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87 Cards in this Set

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