year 13 physical Flashcards
total pressure
sum of all partial pressures (in pascals/kPa)
mole fraction
no of moles of one gas ÷ total no of moles
partial pressure
mole fraction x partial pressure
key point (ratios when doing partial pressure)
ratios can be compared when on the same side of the equation but not to the other side
kp
(+ units)
products ÷ reactants
to the power of moles
curly brackets
little pp
units = kpa (same method as Kc)
what effects Kp
temperature
not pressure!!!
what does movement of equilibrium position do to Kp?
shift to right = larger
shift to left = smaller
which direction does equilibrium move when temp increases?
endothermic direction
(counteract the change)
which direction does equilibrium move when temp decreases?
exothermic direction
(counteract the change)
effect of pressure on Kp?
none
effect of catalyst on Kp?
none
kelvin to celcius no.
273
enthalpy of formation
define
enthalpy change when one mole of a substance is formed from standard states
enthalpy of ionisation
1 mole
gains 1 mole of electrons
gaseous
enthalpy of atomisation
1 mole
atoms from standard state
gaseous
bond dissociation enthalpy
1 mole
covalent bonds broken
gaseous
enthalpy of lattice formation
1 mole
solid ionic compound
formed from ions in gas state
enthalpy of lattice dissociation
1 mole
solid ionic compound
dissociates into ions in gas state
why is 1st electron affinity exothermic
attraction between nucleus and electrons
why is 2nd electron affinity endothermic
electron repulsion between nucleus and outer shell
both already negatively charged
effect of ionic charge on lattice enthalpy
larger causes higher enthalpy change
effect of size on lattice enthalpy
smaller causes higher enthalpy change
difference between theoretical/experimental value
perfect ionic model/covalent character
1st2nd electron affinity
enthalpy change
1 mole
gaseous atoms
gain one election
enthalpy change of solution
1 mole of ions
minimum amount of solvent
dissolved
to ensure no further enthalpy change
resultant of enthalpy change of solution
very saturated solvent
ions become hydrated
enthalpy of solution =
enthalpy of dissosiation + enthalpy of hydration
LiCl (s) -> Li+(g) + Cl-(g) -> Li+(aq) + Cl-(aq)
entropy
measure of disorder (more disorder = higher entropy)
two factors which effect entropy
no. of moles (more moles = more disorder)
state of molecules
calculation entropy
S(products)- S(reactants)
positive entropy value (delta S)
entropically favourable
feasible
Gibbs free energy equation
units
G-TS
J/mol
why might a reaction not occur even if Gibbs works?
activation energy too high
rate too slow
value of k relation with rate
larger = faster rate
initial rate (calculate from graph)
gradient of tangent at 0 mins
finding rate equation
repeat one experiment several times but changing the conc of one reactant at a time
find initial rate using graph (measure product formed/change in pH/colour change/reactant used)
place in table and calculate orders
horizontal line on rate time graph
from straight line down on conc time graph
zero order
where does a rate time graph come from
derivative of conc time graph
positive gradient straight line time graph
from curving down line on conc time graph
first order
positive gradient curved line time graph
from steep curving down line on concrete time graph
second order
rate equation and rate determining step
reactants in rate determining step are always in the rate equation
intermediate in rate determining step (with respect to rate equation)
reactants which made intermediate must be in rate equation
x and y axis of Arrhenius plots
x = 1/T
y = ln K
m and c in Arrhenius plot
m = activation energy
c = Ln A
y = mx + c in Arrhenius
Ln k = Ln A - Ea/RT
explain why increasing temp is better for increasing reaction than increasing concentration of one species
reaction occurs when all molecules have energy above activation energy
doubling temp affects this of all molecules
whereas doubling conc only affects one species
change to rate equation when one species largely in excess
species removed from rate equation
conc virtually constant
units of k
(mol dm-3)n s-1
what should be done to samples of reactions when finding orders
stop the reaction
quench
