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CH 201: EXAM 2
Enthalpy
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ΔH = ΔE + PΔV
For constant pressure processes only
At a constant pressure, ΔHsys = energy flow as heat (and work).
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Enthalpy Equation
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ΔH°rxn = ΣnΔH°f (products) - ΣnΔH°f (reactants)
Standard molar enthalpy (1 mol formed)
ΔH°rxn = Σ bond energies (reactants) - Σ bond energies (products)
Gas phase reactions only
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Hess' Law
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Enthalpic change (ΔH) for a process is a constant whether the process occurs in one step OR a series of steps.
ΔH°rxn = ΔH°1+ ΔH°2
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ΔH properties
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If ΔH = negative, process is exothermic.
If ΔH = positive, process is endothermic.
Sign of ΔH does not necessarily predict spontaneity of a reaction.
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Entropy
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Measure of disorder/randomness of a system
2nd Law of Thermodynamics dictates that ΔSuniv = ΔSsys + ΔSsurr
At equilibrium, ΔSuniv = 0.
ΔSgas >> ΔSliquid > ΔSsolid
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Entropy Equations
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ΔS° = ΣnΔS° (products) - ΣnΔS° (reactants)
ΔSsurr = (-ΔHsys) / T
Entropy of a pure crystal = 0
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Gibbs Free Energy (ΔG)
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ΔG = ΔHsys - T(ΔSsys)
When ΔG = negative, processes are spontaneous; when ΔG = positive, processes are spontaneous in the reverse direction.
When ΔG = 0, processes are at equilibrium.
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ΔG° (ΔG at standard conditions)
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ΔG°rxn = ΣnΔG°f (products) - ΣnΔG°f (reactants)
ΔG°rxn = ΔH°rxn - TΔS°rxn
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ΔG at non-standard conditions
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ΔG = ΔG° + RTln(Q) where
R = 8.3145 J/K mol
T = temperature in K (C + 273.25)
Q = ratio of products/reactants
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How to determine boiling point
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ΔHvap = TΔSvap
T = ΔHvap / ΔSvap
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Interpreting K Values
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K > 1: large, ΔGf is negative, spontaneous
K < 1: small, ΔGf is positive, non-spontaneous
K = 0: slope is 0, ΔGf is 0, reaction at equilibrium
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Equilibrium
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Point at which opposing processes (forward and reverse reactions) occur at the same rate.
Reversible processes only
Dynamic
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Kc
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[A]a[B]b/[C]c[D]d
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Kp
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Kp = Kc(RT)Δn
Where R = .08206, T = temperature in K, Δn = change in # moles of gas
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Reversing an Equation
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Knew = 1/Kold
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