paper 1 Flashcards
(181 cards)
1
Q
define an ion
A
different number of electrons than proton result in positive charge
2
Q
define isotope
A
same number of protons different number of neutrons
3
Q
step 1 in time of flight mass spectrometry
A
1) ionisation -
-electrospray sample dissolved in volitile solvent and then injected through hyperdermic needle attached to high voltage positive terminal this results in particles to gain protons become positive
X (g) + H+ → XH+ (g)
-electron impact sample vaporized and high energy electrons shot at it from and electron gun knocks off electron creating positive ion
X (g) → X+ (g) + e-
4
Q
step 2 in time of flight mass spectrometry
include ke calc
A
2) acceleration
positive ions are attracted to negatively charged plate
they all have the same kenetic energy
ke = 0.5 x mass x velocity²
5
Q
step 3 in time of flight mass spectrometry
A
3) ion drift
travel past the negatively charged plate into flight tube
each particles velocity depends on its
(mass lighter will move faster)
6
Q
step 4 in time of flight mass spectrometry
A
4) detection
-positive ions hit negatively charged detection plate
generate electric current
-size of current correlate to number of ions of specific m/z hitting plate
7
Q
calculations for tof
A
time = distance / velocity
ke = 0.5 x mass x velocity²
time = second
distance = meter
velocity = ms⁻¹
kenetic energy = joules
mass = kg
To get m you use atomic number of ion /1000 to get kg and then /6.022x10^23 to get m of one particle
8
Q
ram of isotopes
A
= (abundance A x m/z A)+(abundance B x m/z B)
all over total abundance or %
9
Q
first ionisation energy
A
amount of energy needed to remove 1 mole of electrons from 1 mole of gaseous atoms
M (g) → M + (g) + e-
10
Q
electron shells
A
1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰
11
Q
exceptions of electron configuration
A
Chromium (Cr): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵ (remove a 4s give to the 3d4 inc stability)
Copper (Cu): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰ (remove a 4s give to the 3d9 inc stability)
12
Q
relitive charge of sub atomic particles
A
proton = +1
neutron = 0
electron = -1
13
Q
relative masses of sub atomic particles
A
proton = 1
neutron = 1
electron = 1/1836
14
Q
atomic number
A
atomic number = is the number of protons in the nucleus of an atom
- number at bottom
15
Q
mass number
A
mass number = total number of protons and neutrons in the nucleus of an atom
- number at top
16
Q
factors affecting first ionisation energy
A
1) Size of the nuclear charge
-more protons in nucleus the larger the attraction for the outer e-
2) atomic radius
- less energy required to remove outermost electron if more shells result in weaker attraction
3) Shielding effect of inner electrons
- more electrons shells result in less energy required to remove it as e- repel each other
17
Q
trends of 1st ie in group 2
A
atomic radius increases as you go down the group so less energy required to remove outermost e-
18
Q
trends of 1st ie in period 3
A
Increased num of protons so greater attraction to outermost e- so more energy required to remove outermost e-
also aluminium evidance for subshells as it is the first orbital that has 2 e- in one subshell and is happily removed as e- repel so dips
19
Q
relative atomic mass of a compound
A
relative atomic mass (Ar) of an element is the ratio of the average mass of the atoms of an element to the unified atomic mass unit
20
Q
empirical formula
how would you solve a q
A
simplest whole number ratio
- write the % as grams
- g/ mr = moles
- divide all by smallest moles
21
Q
number of particles
A
moles x avogadros constant
22
Q
number of moles
A
mass / mr
23
Q
concentration
A
moles/ volume
24
Q
atom economy
A
(mr of desired product / mr of all reactant include stoichimetric value ) x 100
25
% yeild
(actual / theoretical) x 100
26
limiting reagent
-find moles of all reactant
- biggest value / smallest value
- if it doesn't match the stoichiometric values then it will be in excess / limiting reactant
27
hydrated compound working out mol of water
1) total mass of reactant - mass of product = mass of water
2) n=m/mr of both
3) now / all by the smallest n and this gives ratio of water in reactant
28
ionic bond intramolecular bond
type of force between ions
electrostatic attraction between oppositely charged ions in a lattice
29
covalent bond intramolecular bond
A single covalent bond contains a shared pair of electrons.