dilution/cooling/species to react with a reactant/removing catalyst
explaining order from graph
describe gradient
link to order
e.g. used at constant rate
how to find Ea from Arrhenius graph
gradient = -Ea/R
multiply by 8.31, divide by 1000
units for A (Arrhenius constant)
s-1
bronsted lowry acids
proton donors
H+ ions released when aqueous
forms H3O+ in water
HA + H2O <-> H3O+ + A- (equill)
water behaves as a base
bronsted lowry bases
proton acceptors
react with water to form OH-
B + H2O <-> BH+ + OH - (equill)
ethanoic acid equilibrium equation
CH3COOH <-> CH3OO- + H+
lies heavily to left
backwards reaction favoured
hydrochloric acid equillibrium equation
HCl <-> H+ + Cl-
forwards reaction favoured
lies heavily to right
acid base reaction equillibrium
HA + B <-> BH+ + A-
HA = acid
B = base
BH+ = salt
Kw
ionic product of water
constant of water
1 x 10 (-14)
mol2dm-6
pH of strong acid
conc of H+ = conc of strong acid
monoprotic strong acid
[H+] = [acid]
pH = -log 10 [acid]
diprotic strong acid
2[H+] = [acid]
pH = -log [2acid]
pH of strong base
dissociate fully
Kw = [OH-][H+]
[OH-] = [base]
find [H=]
pH = -log10[H+]
acid dissociation constant
HA <-> H+ + A- (lies to left)
Ka = {[H+][A-}}/[HA]
or Ka = [H+}2/[HA]
moldm-3
pH of weak acids
Ka = [H+]2/[HA]
[HA] = conc of acid
assumption with Ka
[acid] (start) = [acid] (equillibrium)
dissociation of acid is greater than dissociation of water (all H+ ions from acid)
calculate Ka/concenctration of a weak acid
find [H+] (from pH)
use Ka equation
pKa
-log10 Ka
equivalence point
fully neutralised acid / base
half neutralisation point
half way between 0 and equivalence point
[HA] = [A-]
cancels out in Ka
Ka = [H+]
pKa = pH
diprotic curves
two equivalence points
H+ ions released separately
acidic buffer
keeps pH below 7
from weak acid and its salt
CH3COOH <=> CH3COO- + H+ (acid, lies to left)
CH3COO-Na <-> CH3COO- + Na (salt, lies to right)
add H+ to acidic buffer
react with negatively charged ions (from salt)
forms undissociated acid (equillibrium lies to left)
add OH - to acidic buffer
react with H+ ions
equillibrium counteracts change
basic buffer
weak base and its salt
pH above 7
NH3 + H2O <-> NH3+ + OH- (base, lies to left)
NH4+Cl- <-> NH4+ + Cl- (salt, lies to right)
add H+ to basic buffer
reacts with OH-
makes water
equillibrium re-established
add OH - to basic buffer
reacts with positive ions in solution
(high conc from salt)
shifts base equillibrium to the left
calculating pH of a buffer
find Ka expression
use equillibrium concentrations not initial
[salt] = A-
find concs
input to find [H+] then pH
calculating pH change of a buffer
use appropriate equillibrium
use ICE to make changes
why is water with pH of near 7 not acidic
[H]+ = [OH]-
why do buffer solutions have a constant pH even when diluted
ratio [HX]/[X-] remains constant
difference in enthalpy of hydration
size to charge ratio
weaker attraction between polar water molecule and ion
difference in enthalpy of lattice formation reasons
smaller ion - greater size:charge ratio
ratio of metal:non metal
high pKa
weaker acid
postive electrode
anode
negative electrode
cathode
positive ion
cation
negative ion
anion
molten electrolysis
only one anion and cation
which cations are more easily reduced
more positive cations
which anions are more easily oxidised
more negative anions
standard conditions of reference half cell (S.H.E.)
1 mole H+ ions
hydrogen gas
platinum metal
298 K
1 atm/ 100kPa
requirements of voltmeter
high resistance
prevent current flowing
use of salt bridge
allows ions to pass from solutions
completes circuit
why can’t metal be used as salt bridge
create its own equillibrium
prevent ions flowing
suitable salt bridges
filter paper soaked in ions
agar jelly soaked in ions
KNO3 used commonly
generally inert highly soluble ions
where is p.d. measured by electrochemical cells
between metal and solutions
change to charge when equillibrium of electrode shifts to left
more electrons
build on electrode
negative charge
change to charge when equillibrium of electrode shifts to right
electrons used up
positive charge
large positive value of electrode
indicates strong oxidising agent
large negative value of electrode
indicates strong reducing agent
