A-LEVEL CHEMIISTRY, KINETICS II.

Question 1

RATE OF REACTION.

Answer 1

THE CHANGE OF CONCENTRATION./.= AMOUNT OF REACTANT OR PRODUCT PER UNIT TIME.

Question 2

RATE OF REACTION, CALCULATION.

Answer 2

RATE.= AMOUNT OF REACTANT USED./.= PRODUCT MADE / TIME.

Question 3

ELECTRICAL CONDUCTIVITY.

Answer 3

DURING A REACTION THERE MAY BE A CHANGE IN THE NUMBER OF IONS.

THIS CHANGE WILL AFFECT THE LEVEL OF ELECTRICAL CONDUCTIVITY.

Question 4

AMOUNT OF MASS LOST.

Answer 4

FOR A REACTION THAT PRODUCES A GAS, PLACE REACTION ON BALANCE AND MEASURE THE MASS LOSS AS A GAS IS LOST.

FAIRLY ACCURATE METHOD HOWEVER USE A FUME CUPBOARD IF GAS IS HARMFUL OR TOXIC.

Question 5

VOLUME OF GAS PRODUCED.

Answer 5

ANOTHER WAY OF MEASURING THE RATE IF A GAS IS PRODUCED.

MEASURE THE AMOUNT OF GAS PRODUCED USING A GAS SYRINGE.

MEASURE THIS OVER A SPECIFIED TIME.

Question 6

THE CHANGE IN pH OF A REACTION.

Answer 6

THE pH OF A REACTION MAY CHANGE OVERTIME IF H+ IONS ARE USED UP OR PRODUCED.

A pH METER CAN BE USED TO MEASURE THE pH OF A REACTION AT REGULAR INTERVALS.

YOU CAN THEN CALCULATE THE H+ ION CONCENTRATION.

Question 7

TITRATION.

Answer 7

WE CAN MONITOR THE CHANGE IN CONCENTRATION OF A REACTANT OR PRODUCT BY TAKING SMALL SAMPLES, ALIQUOT, AT REGULAR TIME INTERVALS AND TITRATING THEN.

WHEN WE TAKE ALIQUOT WE MUST SLOW THE REACTION DOWN IMMEDIATELY.

IF WE DID NOT THE REACTION WOULD CONTINUE AS NORMAL AND THE CONCENTRATION WOULD CHANGE AS WE TRY TO CONDUCT THE TITRATION.

TO SLOW THE REACTION DOWN WE CAN EITHER DO:

DILUTION WITH DEIONISED WATER,

COOL IT DOWN,

ADD A CHEMICAL TO STOP THE REACTION, QUENCHING.

Question 8

COLORIMETER.

Answer 8

IF A REACTION HAS A COLOUR CHNAGE THEN THIS CAN BE MEASURED USING A COLORIMETER.

A COLORIMETER MEASURES THE ABSORBCANCE OF LIGHT BY A COLOURED SAMPLE.

THE MORE CONCENTRATED A SAMPLE IS, THE DARKER IT’S COLOUR AND HENCE MORE LIGHT IS ABSORBED.

FIRST WE HAVE TO PLOT A CALIBIRATION CURVE.

A COMMON REACTION IS BETWEEN POPANONE AND IODINE:

I2 + CH3 CO CH3 -> CH3COCHI + I- + H+

ALL AQEOUS, SO THE STATE SYMBOL IS (aq.).

THIS IS CREATED BY MAKING UP A RANGE OF KNOWN DIFFERENT CONCENTRATION OF IODINE.

THE ABSORPTION IS MEASURED FOR EACH ONE OF THE CONCENTRATIONS AND THE RESULTS ARE PLOTTED.

THE X AXIS, CONCETRATION.
THE Y AXIS, ABSORBANCE.

AN EXPERIMENT LIKE THE ONE ABOVE IS ET AWAY AND SAMPLES ARE TAKES REGULARLY.

THESE SAMPLE ARE TESTED FOR ABSORBANCE.

WE THEN USE THE CALIBIRATION GRAPH AND ABSORBANCE TO FIND OUT THE I2.