30
A co-ordinate (dative covalent) bond intramolecular bond
A co-ordinate (dative covalent) bond contains a shared pair of electrons with both electrons supplied by one atom.
31
metallic bond intramolecular bond
attraction between delocalised electrons and positive ions arranged in a lattice.
32
properties of ionic lattices
1)alternating cations (+ve) and anions (-ve)
2) conduct electricity when molten or dissolved
3) high melting point
33
properties of giant covalent molecular lattices of carbon
1)graphite
- 3 bond per carbon so one free delocalised e- this conduct electricity
- layers also slide over eachother
- high melting point
2)diamond
- 4 bond per carbon
- high mp
- no delocalised e- this cant conduct electricity
34
Properties of Metallic bonding
- metal ions are surrounded by a ‘sea’ of delocalised electrons
- conduct heat + electricity
- high mp
35
bond shapes
2bp 0lp linear 180°
3bp 0lp trigonal planar 120°
4bp 0lp tetrahedral 109.5°
5bp 0lp trigonal bipyramid 90° & 120°
6bp 0lp octahedral 90°
36
shapes of lone pairs
closer so more repulsion -2.5
3bp 1lp pyramidal 107°
2bp 2lp bent seesaw 104.5°
3bp 2lp trigonal planar 120°
4bp 2lp square planar 90°
37
define electronegativity
power of an atom to attract the pair of electrons in a covalent bond
37
finding shape
1) group of central atom
2) num of bonds
3) if positive -1 if negative +1
4) /2 = amount of bonds
38
types of intermolecular force
hydrogen bond = only occur if H directly bonded to N, O, F
dipole-dipole = indifference in electonegativity result in permanent dipole made this means molecule is polar
van der Waals forces = random movement of electrons around an atom results in temporary dipoles being created
39
exothermic
Energy is released from the reaction to the surroundings when new bonds are formed
-ve enthalpy
40
endothermic
energy taken in from surroundings when breaking bonds
+ve enthalpy
41
define standard enthalpy of combustion (∆c H Ɵ )
enthalpy change when 1 mole of substance is burnt in excess oxygen under standard conditions in standard state
42
define standard enthalpy of formation (∆f H Ɵ ).
ethnalpy change when 1 mole of substance produces from elements in standard states under standard conditions
43
calorimetry definition and equation
Calorimetry is the measurement enthalpy changes in chemical reaction
q = mc ∆T
∆H = (q/1000) / n
m = mass of thing getting heated (grams)
c = specific heat cap
∆T = temp change units dont matter still same change
n = moles
44
define hess law
The enthalpy change of the reaction is independant to the route taken
45
enthalpy of formation
hess cycle arrow direction bottom of arrows
arrows going up from c(s) 02(g) h2(g)
46
enthalpy of combustion hess cycle arrow direction and elelment
arrows going down to co2(g) h2o(l)
47
define mean bond enthalpy
amount of energy required to break one mole of a specific covalent bond in the gas phase
48
calculating mean bond enthalpy
all of reactant - all of product
49
rate of reaction increased when
1) increased surface area
2) increased concentration
3) increased pressure
4) increased temperature (this increase collision energy and successful collision)
5) catalyst (decrease activation energy and increase num of successful collisions)
50
maxwell boltzman distribution curve
- num of particles on y
- particle energy on x
- area under graph is num of particles
- never touch 0 right side
51
define Le Chatelier's principle
when a change is made to a reaction the system attempts to counteract it by shifting the position of equilibrium
52
changing conc temp press
dec pressure eq shift to side of most mol of gas
inc pressure eq shift to side of least moles
of gas
dec conc eq shift to same side
inc conc eq shift to opposite side
dec temp shift to exo as temp taken in from surroundings
inc temp shift to endo as temp given off to surroundings
53
equilibrium constant kc expression
( [ C ]^c * [ D ]^d )
-------------
( [ A ]^a * [ B ]^b )
