Classical and Sta,s,cal Thermodynamics

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BIOCH 590: Biomacromolecules Spring 2010

Copyright: Jianhan Chen

Key references: 1. Chapter 2 of van Holde (2nd Ed.)

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"E = q # w dE =$q -$w

!

dE ="q - PdV="q if constant V

#E = qv

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dE ="q - PdVd(E +PV ) = dE +PdV +VdP

="q - PdV +PdV +VdP= "q+VdP

dH = "q+VdP= "q if constant P

#H = qp

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Given:

B2O3(s) + 3H2O(g) ! 3O2(g) + B2H6(g) "H = +2035 kJ H2O(l) ! H2O(g) "H = +44 kJ H2(g) + (1/2)O2(g) ! H2O(l) "H = -286 kJ 2B(s) + 3H2(g) ! B2H6(g) "H = +36 kJ

Find the "H of:

2B(s) + (3/2)O2(g) ! B2O3(s)

Answer: "H = -1273 kJ

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!

S = "kB pi ln pii#

!

W =N!

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lnW = (N lnN " N) " (ni lnni " nii# )

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dS = kBd lnW = "kBd ni lnnii#

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i# = 0)

ni = n1 exp("ei /kBT) (in reference to e1 = 0)

lnni = lnn1 "eikBT

dS = kBeikBTi

# dni =1T

eii# dni =

dqrev

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S = "kB pi ln pii#

= "kBe"ei / kBT

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Qi#

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E = "d lnQd#

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A = "kBT lnQ

A: Helmholtz free energy

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1. Classical approach

!

dE = dq " dw = 0

dq = dw = PdV = RT dVV

#S =qrevT

= R dVV

= R ln V2V1

$

% &

'

( ) V1

V2*

2. Statistical approach: V1 = n1 V0 V2 = n2 V0

!

W1 " n1N W2 " n2

N

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n1N

%

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dS = 0

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EMBO reports 8, 1, 63–69 (2007)

the yeast exosome

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Nature Protocols 1, - 186 - 191 (2006)

Weak inhibitor strong inhibitor (Ka ~ 1012 M-1)

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Ref: “Calorimetry and Thermodynamics in Drug Design” by J. B. Chaires, Annu. Rev. Biophys. 2008. 37:135–51 .