Question 9

CALCULATE RATE FROM A GRAPH.

Answer 9

RATE CAN BE FOUND FROM THE GRADIENT.

GRADIENT.= CHANGE IN Y/ CHNAGE IN X.
WE USE THE TRIANGLE METHOD FOR THIS.

Question 10

THE RATE EQUATION.

Answer 10

THE RATE EQUATION LINKS RATE WITH CONCENTRATIONS OF SUBSTANCES.

RATE.= K [A]^a [B] ^b.

RATE,
moldm^-3 s ^-1.

RATE CONSTANT K,
UNITS VARY.

A AND B,
CONCENTRATION OF SUBSTANCE.

a AND b,
ORDERS OF THE REACTION.

Question 11

ORDERS OF REACTION.

Answer 11

AN ORDER IS THE POWER TO WHICH A CONCENTRATION IS RAISES TO IN THE RATE EQUATION.

IT TELLS US HOW THE CONCENTRATION OF THE SUBSTANCE AFFECTS THE RATE.

Question 12

ZERO ORDER.

Answer 12

CHANGES IN CONCENTRATION HAS NO EFFECT ON RATE.

FOR EXAMPLE, IF [A] DOUBLES THEN THE RATE DOES NOT CHANGE.

SO IF a OR b EQUAL ZERO,0.

Question 13

FIRST ORDER.

Answer 13

CHANGES IN CONCENTRATION HAS A PROPORTIONAL CHANGE ON RATE.

FOR EXAMPLE IF A DOUBLES THEN RATE DOUBLES.

Question 14

SECOND ORDER.

Answer 14

CHANGES IN CONCENTRATION HAS A SQUARED PROPORTIONAL CHANGE ON RATE.

FOR EXAMPLE IF A DOUBLES THEN RATE QUADRUPLES.

Question 15

ORDERS DETERMINED.

Answer 15

ORDERS CAN ONLY BE DETERMINED BY EXPERIMENT.

YOU CAN NOT WORK THEM OUT BY LOOKING AT AN EQUATION.

Question 16

RATE CONSTANT.

Answer 16

A CONSTANT IS A NUMBER THAT ALLOWS US TO EQUATE RATE AND CONCENTRATION.

THE RATE CONSTANT IS ONLY FIXED AT A PARTICULAR TEMPERATURE.

IF THE TEMPERATURE CHANGES SO DOES THE RATE CONSTANT.

K INCREASES WHEN TEMPERATURE INCTREASES.

Question 17

K VALUE.

Answer 17

THE LARGER THE VALUE OF K THE FASTER THE RATE OF THE REACTION.

AS WE INCREASE THE TEMPERATURE THE PARTICLES HAVE MORE KINETIC ENERGY AND COLLIDE MORE OFTEN.

THIS INCREASES THE RATE.

BUT THE CONCENTRATIONS OF THE SUBSTANCES REMAIN CONSTANT.

TO MAKE THIS EQUATION BALANCE THEN THE VALUE OF K MUST INCREASE.

Question 18

INITIAL RATE, CALCULATING IT FROM A GRAPH.

Answer 18

THE INITIAL RATE IS THE RATE RIGHT AT THE START OF THE REACTION.

WE TAKE THE GRADIENT OF THE TANGENT BUT WITH THE INITIAL RATE WE TAKE IT AT ZERO,0, MINUTES.

WE DRAW A LINE STRAIGHT DOWN TO THE X AXIS.

Question 19

CLOCK EXPERIMENTS.

Answer 19

A CLOCK REACTION IS WHERE YOU CAN TIME HOW LONG IT TAKES FOR A REACTION TO OCCUR.

THE CLOCK REACTION IS USED TO SIMPLIFIY THE INITIAL RATE REACTION.

MOST REACTIONS LIKE IODINE CLOCK CAN BE MONITORED BY SITTING THE REACTION VESSEL ON SOME PAPER WITH A CROSS ON, WE TIME HOW LONG IT TAKES UNTIL WE CAN NOT SEE THE CROOS THROUGH THE BEAKER. THERE IS A COLOUR CHANGE. THIS IS KNOWN AS THE END POINT.