54
kc calculations
1) convert all to n = c * (v / 1000)
2) ice box of these values
3) c = n / v
4) put into equations
5) units are mol dm-3 but cancel
55
kp calculation
method of working out the par press
1) work out mol fraction
mol / total mol
2) partial pressure =
mol fraction * given pressure
3) units are kPa but cancel
56
kp expression
( p(C)^c * p(D )^d) /
( p(A)^a * p(B)^b)
57
oxidation is
loss of electron they go on right side of equation
58
reduction is
gain of electrons they go on the left side of equation
59
balancing redox reactions
every elements
one oxygens add h2o
hates hydrogens add h+
chemistry charge add e-
60
oxidising agents
gets reduced (gain e-)
61
reducing agent
gets oxidised (loses e-)
62
first ie
(ΔHieꝋ) the amount of energy required to remove 1 mole of electrons from 1 mole of gaseous atom to form a gaseous 1+ ion under standard conditions
63
second ie
the amount of energy required to remove 1 mole of electrons from 1 mole of gaseous 1+ ions to form a gaseous 2+ ion under standard conditions
64
atomisation
enthalpy change when 1 mole of gaseous atoms is formed from its element under standard condition
65
lattice enthalpy of formation
the enthalpy change when one mole of an ionic solid is formed from its gaseous ions.
66
lattice enthalpy of dissociation
the enthalpy change when one mole of an ionic solid compound is completely dissociated into its gaseous ions.
67
bond diss
Bond dissociation enthalpy is the enthalpy change one mole of a particular covalent bond is broken into separate gaseous atoms
68
1st ea
First electron affinity of an element is the enthalpy change when one mole of gaseous atoms gains one mole of electrons to form a mole of gaseous 1- ions.
69
2nd ea
Second electron affinity of an element is the enthalpy change when one mole of gaseous 1- ions gains one mole of electrons to form a mole of gaseous 2- ions.
70
Enthalpy of solution
the enthalpy change when one mole of a compound dissolves completely in water so that the ions do not interact.
71
Enthalpy of hydration
Enthalpy change when one mole of gaseous ions forms aqueous ions
72
direction of arrows in born haber cycle
up = breaking of bonds endo
down = making of bonds exo
73
assumptions of theoretical and experimental values of le form
theoretical - assumes the perfect ionic model as ions are perfect spheres and have charge equally distributed throughout molecule
74
factors influencing covalent characters
high charge density of the +ve ion
75
what is the salt and how is it represented
Salt is formed when the H+
ion of an acid is replaced
by a metal ion or an ammonium ion
[ A- ]
76
Reactions of Acid with Metals produce
how can it be tested
Metal + acid ~~~~~~~
~~~~~~> salt + hydrogen
h2 gas produced meaning will pop when placed near lit splint
77
reactions of Acids that are neutralised by alkalis
base + acid ~> salt + water
78
reactions of acids and carbonates
testing also
Acid + Carbonate ~~~~
~~~>Salt + Water + Carbon Dioxide
carbonate is a (metal)CO3
turn lime water cloudy as it produce co2 gas
79
Brønsted–Lowry acid–base definitions
acid = proton / h+ donor
base = proton / h+ acceptor
80
what is the name of a h3o+ ion
hydroxonium ion
81
define strong and weak acid
strong = fully dissociate in water
weak = slightly dissociate in water
82
calculation for pH and
[ H+ ]
[ H+ ] = 10-pH
ph = -log( H+ )
83
define strong and weak base
strong = completely protonated in water
weak = slightly protonated in water
84
whats the ionic product of water and why is water not included
Kw = [H+(aq) ][OH-(aq) ]
in large concentration so remains essentially constant
85
what is the kw equation for pure water
and why
Kw = [H+(aq) ]2
as ph of 7 means that
[H+] = [OH-]
86
whats the ka expression and what 2 assumptions
Ka = [H+(aq)][A-(aq)]
-----------------------
[HA (aq)]
or
Ka = [H+(aq)]2
------------
[HA (aq)]
1) [H+(aq)]eqm = [A-(aq)] eqm because they have dissociated according to a 1:1 ratio.
