paper 3

Question 1

ehats general rate eq

Answer 1

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.

Question 2

working out orders from the list of numbers

Answer 2

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

Question 2

how are the orders represented in the rate eq using [ a ]

Answer 2

1st order = [ a ]

2nd order = [ a ]2

0 order = not written in rate eq

Question 2

what are the only values for the orders and what do they look like on graphs
and what are the values on axis

Answer 2

1st order linear

2nd order exponential

0 order a strainght horizontal line

y axis = rate
x axis = conc of reactant

Question 2

calc k in rate eq

Answer 2

k = rate / conc of reactants to their powers
rate = mol dm-3s-1
conc = mol dm-3

overall units cancel out

Question 2

A co-ordinate (dative covalent) bond intramolecular bond

Answer 2

A co-ordinate (dative covalent) bond contains a shared pair of electrons with both electrons supplied by one atom.

Question 2

ionic bond intramolecular bond

type of force between ions

Answer 2

electrostatic attraction between oppositely charged ions in a lattice

Question 2

covalent bond intramolecular bond

Answer 2

A single covalent bond contains a shared pair of electrons.

Question 3

identifying orders when both conc are changing

Answer 3

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

Question 3

properties of ionic lattices

Answer 3

1)alternating cations (+ve) and anions (-ve)
2) conduct electricity when molten or dissolved
3) high melting point

Question 3

metallic bond intramolecular bond

Answer 3

attraction between delocalised electrons and positive ions arranged in a lattice.

Question 3

properties of giant covalent molecular lattices of carbon

Answer 3

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

Question 3

define electronegativity

Answer 3

power of an atom to attract the pair of electrons in a covalent bond

Question 3

Properties of Metallic bonding

Answer 3

• metal ions are surrounded by a ‘sea’ of delocalised electrons
• conduct heat + electricity
• high mp

Question 3

define Le Chatelier’s principle

Answer 3

when a change is made to a reaction the system attempts to counteract it by shifting the position of equilibrium

Question 4

changing conc temp press

Answer 4

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

Question 4

step 2 in time of flight mass spectrometry
include ke calc

Answer 4

2) acceleration
positive ions are attracted to negatively charged plate
they all have the same kenetic energy
ke = 0.5 x mass x velocity²

Question 4

equilibrium constant kc expression

Answer 4

( [ A ]^a * [ B ]^b )

Question 4

kc calculations

Answer 4

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

Question 5

kp calculation

method of working out the par press

Answer 5

1) work out mol fraction
mol / total mol
2) partial pressure =
mol fraction * given pressure
3) units are kPa but cancel

Question 5

kp expression

Answer 5

( p(C)^c * p(D )^d) /
( p(A)^a * p(B)^b)

Question 5

trends of 1st ie in period 3

Answer 5

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

Question 5

step 1 in time of flight mass spectrometry

Answer 5

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-

Question 6

calculations for tof

Answer 6

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

Question 7

step 3 in time of flight mass spectrometry

Answer 7

3) ion drift travel past the negatively charged plate into flight tube each particles velocity depends on its (mass lighter will move faster)

Question 7

step 4 in time of flight mass spectrometry

Answer 7

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

Question 8

factors affecting first ionisation energy

Answer 8

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

Question 9

trends of 1st ie in group 2

Answer 9

atomic radius increases as you go down the group so less energy required to remove outermost e-

Question 10

oxidation of alcohol

Answer 10

primary alc ~> aldehyde ~> carboxylic acid secondary alc ~> ketone tertiary alc ~> no reaction ox agent is potassium dichromate cr2o7^-2 turn from orange to green

Question 11

colours of metal aqua complexes

Answer 11

1) [Cu(h2o)]2+ =blue solution 2) [Fe(h2o)]2+ = green solution 3) [Fe(h2o)]3+ = yellow solution 4) [Al(h2o)]3+ = colourless solution

Question 12

Reactions of oh- (aq) excess only happens with one - what word describes this reaction

Answer 12

[Al(h2o)3(oh)3] (s) + oh- ~~ ~~> [Al(oh)4]- (aq) + 3h2o l colourless sol - amphoteric

Question 12

Whats autocatalysis

Answer 12

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

Question 12

Reactions of H+ (aq) excess only happens with one - what word describes this reaction

Answer 12

[Al(h2o)3(oh)3] (s) + 3H+ ~~ ~~> [Al(h2o)6]3+ (aq) colourless sol - amphoteric

Question 13

reactions of every metal aqua ion ([M(h2o)6]3+ (aq)) and limited oh- (aq) colours also

Answer 13

[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

Question 14

Reactions of NH3 (aq) excess only happens with one

Answer 14

[Cu(h2o)4(oh)2] (s) + nh3 ~ ~> [Cu(nh3)4(h2o)2] + 2h2o +2oh- deep blue sol

Question 15

Reactions of NH3 (aq) limited and metal aqua ion ([M(h2o)6]x+ (aq))

Answer 15

[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

Question 15

Reactions of co3 2- (aq) and metal aqua ion ([M(h2o)6]2+ (aq))

Answer 15

[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

Question 16

Reactions of co3 2- (aq) and metal aqua ion ([M(h2o)6]3+ (aq)) how to test what ion is present

Answer 16

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

Question 17

key parts of tlc chromatography

Answer 17

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

Question 18

how to calcullate rf value in tlc

Answer 18

rf= dist moved by spot / dist moved by solvent

Question 19

What are quaternary ammonium salts used for

Answer 19

Shampoos and laundry detergent as they attract -ve ions

Question 19

Why are aromatic bases weak

Answer 19

As the aromatic ring results in the line pairs being drawn in and partially delocalised less available for donation