ESSENTIALLY, THE QUICKER THE CLOCK REACTION, THE FASTER THE INITIAL RATE OF REACTION IS.

Question 20

THE CLOCK EXPERIMENT, ASSUMPTIONS.

Answer 20

THE TEMPERATUE OF THE REACTION REMAISN CONSTANT.

THE CONCENTRATION OF THE REACTANTS DOES NOT CHANGE SIGNIFICANTLY DURING THE TIME PERIOD OF THE REACTION.

REACTION HAS NOT PROCEEDED TOO FAR WHEN THE END POINT IS SEEN.

ON THE BASIS OF THEESE ASSUMPTIONS WE CAN SAY THE RATE OF REACTION REMAINS CONSTANT DURING THE TIME PERIOD YOU ARE MEASURING.

Question 21

THE IODIN CLOCK REACTION.

Answer 21

HARCOURT-ESSON.

THE IODINE CLOCK REACTION IS:

H2O2 + 2H+ + 2I- -> 2H20 + I2

ALL AQEOUS BESIDES FROM WATER, H2O, WHICH IS A LIQUID.

ALL WE DO IS ADD SODIUM THIOSULFATE AND STARCH, WHICH ACTS AS AN INDICATOR, TO EXCESS HYDROGEN PEROXIDE.

THE SODIUM THIOSULFATE REACTS IMMEDIATELY WITH THE IODINE, I2, THAT IS PRODUCED IN THIS REACTION.

2S2O3 2- + I2 -> 2I- + S406 2-

ALL AQEOUS.

WHEN THERE IS NO MORE SODIUM THIOSULFATE LEFT THEN THE IODINE, I2, REACTS WITH THE STARCH.

THIS GIVES A DEEP BLUE./.= BLACK COLOUR.

Question 22

THE IODINE CLOCK REACTION, CONCENTRATIONS.

Answer 22

VARYING THE CONCENTRATION OF IODINE, I2, AND./.= OR HYDROGEN PEROXIDE, H2O2, AND KEEPING EVERYTHING ELSE CONSTANT WILL RESULT IN THE TIME TAKEN FOR THE BLUE./.= BLACK COLOUR TO APPEAR CHANGING.

WE CAN THEN WORK OUT THE ORDER OF THE REACTION.

Question 23

HALF-LIFE.

Answer 23

HALF-LIFE, t 1/2, IS THE TIME IT TAKES FOR HALF OF THE REACTANT TO BE USED UP.

Question 24

IDENTIFYING ORDER.

Answer 24

ZERO ORDER, THE RATE ON A STRAIGHT LINE GRAPH IS CONSTANT.

CHANGING CONCENTRATION DOES NOT CHANGE THE RATE.

HALF-LIFE DECREASES.

FIRST ORDER, THE RATE ON A SHALLOW CURVE GRAPH IN EQUALS AOUNTS.

CHANGING CONCENTRATION CHANGES THE RATE EQUALLY.

HALF-LIFE IS CONSTANT.

SECOND ORDER,
THE RATE ON A STEEP CURVE GRAPH CHANGES IN UNEQUAL AMOUNTS.

CHANGING CONCENTRATION CHANGES THE RATE SQUARED.

HALF-LIFE INCREASES.

Question 25

THE RATE DETERMINING STEP.

Answer 25

THE RATE DETERMINING STEP IS THE SLOWEST STEP IN A MULTI-STEP REACTION.

THE WHOLE REACTION RATE DEPENDS ON HOW QUICK THE RATE DETERMINING STEP IS.

Question 26

THE RATE DETERMINING STEP, CATALYST AND TEMPERATURE.

Answer 26

TO SPEED UP THE RATE DETERMINING STEP WE COULD USE A CATALYST OR CHANGE THE TEMPERATURE.

Question 27

THE RATE DETERMINING STEP, REACTANTS.

Answer 27

REACTANTS THAT APPEAR IN THE RATE EQUATION AFFECT THE RATE OF REACTION.