2) As the amount of dissociation is small we assume that the initial
concentration of the undissociated acid has remained constant.
87
define conjugate base and acid
conjugate base = acid that has lost its h+ an will accept h+
conjugate acid = base that has accepted h+ and will donate it
88
factors affecting vdw forces strength
1) area of contact more surface contact more strength
2) more e- greater strength of temporary dipole
89
pH of neutrilisation reaction calc
(excess acid)
strong base strong acid
1) calculate moles of acid and base
2) ice box subtract smallest from them
3) convert back to conc
c = n / tot v
4) pH = -log[ H+ ]
90
pH of neutrilisation reaction calc
(excess base)
strong base strong acid
1) work out moles of everything
2) icebox take away smallest value
3) convert back to conc
c = n / tot v
4) [ H+ ] = KW / OH-
5) pH = -log[ H+ ]
91
pH of neutrilisation reaction calc
(excess base)
weak acid
1) work out mol of HA and OH-
2) ice box to work out n of A-
3) n / tot vol = c for HA and A-
4) [ H+ ] = ka * [ HA ]
--------------
[ A- ]
5) pH = -log[ H+ ]
92
define a buffer solution
solution is one where the pH does not change
significantly if small amounts of acid or alkali are added to it.
93
explain how an acidic buffer is created
An acidic buffer solution is made from a weak acid and a salt of that weak acid
94
explain how an basic buffer is created
A basic buffer solution is made from a weak
base and a salt of that weak base
95
what will happen if small amounts of acid are added to buffer
equilibrium will shift to the left removing nearly all the H+ ions added,
CH3CO2-(aq) + H+(aq) ~~~ ~~~> CH3CO2H (aq)
96
what will happen if small amounts of base are added to buffer
small amounts of alkali is added to the buffer. The OH- ions will react with H+
ions to form water.
H+ + OH- ~> H2O
The Equilibrium will then shift to the right to produce more H+
ions
97
describe method of acid titrations
1. add 25cm3 of to a conical flask with a pipette
2. set up biurette above the conical flask with other chemical
3. Add indicator to the conical flask
4. note vol in biurette
5. slowly add solution from biurette to the conical flask
6. stop titration when there is a permantent colour change
7. record final vol on the burette
8. titre = start vol - end vol
9. repeat till concoordant results produced (0.10cm3)
98
what are the 4 types of curve and describe how they look
acid base titrations
1. Strong acid and strong base
- start low at acidic end high at alkali
2. Weak acid and strong base
- dont start as low at acidic end high at alkali
3. Strong acid and weak base
- start low at acidic end not as high at alkali
4. Weak acid and weak base
- dont start as low at acidic and dont end as high at alkali
99
calculating half equivelence
equivalence vol / 2
100
how can you pick a correct indicator
colour change must occur at equivelence
101
when should you use phenolphthalein
Only use phenolphthalein in titrations with strong
bases
start with acid in beaker
Colour change:
colourless acid ~> pink alkali
102
when should you use methyl orange
Use methyl orange with titrations with
strong acids
start with acid in beaker
Colour change:
red acid ~> yellow alkali
(orange end point)
103
calculaing bond enthalpy
state units
reactants - products
KJ mol-1
104
calculating entropy
state units
products - reactants
J K-1 mol-1
105
factors affecting entropy
1) state change
s ~> L ~> G
2) more moles on products side more entropy
3)
106
gibbs free energy
units
∆G = ∆H - T (∆s/1000 get kj)
KJ mol-1