Question 19

Classification of amines

Answer 19

1° - 1 c-n. 2n-h. 1 lp 2° - 2 c-n 1n-h 1lp 3° - 3c-n. 0n-h. 1lp 4° - 3c-n. 0n-h. 0lp. And a +ve and the ion making salt

Question 20

Factors affecting amine base strength

Answer 20

Positive inductive effect of the alkyl groups bonded result in e- being more available for donation

Question 21

Synthesising amines

Answer 21

1) React halogenoalkane with excess ammonia Nucleophillic substitution nh3 attack the c-g7 and then g7 leaves create c-nh3+ another ammonia comes and takes h+ form c-nh2 and (nh4+cl-) This reacts further making 2°,3°4° and amine can act as nucleophile 2)reducing nitriles a}Ni catalyst and h2 gas catalytic hydrogenation Nitrile + h2 -> amine b} lialh4 strong reducing agent and dry ether solvent Nitrile + 4[H]-> amine

Question 21

Amines use as bases

Answer 21

Form dative covalent bonds with h+ giving both e- in line pair become +ve charged

Question 22

define a transition metal

Answer 22

1 or more stable ion with an incomplete d subshell

Question 23

properties of transition metals

Answer 23

1) formmation of complexes 2) colourful ions 3) variable oxidation states 4) used as catalyst

Question 23

Anti cancer drugs

Answer 23

Cis-platin Central pt Nh3 x2 on same side Cl x2 on same side Square planar shape Cl- bonds easily break off and the pt make two bonds to guanine bases Binds to the sugar phosphate backbone and makes it no longer complementary and prevents the replication of the dna preventing mitosis

Question 24

define complex ion

Answer 24

central transition metal ion surrounded by ligands bonded by coordinate bonds

Question 24

why arent scandium and zinc classed as transition metals

Answer 24

scandium only 3+ so empty d subshell 4s2 and its only 3d1 zinc only 2+ full d sub shell as loses 4s2

Question 25

complex shapes sizes of ligands

Answer 25

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

Question 25

what does a ligand do how are they classified examples

Answer 25

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-)

Question 26

whats coordination number

Answer 26

number of coordinate bond in the complex

Question 27

whats the equation for working out energy change in absorbance of light

Answer 27

ΔE = h * v = (h*c) / λ λ = wave length h = planks constant c = 3x10^8 ΔE = energy change v = frequency of light

Question 28

whats the chelate effect

Answer 28

when monodentate ligands are substituted for a bidetate/ multidenntate ligand result in higher stability due to more particles on products more entropy

Question 28

tollens reagent formula and use

Answer 28

[Ag(NH3)2]+ form silver mirror in presence of aldehyde

Question 29

production of alcohol advantages

Answer 29

1) hydrogenation of ethene - e- rich c=c bond in ethene gets attacked by the H+ - carbocation is formed h20 bod to this and the o has a +ve charge one of the H's donate its e- pair and break off forming ethanol and H+ - h3po4 catalyst - 300°c - 70 atm of pressure - use steam advantages - fast - continous process - pure product disadvantages ethene from crude oil non-renewable resource will run out - very expemsive 2) fermentation of sugars glucose ~> ethanol + co2 yeast anaerobic coditions no air 30-50°c for yeast to work advantages - sugar is renewable - low technology used in low income countries disadvantages - batch process slow labourous - impure need to be further distilled

Question 30

elimination reactions w an alcohol or dehydration

Answer 30

conc h2so4 under reflux - h+ bond w lone pair on the oxygen and then this h2o molecule breaks off due to being +vely charged - carbocation produced and the h on adjacent c break off give e- pair to the c-c and this forms c=c give alkene + h+ + h2o

Question 30

ethanol use as bio fuel how its "carbon neutral"

Answer 30

photosynthesis takes in 6 molecules of co2 6co2 + 6h2o ~> c6h12o6 + 6o2 and in the fermentation c6h12o6 ~> 2c2h5oh + 2co2 in combustion 2C₂H₅OH + 6O₂ → 4CO₂ + 6H₂O so overall the co2 cancel however this wont be true as the transport will produce co2

Question 31

calorimetry definition and equation

Answer 31

experimental method of working out enthalpy change of a reaction q=mcΔt /1000 = kJ Δh = q/ n if temp increase then its exothermic so put -ve infront

Question 32

calorimetry practical method

Answer 32

1) fuel burner with known mass and known mr add beaker over it of a known volume 2) use thermometer to measure start temperature 3) light fuel and measure temp change of water 4) use calculation to work out ΔH

Question 33

factors affecting ΔH value

Answer 33

heat loss to surroundings heat absorbed by container INCOMPLETE COMBUSTION evapouration of volitile fuel

Question 34

Making standard solution

Answer 34

1) Weigh out sample in boat 2) Add sample to beaker and rewight empty boat 3) mass used = initial -final 4) dissolve using distilled water and use stirring rod 5) use funnel to pour into 250cm volumetric flask and wash equipment used into the flask 6) fill so bottom of meniscus is on grad mark 7) add stopper and invert 10 times

Question 35

Organic functional group tests

Answer 35

Aldehyde - tollens silver mirror - feilings blue-> brick reds ppt Alkene - decolourise bromine water Alcohol 1°/ 2° - orange sol-> green sol

Question 36

Enthalpy change of solution

Answer 36

When one mole of compound dissolves completely in water so that it’s ions do not interact