THESE REACTANTS, OR SUBSTANCES DERIVED FROM THEM, MUST APPEAR IN THE RATE DETERMINING STEP.

SUBSTANCES NOT IN THE RATE EQUATION WILL NOT BE IN THE RATE DETERMINING STEP.

Question 28

RATE DETERMINING STEP, MULTI-STEP REACTION.

Answer 28

A + B -> 2C FAST.

2C -> D SLOW.

D + E -> F+G FAST.

OVERALL EQUATION:

A+B+E -> F+G.

EQUATION TWO,2, IS THE RATE DETERMINING STEP, AS IT IS THE SLOWEST STEP.

ONE OF THE REACTANTS IS C, SO THIS MUST APPEAR IS THE RATE EQUATION.
BECAUSE THE C HAS A TWO,2, IN FROT OF IT, THIS MEANS THAT IT IS TO THE ORDER TWO,2.

C IS AN INTERMEDIATE WHICH IS FORMED, DERIVED, FROM A +B. FOR THIS REASON BOTH A AND B MUST BE IN THE RATE EQUATION TO.

THERE ARE ONLY ONE,1, OF A AND B SO THE ORDER IS FIRST ORDER WITH RESPECT TO A AND B.

THE RATE EQUATION.= K[A][B][C]^2.

C IS NOT A REACTANT AS IT DOES NOT APPEAR IN THE OVERALL EQUATION.

C IS A CATALYST.

THIS REACTION IS AN EXAMPLE OF AUTOCATALYSIS.

Question 29

FINDING THE RATE DETERMING STEP, FROM THE RATE EQUATION.

Answer 29

NO + NO ->N2O2.

N2O2 + O2 -> 2NO2.

RATE EQUATION.= K[NO]^2[O2].

LOOK FOR THE RATION OF MOLECULES WE ARE LOOKING FOR:

TWO,2, NO.
ONE,1, O2.

LOOKING AT THE REACTION STEPS THERE IS NOT ONE WITH THIS IN.

SO WE NEED TO START FROM STEP ONE,1, AND SCOR OFF THE MOLECULES./.= ATOMS NEEDED TO MATCH THE RATIO IN THE RATE EQUATION.

WE STOP WHEN WE HAVE COUNTED FOR THEM ALL.

THIS STEP WILL BE THE RATE DETERMINING STEP.

SCORE OFF THE TWO,2, NO’S IN STEP ONE AND THE O2 IN THE STEP TWO,2, REACTION.

SO OUR RATE DETERMING STEP IS STEP TWO,2, AS THIS IS WHERE WE HAVE SCORED OFF ALL OF OUR REACTANTS IN THE RATE REACTION.

Question 30

THE RATE MECHANISM, THE RATE DETERMING STEP.

Answer 30

THE RATE REACTION TELLS US ONLY THE REACTANTS IN THE RATE DETERMING STEP.

Question 31

HALOGENOALKANES HYDROLYSIS.

Answer 31

HALOGENOALKANES CAN BY HYDROLYSED, SPLIT, BY HYDROXIDE IONS.

THEY FOLLOW DIFFERENT MECHANISMS DEPENDING ON IF THEY ARE PRIMARY, SECONDARY, OR TERTIARY.

Question 32

PRIMARY HALOGENOALKANES.

Answer 32

ONLY ONE,1, AKYL GROUP ATTACHED TO CARBON,C, WITH HALOGEN ON.

HYDROGEN,H, DOES NOT COUNT AS AN ALKYL GROUP.

Question 33

SECONDARY HALOGENOALKANES.

Answer 33

TWO,2, ALKYL GROUPS BONDED TO CARBON,C, WITH HALOGEN ON.

Question 34

TERTIARY HALOGENOALKANES.

Answer 34

THREE,3, ALKYL GROUPS ATTACHED TO CARBON,C, WITH HALOGEN ON.

Question 35

HALOGENOALKANES BONDS.

Answer 35

HALOGENOALKANES HAVE A POLAR BOND AND ARE ATTACKED BY NUCLEOPHILES.