107
finding critical temp of a reaction
T= ∆H / ∆S
108
what will happen when ∆H is +ve and ∆S is +ve
feasible at high temperatures
109
what will happen when ∆H is +ve and ∆S is -ve
positive ΔGꝋ so the reaction is not feasible
never be feasible
110
what will happen when ∆H is -ve and ∆S is +ve
negative ΔGꝋ so the reaction is feasible
always be feasible
111
what will happen when ∆H is -ve and ∆S is -ve
at low temperatures ΔGꝋ is negitive
the reaction is at low temp feasible
112
measuring change gas produced
use gas syringe
113
what does collision theory state
how can this effect rate of reaction
for a chemical reaction to take place the particles need to collide with each other in the correct orientation and with enough energy
if enough energy and correct orientation then there will be a successful collision
113
define rate of reaction
give units
the speed at which a chemical reaction takes place
mol dm-3 s-1
mol dm-3 min-1
114
factors affecting rate of reaction
temperature increase more particles with higher kinetic energy more frequent collisions and therefore higher probability of more successful collisions
concentration increase
result in an increased collision frequency and therefore an increased rate of reaction
pressure is increased
molecules have less space in which they can move
number of successful collisions increases due to an increased collision frequency
adding catalyst
decrease the ea of a reaction by providing alternative energy pathway
114
what do maxwell-boltzman distribution curves display
a graph that shows the distribution of particle energies at a certain temperature
115
describe key features of maxwell-boltzman distribution curves
large peak at lower energy this is the most probable energy
the right never touches x axis as its impossible for 0 particles can have high energy
the higher the temp the more particles with sufficient ea
116
whats the general rate equation
Rate = k * [A]m * [B]n
where m and n are the orders of reaction with respect to reactants A and B
k is the rate constant.
117
how are the orders represented in the rate eq using [ a ]
1st order = [ a ]
2nd order = [ a ]2
0 order = not written in rate eq
117
what are the only values for the orders and what do they look like on graphs
and what are the values on axis
1st order linear
2nd order exponential
0 order a strainght horizontal line
y axis = rate
x axis = conc of reactant
118
calculating k in rate equation
k = rate / conc of reactants to their powers
rate = mol dm-3s-1
conc = mol dm-3
overall units cancel out
119
working out orders from the list of numbers
1) identify a test where only one reactant changes
2) find out how much it changed by(2x 3x 4x 0.25x)
3) rate = multiple of change to the power of x
4) find how you would get from order to rate number
120
identifying orders when both conc are changing
1) identify the one condition where only one changes and work out the order for that reaction
2) look where both the conc change and then find the multiplication of both
3) multiply the known value factor by its power and then divide the rate factor by this value
4) now left with rate = [ B ]x figure out the x
121
two ways of expressing Arrhenius equation
k = Ae-EA/RT
ln k = ln A – EA/(RT)
122
what are the units for Arrhenius constant
the exact same as k
122
how to identify the rate determining step
Mechanism
Step 1 A + B ~> X + D fast
Step 2 X + C ~> Y slow
Step 3 Y + B ~> E fast
its the slow step so the rate equation would be
r = k[A] [B] [C]
