilibrium dG = VdP - SdT in equilibrium dG(L) = dG(g) V(l)dP - S(l)dT = V(g)dP - S(g)dT So dP/dT = (S(g) - S(l))/(V(g)/V(l)) = deltaSvap / deltaVvap
deltaGvap = deltaHvap - TdeltaSvap = 0 deltaSvap = deltaHvap / T dP/dT = deltaH / TdeltaV
when T is boiling point at pressure under considersation
(dP/dT)eq = deltaH / TdeltaV
what is the clausius-clapeyron equation
how is it found
ln(P) = (-deltaHvap / RT) + C
relates temp dependance of vapour pressure of a liquid to deltaHvap enthalpy change per mole on vaporisation.
dP/dT = deltaHvap/TdeltaV deltaVvap = V(g) - V(l) deltaT/deltaP= TdeltaV/deltaH
at rtp and 1atm V(g)=24000 cm3 V(l) = 100 cm3 for 1mol substance V(g)>>V(l) V(g) = RT/P
dP/dT = (deltaH/RT2)P
dln(P)/dT = deltaH/RT2
integrate in terms of dT
what is the equation for the vapour pressure of liquids
delta_stdGvap = -RT ln(P) delta_stsGvap = 0 at boiling point
free energt 1mol of vapour:
G(g) = stdG (g) + RTln(P)
when 1mol l produce 1mol g at equilibrium pressure:
deltaG = G(g) - G(l) =
stdG(g) - (stsG(l) + RTln(P))
liquid and vapour in equil no change in free energy when quantity of liquid is vaporised G(g) = G(l)
delta_stdGvap = -RT ln(P)
how is clausius-clapeyron equation used to find delta_stdHvap of vapourisation
ln(P) = -deltaHvap/RT + C plot ln(P) againt 1/T gradient = -delta_stdHvap/R
show the equation for dG including chemical potential
dG = VdP - SdT + sum mui x dni mui = dG/dn1
what is chemical potential
the increase in free energy of the system when one mole of component is added to an infinitely large quantity og mixture so that it doesn’t significantly change overll composition
if transferring dni moles from a to b, what is dG
dG = [mui(b) - mui(a)]dni
at equilibrium dG=0 so mui(b) - mui(a) is a condition of equilibrium
