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87 Cards in this Set
- Front
- Back
When is a reaction one way?
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It is unlikely 3 products will combine again. So when it is unlikely so many products will combine again.
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dynamic equilibrium
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reversible rxns in a closed system. The rate of the fwd rxn = rate of reverse rxn
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How can you tell when equilibrium is reached?
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All macroscopic properties (vol, temp, press) are constant
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Three types of equilibrium systems
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Phase Equilibrium
Solubility EqM Chemical |
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Phase Eqm
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Two or more states of a pure substance are in eqm. H2O(l) <=> H2O(g)
Rate evaporation=rate condensation For every molecule that evaporates, one condenses |
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Solubility Eqm
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Solute in a saturated solution of a solvent. Rate dissolving = rate crystallyzation.
CuSO4(s) <=>Cu2+(aq) + SO42-(aq) Solid solute in eqm with its ions |
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solute
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the dissolved matter in a solution; the component of a solution that changes its state
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solvent
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a liquid substance capable of dissolving other substances; "the solvent does not change its state in forming a solution"
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saturated
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being the most concentrated solution possible at a given temperature; unable to dissolve still more of a substance; "a saturated solution"
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Solubility eqm example
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When I2 solid dissolved, I2 aq come out with little I2 solid at bottom. If you add I2 solid radioactive, and filter, after several hours radioactivity can b detected in solution. "For every iodine in solution, one comes out"
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Chemical eqm
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H2(g) + I2(g) <=> 2HI(g)
colourlesspurple colourless It will b lite purple in closed container. |
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Le Chatelier's Principle
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When a stress is added to a system at eqm, the system readjusts so as to relieve or offset the stress
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Haber Process 4 formation of ammonia
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N2(g) + 3H2(g) <=> 2NH3(g) + 92.5 KJ
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Draw rate vs time graph 4 chem eqm
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Draw conc vs time graph 4 chem eqm
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Adding a catalyst...
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decreases activation energy for the fwd and rev rxns by the same amount so that the fwd and reverse rxn both increase by the same amount. This has no effect on eqm except that this state is established faster
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When something is added in eqm to react with a reactant....
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the concentration of this reactant decreases
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Review limiting reactants
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Keq = what ks?
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Kfwd / Krev
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k >> 1
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large value.
goes to completion [PRODUCTS] / [reactants] |
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k << 1
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Little rxn at all
[products] / [REACTANTS] |
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k ~ 1
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intermediate value
Measurable amounts of product and reactant at eqm |
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Pure solids or liquids
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have constant []'s so are omitted from Keq calc
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The only things that affect Kc
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temp and pressure
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To convert from conc to mol
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multiply by number of Litres
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For (1-2x) simplifying assumption to check % is:
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2x / 1
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For (2-3x) simplifying assumption to check % is:
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3x / 2
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if Q>K
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shifts left
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if Q=K
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at eqm
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If Q<K
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shift rite
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At constant temp, [] is directly related to the
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partial pressure of the gas
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[] of gases can b expressed in terms of
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partial pressures
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P=
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(n/v)RT
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Partila pressure formula =
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(n / v) RT = []RT
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R in pressure formulas =
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.0821 atm *L / mol*K
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Kp =
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Kc (RT)^dn
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Kc = Kp only when
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same # moles on each side
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When excess solid is present in a saturated solution...
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there is a SOLUBILITY EQM formed between the solid and its ions
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solubility constant is
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Ksp = equilibrium product constant = ion product constant
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Ksp is dependent on
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temp. not pressure bcs not gas
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Ksp only written for
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ionic compounds w/low solubility
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If solubility is high....
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ionic compound dissolves completely
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Concentration X # litres = new
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concentration! ex; [Ag+]
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precipitate
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Precipitation is the formation of a solid in a solution during a chemical reaction. When the reaction occurs, the solid formed is called the precipitate, and the liquid remaining above the solid is called the supernate.
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The precipitate will precipitate until
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Qsp = Ksp
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100 rule
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assumption valid as long as [ion] is 100x > Ksp
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reread "effect of a common ion on solubility"
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entropy is
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a measure of molecular randomness or disorder
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nature spontaneously proceeds towards...
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the states that have the highest probability of existing.
Probability favours a disordered state |
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Positional entropy
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Entropy increases if the # of configurations in space (positional microstate) increase. That is, more possible positions.
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1st law of thermodynamic
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Energy is never created/destroyed. Energy of universe is constant
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2nd law of thermodynamic
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Spontaneous processes will increase the entropy of the universe. The entropy of the universe is increasing.
dSuniv = dSsyst + dSsurr |
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If dSuniv>0
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rxn is spontaneous
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if dSuniv <0
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rxn spontaneous in opp direction
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if dSuniv=0
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equilibrium
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Exothermic systems increase molecular motions in the surroundings so
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dSsurr = +
and dH = - |
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For endothermic
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dH = +
and dSsurr = - |
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if dSsys > 0
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spontaneous fwd
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If dSsurr < 0
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spontaneous rev
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What actually occurs (fwd or rev) depends on
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temp
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The impact of a transfer of energy as heat to the surroundings will be greater at
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lower temps (a greater % change in randomness)
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dSsurr =
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-dH / T (J/K) At constant temp and pressure. Point of view of surroundings are opp to system
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Gibb's Free Energy deals with
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the temp factor.
dG = dH - TdS All quantities refer to system. |
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dSuniv = -
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-dG / T
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Rxns at constant temp and pressure will b spontaneous only if
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dG is neg
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Spontaneity requires that free energy must
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decrease
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3rd Law of Thermodynamics:
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The entropy of a perfect crystal is zero at absolute zero.Theoretical. Every substance has a measurable absolute entropy.
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dS*rxn =
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sigma np S*p - sigma nr S*r
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No instrument can measure
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free energy
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three methods of calcing dG*: 1
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dG* = dH* - TdS*
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three methods of calcing dG*: 2
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dG is a state function (indep of pathway). thrfr we can determine using a procedure similar to Hess's Law and dH
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Standard Free Energy of Formation Method 3
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dG* = sigma np dG*f(prod) -- sigma nr dG*f(reactats)
at 25 C and 1 atm |
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To calc dG not at STP use
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dG = dG* + RT lnQ
in J and K and stuff where K=8.314 |
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The lowest possible free energy that chemical systems always seek occurs at eqm where
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dG = 0
Q = K (Kp or Kc) |
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Therfore, rearranging olden equaiton with zeros and Qs....
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dG* = -RT lnK
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When Q = K, free energy is
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at a min
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Is it the sign of dG or dG* that determines the direction of the rxn's spontaneity?
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dG
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What does dG* tell us?
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The relative amounts of products and reactants when eqm is reached.
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If K>1, ln K, dG* and comments
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ln K = +
dG* = - Comments = products favored over reactants at eqm |
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If K=1, ln K, dG* and comments
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ln K = 0
dG* = 0 Comments = products and reactants equally favored at eqm |
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If K<1, ln K, dG* and comments
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ln K = -
dG* = + Comments = reactants favored over products and eqm |
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neg dH favours
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fwd
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neg dS favours
(of sys) |
rev
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When pressure is lower
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vol is bigger. So lowering pressure increases entropy.
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dG and dH units
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kJ / mol
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dS units are
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J / K mol
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propane
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C3H8(g)
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