as they appear once in the reactant side so will be first order and d appears once on each side so cancels out
123
trends across period 3
atomic radius - decrease as number of protons + electrons increase same number of shells as you go across nucleus attracts the electrons more strongly pulling them closer to the nucleus result in smaller atomic rad
melting point - Na donate 1 e- to delocalised e- sea and Mg = 2 Al = 3 so this inc in mp
Si has the highest melting point due giant molecular structure
and P, S, Cl and Ar are non-metallic elements and exist as simple molecules so have low mp
first ionisation energy - increase across a period
The nuclear charge increases The atomic radius decreases so stronger attractive forces between the nucleus and outer electrons
It therefore gets harder to remove any electrons
124
trends in group 2
atomic rad - increase in atomic radius going down the group as there are more shells
first ie - decreases as less energy is required to remove outer most e- as it has less attraction
melting point - decreases going down as the outer electrons get further away from the nucleus weaker attraction between for bonding electrons decreases causing a decrease in melting point
125
group 2 react w water
forming metal hydroxides
1) no rection w berilyum
2) mg react slowly w cold water but vigorously w steam produce MgO
3) the rest
g2 s + 2h2o L ~> g2(OH)2 aq + h2 g
126
solubility of group 2 metal hydroxides
increases as you go down the table
127
solubility of group 2 metal sulfates
decreases as you go down the table
128
testing for sulfates
1) add hcl to remove any carbonates that may lead to false result
2) add barium chloride
3) if positive BaSO4 S/ white preciptiate
4) this is insoluable
129
uses of metal hydroxides
Ca(OH)2 - calcium hydroxide / slaked lime
- used to neutrilise the soil increasing pH less acidic
Mg(OH)2 - magnesium hydroxide / milk of magnesia
- neutralise stomach acid
130
uses of metal sulfates
how doesnt this harm us
BaSO4 - barium sulfate / barium meal
- used to identify blockages in the digestive tract as it absorbs xrays
very toxic but doesnt get absorbed into the blood
131
extracting titanium
1) TiO2 (s) + 2C (s) + 2Cl2 (g) → TiCl4 (g) + 2CO (g)
TiO2 is the natural state of the ore and gets heated w carbon and chlorine gas
2) TiCl4 is ran through fractional distillation column to inc its purity
3) purified TiCl4 is then reduced using magnesium in 1000°C furnace
TiCl4 (g) + 2Mg (l) → Ti (s) + 2MgCl2 (l)
132
The use of CaO or CaCO3 to remove SO2 from flue gases.
CaO + SO2 ~> CaSO3
react on to make CaSO4 gypsum used in plastering
133
appearance of the halogens
1) flourine- pale yellow gas
2) chlorine- pale green gas
3) bromine- orange/brown liquid
4) iodine- grey solid
134
trends in group 7
1) boiling point
increases as there are more e- so more van der waals forces of attraction
2) electronegativity
decreases as you go down as atomic rad increases weaken strength of nucleus attracting the e- pair
135
displacement reactions of the halogens
reactivity decreases as you go down group
more upwards on the group displaces out of the Nag7 / Kg7
chlorine will be best
iodine will be worst
cl2 aq = pale green
br2 aq = orange
i2 = brown
136
mixing NaOH and chlorine
2 NaOH aq + Cl2 g ~> NaClO aq + NaCl aq + h2o l
disproportionation as the NaClO is +1 and teh NaCl is -1
NaClO is used to treat water or bleach fabrics
136
mixing water to chlorine absence of sunlight
Cl2 aq + h2o l ~> hcl aq + HClO
ClO- used to kill bacteria
137
mixing water to chlorine presence of sunlight
