Lectures 26-35 Flashcards
chemical equilibrium
when molecules form a state in which the composition of the reaction mixture remains constant and concentrations/partial pressures of reactants and products no longer change
at equilibrium, forward reaction rate =
reverse reaction rate
K(p)
p(products)^coefficients/p(reactants)^coefficients
K(c)
[products]^coefficients/[reactants]^coefficients
relationship between forward and reverse reactions for K
K(reverse) = 1/K(forward)
doubling reactions
K^2
activities
describe ‘effective’ concentration
for pure liquids and solids, concentrations of these components
do not change and are given an activity of 1 (not included in calculations)
solubility product
normal K(c) with solid not included; only included solvated ions/molecules
smaller K(sp), harder it is
to dissolve a substance
product-favoured
K > 1
reactant-favoured
K < 1
reaction quotient
tells us how far we are from equilibrium
at equilibrium concentrations, Q =
K
if Q < K, then
forward reaction is faster until equilibrium attained
if Q > K, then
reverse reaction is faster until equilibrium attained
Q = 1
standard-state reaction mixture
equation relating Gibbs free energy and reaction quotient
deltarG = deltarG^o + RTln(Q)
in standard-state reaction mixture, Q = 1 and ln(Q) =
zero, so deltarG = deltarG^o
at equilibrium, deltarG = 0 =
deltarG^o + RTln(K)
at equilibrium, deltarG^o =
-RTln(K)
Le Chatelier’s Principle
when a change is made to a system in dynamic equilibrium, the system responds to minimise the effect of the change
if reactant is added,
equilibrium is re-established by consuming some of the added reactant
if product is added,
equilibrium is re-established by consuming some of the added product
for gaseous reactions, increasing pressure will push equilibrium
in the direction of fewer molecules
if a reaction is exothermic, a temperature increase will favour
reactants
if a reaction is endothermic, a temperature increase will favour
products
linking entropy with equilibrium constants
ln(K) = -deltaH/RT + deltaS/R
van’T Hoff equation
ln(K1/K2) = deltaH/R(1/T1 - 1/T2)
ICE tables
initial, change, equilibrium
equation connecting K(p) and K(c)
K(p) = K(c)(RT)^delta(n)(gas)
other value (not 8.314)
0.0821 L atm/mol K
if K > 1, deltaG is expected to be
negative
if K < 1, deltaG is expected to be
positive
Lewis acids and bases
electron acceptors (acids) and donors (bases)
Bronsted-Lowry acids and bases
proton donors (acids) and bases (acceptors)
all ions/molecules are surrounded by solvent (water) to form a
solvation shell
HA is acid,
A^- is the conjugate base
B is the base
HB+ is the conjugate acid
weak acid-base pair is always
favoured in the equilibrium
water undergoes
self-ionisation and is a weak acid and weak base (reaction lies to left)
ionisation constant, K(w)
extent of ionisation of water; [H+][OH-] = 10^-14
in pure water, [H+] =
[OH-] = 10^-7
solution is neutral when
[H+] = [OH-]
ionisation of water is an
endothermic process
addition of other substances may change [H+] and [OH-], but the
product of their concentrations is always constant
[H+] > 10^-7 M
acidic
[H+] < 10^-7
basic
pH
—log[H+]
pOH
—log[OH-]
pH + pOH =
14
K(a), acid dissociation constant
[H3O+][A-]/[HA]
pK(a)
—log[K(a)]
strong acids have
high K(a) and small pK(a)
K(b), base equilibrium constant
[OH-][BH+]/[B]
pK(b)
—log[K(b)]
strong bases have
high K(b) and small pK(b)
strong acids
ionise almost completely in solution; reaction lies to right
conjugate base of a strong acid is very
weak
for strong acids, the [H3O+] concentration equals
the acid concentration
for strong bases, we can assume the concentration of the base equals
the [OH-] concentration
for weak acid, equilibrium lies well to
left
calculating pH of solutions of weak acids
[H3O+] = (K(a)[HA])^1/2 and go from there
[weak acid initially in solution ] = very close to
[weak acid concentration at equilibrium]
5% Rule
if dissociation is <5%, approximation is valid
% dissociation
[conjugate base]/[acid] x 100
calculating pH of solutions of weak bases
write equation, write basicity constant, write ICE table (if only base added, [BH+] = [OH-]), decide if base can be assumed weak and solve for x ([OH-]), check assumption is correct
buffered solution is one that
resists a change in pH when either OH- or H+ ions are added
good buffers will have close to equal
proportions of acid and its conjugate base
if too much acid or base is added to the buffer, it will
fail
buffer strength is at its strongest when
pH = pK(a)
Henderson-Hasselbach equation
pH = pK(a) + log([A-]/[HA])
solving buffer problems without using Henderson-Hasselbach equation
write out the chemical equilibrium, use an ICE table, write out the equilibrium constant and define in terms of x, solve for x
buffers may be prepared by adding a weak acid
and its conjugate base to a solution
buffers may be prepared by adding enough strong base to
neutralise the weak acid
polyprotic acids
molecules able to be ionised more than once
negative charge on an acid makes it
harder to ionise
lewis acids are
cations, incomplete octet, polar bonds
lewis bases are
anions, H2O, lone pairs
conjugate acids and bases in a buffer solution should have similar concentrations
within a factor of 10
when creating buffers, if not adding conjugate acid/base and simply adding a random STRONG acid/base, only add
HALF original concentration
factors deciding whether an acid is strong or not
electronegativity, hybridisation, resonance stabilisation of anions
more electronegative the non-hydrogen atom,
the more acidic
the larger the ion,
the more the products are favoured and the stronger the acid (charge is stabilised)
loss of charge on dissociation
favours that reaction
increasing s orbital contribution
sp^3 < sp^2 < sp
acidity is judged by stability of
anion formed after releasing proton
increasing acidity (orbitals)
sp^3, sp^2, sp
stabilised anion -> resonance
stronger acid
cations of bases NOT HAVING resonance
stronger base
Mesomeric
+M or -M
electron-donating groups
+M; have lone pair of electrons; stabilise positive charges and destabilise conjugate base
electron-withdrawing groups
-M; have full/partial positive charge and stabilise negative charges; stabilise conjugate base
if Ksp is very small,
solid is only sparingly soluble
solubility
amount of a salt in a specified volume of a saturated solution. Measured in M. Find concentration of salt at equilibrium!
presence of a common ion
decreases solubility of the salt
if Q = Ksp
saturated; at equilibrium
if Q < Ksp
unsaturated; more salt can dissolve
Q > Ksp
supersaturated; precipitation will occur to bring system to equilibrium
equilibrium constant for soluble reactions
> 1
equilibrium constant for insoluble reactions
< 1
precipitation reaction equation
creating the solid
higher value for solubility,
less soluble (see Ksp)
finding pH for weak acid
pH = 1/2pK(a) -1/2log[HA]
K(a) x K(b) =
10^-14
if Ka(2) for second reaction is very small, assume
this reaction is negligible when calculating pH and [conjugate base] = [overall H3O+]
diluting a buffer solution does not change [base] : [acid] ratio, so
pH of solution will not change
pK(a) + pK(b)
= 14