HALOGENS ARE MORE ELECTRONEGATIVE THAN CARBON,C, SO THEY PULL ELECTRONS TOWARDS THEMSELVES IN A COVALENT BOND.

THIS LEADS TO A POLAR BOND.

THIS POLAR BOND MEANS THAT HALOGENOALKANES CAN BE ATTACKED BY NUCLEOPHILES.

Question 36

NUCLEOPHILES.

Answer 36

A NUCLEOPHILE IS A SUBSTANCE THAT IS AN ELECTRON PAIR DONOR.

THEY ARE USUALLY NEGATIVE OR NEURTRAL CHARGE.

AN EXAMPLE OF A NUCLEOPHILE IS A HYDROXIDE ION, OH-.

Question 37

HALOGENOALKANES REACTION WITH HYDROXIDE IONS.

Answer 37

HALOGENOALKANES REACT WITH HYDROXIDE IONS VIA NUCLEOPHILIC SUBSTITUTION.

THE CONDITIONS:

WARM AQEOUS SODIUM HYDROXIDE, SOURCE OF OJ- IONS.

CARRIED OUT UNDER REFLUX.

A NUCLEOPHILE WILL ATTACK THE DELTA POSITIVE CARBON,C AND WILL REPLACE THE HALGOEN ON THE HALOALKANE.

THE C-X, X REPRESENT A HALOGEN BREKS, ELCTRONS MOVE FORM THE BOND TO THE HALOGEN IN HETERLOYTIC FISSION.

A NEW BOND IS FORMED BETWEEN OH- ION AND CARBON,C.

OVERALL EQUATION:

R-X + NaOH -> ROH +NaX.

R IS THE ALKYL GROUP, SUCH AS -CH3.
X IS THE HALOGEN, SUCH AS -Br AND -Cl.

AN ALCOHOL IS FORMED.

Question 38

NUCLEPHILIC SUBSTITUTION.

Answer 38

THERE ARE TWO,2, DIFFERENT TYPES OF NUCLEOPHILIC SUBSTIUTION THEY ARE SN1 AND SN2.

SN1 REACTIONS HAVE ONLY ONE,1, MOLECULE OR ION IN THE RATE DETERMINING STEP.

SN2 REACTIONS HAVE TWO,2, MOLECULES OR IONS OR BOTH IN THE RATE DETERMINING STEP.

SN1 MECHANISM, SECONDARY AND TERTIARY HALOGENALOKANES.

SN2 MECHANISN, PRIMARY HALOGENOALKANES.

Question 39

ARRHENOUS EQUATION.

Answer 39

THE ARRHENIUS EQUATION LINKS ACTIVATION ENERGY AND TEMPERATURE TO THE RATE CONSTANT, K.

K.=Ae ^ ( -Ea/RT)

K,
RAT CONSTANT.

Ae,
ARRHENIUS CONSTANT.

Ea,
ACTIVATION ENERGY, JULES,J.

R.
GAS CONSTANT,
8.31 JK^-1mol^-1.

TEMPERATURE,
KELVINS,K.

Question 40

GAS CONSTANT.

Answer 40

R,
8.31 JK^-1 mol^-1.

Question 41

ARRHENIUS EQUATION, ACTIVATION ENERGY.

Answer 41

AS THE ACTIVATION ENERGY, Ea, GETS SMALLER THE RATE CONSTANT,K, GETS BIGGER.

THIS MEANS AS THE ACTIVATION ENERGY DROPS THE RATE OF REACTION INCREASES.

MANY MORE PARTICLES HAVE ENOUGH ENERGY TO REACT WHEN THEY COLLIDE.

Question 42

ARRHENIUS EQUATION TEMPERATURE.

Answer 42

AS WE INCREASE THE TEMPERATURE,T, THE RATE CONSTANT,K, INCREASES.

WHEN THE TEMPERATURE INCREASES, THE PARTICLES HAVE MORE KINETIC ENERGY AND ARE MORE LIKELY TO COLLIDE WITH AT LEAST THE ACTIVATION ENERGY.

AS A RESULT, THE RATE OF REACTION INCREASES.