Cl2 aq + 2h2o l ~> 4h+ aq + 2Cl- + o2
138
evaluating chlorinating drinking water
+ sterilise water kill bacteria
- cl2 gas is toxic mayresult in respiritory issues
- liquid cl cause severe chemical burns
+ve outweigh -ve
139
reducing ability of halide ions
as you go down ionic radius inc so as you go down reducing ability increase
140
define a transition metal
1 or more stable ion with an incomplete d subshell
141
why arent scandium and zinc classed as transition metals
scandium only 3+ so empty d subshell 4s2 and its only 3d1
zinc only 2+ full d sub shell as loses 4s2
142
properties of transition metals
1) formmation of complexes
2) colourful ions
3) variable oxidation states
4) used as catalyst
143
why do they have variable oxidation states
vo2 ^+ = yellow (5)
vo2 ^2+ = blue (4)
vo^3+ = dark green (3)
vo^2+ = purple (2)
get reduced by zinc
Zn ~> Zn^2+ + 2e-
144
define complex ion
central transition metal ion surrounded by ligands bonded by coordinate bonds
145
what does a ligand do
how are they classified
examples
provide an electron pair
monodentate - provide one e- pair per molecule
(H2O: , :NH3 , :Cl-)
bidentate - provide two e- pair per molecule
(1 ethane dioate 2- , a double carboxyl without the h on the oh just a :O
ethane 1,2 diamine an ethane w 2 :NH2 per carbon)
multidentate - provide many e- pair per molecule
(EDTA^4-)
146
whats coordination number
number of coordinate bond in the complex
147
complex shapes sizes of ligands
small ligands such as oh- / h2o are small six of these can bond to central metal ion octahedral
medium size = Cl- only 4 of these can bond per metal ion tetrahedral
larger ligands = bidentate ligands only bond three times octahedral still coordination number of 6
148
what is cis platin
[Pt(NH3)2(Cl)2]
square planar used to treat cancer
149
what is [Ag(NH3)2]+
tollens reagent
150
why do transition metals show colour
when aligand binds to a metal ion it splits the d orbital
when e- absorb light energy they may jump to the higher energy level absorbing specific frequency of the white light show all non absorbed colours
151
whats the equation for working out energy change in absorbance of light
ΔE = h * v
= (h*c) / λ
λ = wave length
h = planks constant
c = 3x10^8
ΔE = energy change
v = frequency of light
152
using colourimeter
1) zero it using a clear sample usually just water and measure absorbence of light
2) put sample in and it will compare absorbance to the clear sample
30 zero with clear sample again after use
153
whats the chelate effect
when monodentate ligands are substituted for a bidetate/ multidenntate ligand result in higher stability due to more particles on products more entropy
154
redox titrations
self determining end point and MnO4- is purple and will be seen when it becomes in excess
MnO4- + 8H+ + 5Fe2+ ~~~ ~~> Mn2+ + 5Fe3+ + 4H2O
155
titration questions
1) write out full eqaution
2) work out moles from what you have been given
3) use mol ratio to find out the number of moles of desired thing
4) calculate conc = n/v
156
two types of catalyst
heterogeneous catalyst- different phase to reactants
homogeneous catalyst- same phase to reactants
157
factors effecting the hetergeneous catalyst
increasing surface area of the catalyst will increase rate of reaction
sprayed onto honeycomb mesh
158
process happening at heterogeneous catalyst
contact process
V2O5 + SO2 ~> V2O4 + SO3
V2O4 + 0.5 O2 ~> V2O5
159
whats catalytic poisoning
solid particles that bind to surface of heterogeneous catalyst and are very hard to remove blocks the binding sites for reactants reduce ror
