330 likes | 453 Views
Chapter 19. Chemical Thermodynamics. 19.1 Spontaneous. Spontaneous-opposite of induced Reversible process Irreversible process When a chemical system is in equilibrium, we can go reversible between reactants and products.
E N D
Chapter 19 Chemical Thermodynamics
19.1 Spontaneous • Spontaneous-opposite of induced • Reversible process • Irreversible process • When a chemical system is in equilibrium, we can go reversible between reactants and products. • In any spontaneous process, the path between reactants and products is IRREVERSIBLE!
19.2 Entropy • Entropy-S- randomness of system; • is a state function • ΔS= Products – Reactants • ΔS: Measured in J/mol-K • If: • ΔS: is positive: More random • ΔS: is negative: More orderly
Your room at home. High Entropy!
Phase changes Gas Liquid Solid ΔH = - ∆H = + ∆S = + ΔS = -
1st Law of Thermodynamics • NRG of the universe is constant. • or NRG neither be created nor destroyed.
19.2 2nd Law of Thermodynamic • 2nd law of thermodynamics • S of universe is always increasing • Isolated System-Does not exchange energy with its surrounding
19.3 Molecular Interpretations of S • Translation Motion • Vibrational Motion • Rotational Motion
3rd law of thermodynamics • -the S of a pure crystalline substance at 0 ° K = ZERO
19.4 Calculate S • Look on page 1112. • Same as ∆H.
19.5 Gibbs Free Energy ΔG • ΔG=Products- Reactants • Measured in kJ/mol • Elements are Zero • ΔG=Negative… Spontaneous • ΔG=Zero… At equilibrium • ΔG=Positive… No reaction • ΔG=ΔH - TΔS
ΔG=ΔH - TΔS • Go Home to Supper equation
19.6 Free Energy+ Temperature • ΔG=ΔH - TΔS at 25°C
ΔG will tell us if a reactions is spontaneous but will not tell us how fast the reaction proceeds. • C + O2 CO2 • What is the study of the speed of reactions?
19.7 Free Energy and Equal Constant • ΔG=ΔG°+ RTlnQ • R=8.314 J/mol K • Q= Reaction quotient (like K) • ΔG=-RTlnK • ΔG Negative; K>1 • ΔG Zero; K=1 • ΔG Positive; K<1
Perpetual motion machine
Ideal frogs • initial final • xxx • xxx xxx xxx K = 3/3 = 1 ∆H = 0 ∆S = 0
Initial final • K = 4/2 = 2 • ∆H = - (exo) ∆S = 0 xxxxxx xx xxxx ∆H
Initial final • K = 4/2 = 2 • ∆H = 0 ∆S • ∆S = + more random xx x xxx xxxxxx
Initialfinal • ∆S • ∆H • K = 4/2 = 2 This reaction is ∆S driven. • ∆H = + (endo) • ∆S = + x xxx xx xxxxxx
ΔG=ΔH - TΔS • = + - T(+) • = + - = ? • Neg answer at high temp….and spont • Pos answer at low temp……and N.R.
Initial final • K = 4/2 = 2 • ∆H = - (exo) • ∆S = - more orderly…This reaction is ∆H driven x x xxxxxx ∆H xxxx
ΔG=ΔH - TΔS • = -- T(-) • = - + = ? • Neg answer at low temp….and spont • Pos answer at high temp…and N.R.
Initial final • K = 5/1 = 5 ∆H • ∆H = - (exo) • ∆S = + (more random) ∆S xxxxxx x xxxxx
ΔG=ΔH - TΔS • = --T(+) • = - + - = always neg • Always spont.