160
process happening at homogeneous catalyst
S2O8^-2 + 2Fe+2 ~~~~ ~~~> 2Fe+3 + 2 SO4-2
2I- + 2Fe+3 ~~~~~~~ ~~~> 2Fe+2 + I2
161
Whats autocatalysis
when the product catalyses the reaction between the two reactants
2MnO4- + 16H+ + 5c2o4^2- ~~~~~> 2Mn^2+ + 10co2 + 8H2O
react really slow at start as you are reacting 2 negative ions togeather
gets faster as reaction goes on as the product will catalyse the reaction
162
colours of metal aqua complexes
1) [Cu(h2o)]2+ =blue solution
2) [Fe(h2o)]2+ = green solution
3) [Fe(h2o)]3+ = yellow solution
4) [Al(h2o)]3+ = colourless solution
163
factors affecting acidity of metal aqua ions
higher charge density 3+ means there is less electron density in the O-H bond and is weakened easier to remove the H+
164
Reactions of
oh- (aq) excess only happens with one
- what word describes this reaction
[Al(h2o)3(oh)3] (s) + oh- ~~ ~~> [Al(oh)4]- (aq) + 3h2o l
colourless sol
- amphoteric
165
reactions of every metal aqua ion ([M(h2o)6]3+ (aq)) and limited oh- (aq) colours also
[Cu(h2o)4(oh)2] (s) +2 h2o (l) = blue ppt
[Fe(h2o)4(oh)2] (s) +2h2o (l) = green ppt
[Fe(h2o)3(oh)3] (s) + 3h2o (l) = brown ppt
[Al(h2o)3(oh)3] (s) + 3h2o (l) = white ppt
166
Reactions of
H+ (aq) excess only happens with one
- what word describes this reaction
[Al(h2o)3(oh)3] (s) + 3H+ ~~ ~~> [Al(h2o)6]3+ (aq)
colourless sol
- amphoteric
167
Reactions of
NH3 (aq) excess only happens with one
[Cu(h2o)4(oh)2] (s) + nh3 ~ ~> [Cu(nh3)4(h2o)2] + 2h2o +2oh-
deep blue sol
168
Reactions of
NH3 (aq) limited and metal aqua ion ([M(h2o)6]x+ (aq))
[Cu(h2o)4(oh)2] (s) + 2nh4+
= blue ppt
[Fe(h2o)4(oh)2] (s) + 2nh4+
= green ppt
[Fe(h2o)3(oh)3] (s) + 3nh4+
= brown ppt
[Al(h2o)3(oh)3] (s) + 3nh4+
= white ppt
169
Reactions of
co3 2- (aq) and metal aqua ion ([M(h2o)6]2+ (aq))
[Cu(h2o)6]2+ (aq) + 2co3 2-
~~> CuCO3 (s) + 6h2o
blue green ppt
[Fe(h2o)6]2+ (aq) + 2co3 2-
~~> FeCO3 (s) + 6h2o
green ppt
170
Reactions of
co3 2- (aq) and metal aqua ion ([M(h2o)6]3+ (aq))
how to test what ion is present
2 [Fe(h2o)6]3+ (aq)
+ 3 co3 2-
~~> 2 [Fe(h2o)3(oh)3] (s) + 3 h2o + 3 co2
brown ppt
2 [Al(h2o)6]3+ (aq)
+ 3 co3 2-
~~> 2 [Al(h2o)3(oh)3] (s) +
3 h2o + 3 co2
white ppt
bubble gas produced through lime water and this will turn it cloudy
171
solving titration q
1. work out moles of thing added from burette
2. use equation to work out mol ratio
3. n/v = c
172
key parts of tlc chromatography
stationary phase - part that doesnt move
- this is the silica plate
mobile phase - molecules can move
- liquid hexane in the container
The atmosphere inside the chamber should be saturated with solvent vapour to prevent the mobile phase evaporating from the plate during the separation.
-use a lid on
the chamber
- add some filter paper behind the plate saturated with solvent
the further spot move the lower the affinity for the stationary phase
173
how to calcullate rf value in tlc
rf= dist moved by spot /
dist moved by solvent
174
halide reactions with sulfuric acid
NaCl + H2SO4 ~> NaHSO4 + HCl (g = misty fumes)
H2SO4 + 2h+ + 2Br- ~~>Br2 + SO2 + 2H2O
( coloured halogen gas and choking SO2 gas )
H2SO4 + 6H+ + 6i- ~~> 3i2 + s + 4h2o
(yellow solid = sulfur )
H2SO4 + 8H+ + 8i- ~~> 4i2 + h2s + 4h2o
( rotten egg smelling gas )
175
halide reactions with silver nitrate
further testing
add nitric acid to remove impurities such as carbonates that may react and give positive result
Ag+ + G7- ~> AgG7 (s)
add dilute ammonia
AgCl dissolve
add conc ammonia
AgBr dissolve
AgI will never dissolve
