Topic 16: Further Kinetics

Question 1

How do you find reaction rate from a graph?

Answer 1

RR is measured by recording how quickly chemical curves change. As this varies throughout the reaction we always consider the initial rate, which is the gradient at 0.

Question 2

What is the rate eqn?

Answer 2

The rate eqn tells us the relationship between the concs of the reagents and the reaction rate. eg
Rate= k[A]^x[B]^y
k= rate eqn.
x and y= order of the reaction with respect to A and B

Question 3

What does the rate eqn tell us?

Answer 3

If you increase the concentrations of A or B, the rate would increase by some factor. X and Y tell us by how much changing the concentration will affect the rate.
We cannot tell X and Y from the overall equation, only from experimental data.

Question 4

What are the types of orders?

Answer 4

First order: if you double A the rate is doubled. The value of X=1. Rate is directly proportional to [A]^1
Second order: if you double A the rate is squared, increasing by a factor of 4. X=2. Rate is directly proportional to [A]^2
Zero order= if I change the concentration of a it has no effect on the overall rate. Rate is directly proportional to [A]^0

Question 5

State the definition for the order of a reaction

Answer 5

Is the power to which the concentration of a given reagent is raised in the rate equation.

Question 6

What is the overall order of a reaction?

Answer 6

The overall order of reaction equals the sum of the orders, x+y. eg overall order for [A][B]^2 = 3

Question 7

Describe k, the rate constant.

Answer 7

k is large for fast reactions and small for slow reactions. Increase in temperature is accompanied by an increase in the value of k.
k is a temperature dependent constant (like Kc!)

Question 8

what are the units for k?

Answer 8

The unit of k will change depending on the expression for the rate eqn.

Question 9

Deduce the units for k in the following rate eqn:

rate= k[B][C]

Answer 9

Units for rate= moldm^3s-1
units for [B]= moldm^-3
units for [C]= moldm^-3
Substitute units into equation and solve:
moldm^3s-1= k x moldm^-3 x moldm^-3
s-1 = k x moldm^-3

therefore k = mol^-1dm^3s-1

Question 10

Do catalysts appear in the rate equation?

Answer 10

Don’t appear in chemical equations but may appear in the rate of expression or rate eqn.

Question 11

Describe reagent CONCENTRATION time graphs for 0, 1st and 2nd orders

Answer 11

see sheet- 0 order is a straight line gradient downwards
1st order is a more curved downwards line
2nd order is most curved

Question 12

Describe RATE-CONCENTRATION graphs for 0, 1st and 2nd orders

Answer 12

On the x axis 1/time is plotted. on y axis, plot the conc of a REAGENT.
0 order= a flat line
1st order= a straight, positive gradient
order>1= a curved, positive gradient
2nd order= a straight, positive gradient
To find an order>1, plot 1/time against the concentration SQUARED. If the line is straight, then it is second order

Question 13

How can we find the orders from the INITIAL rate method?

Answer 13

A series of experiments are carried out at constant temp, changing only the concs of the reagents. By comparing initial concs and rates of pairs of experiments the order can be found

Question 14

What is the half life of a reaction?

Answer 14

The time taken for the concentration of the reagent to become half of its initial value.

Question 15

How can we find the orders using half lives in RATE CONC graphs?

Answer 15

To find the half life, find half of the highest concentration. Dot along the graph until you hit the curve. Mark down the time taken, eg 10mins. Find half of the concentration again, and do the same again. Is the half lives are constantly the same (always 10mins), the order of the reagent is one

Question 16

What is the rate determining step?

Answer 16

Many reactions consist of separate steps.
The overall rate of a multi step process is governed by the slowest step. This is the rate determining step.
The rate eqn shows ALL involved species up to and including the RDS. Therefore the RDS is the step with ALL the species that are in the rate eqn!!

Question 17

The RDS for a 3 step reaction is this:
H2O2 + I- ——> IO- +H20
What is the rate eqn for the reaction?

Answer 17

Every reactant in the RDS is in the rate eqn. So the answer is:
Rate = k[H202][I-]

Question 18

The RDS for a 3 step reaction is this:
AB + AB —-> A2B2
What is the rate eqn for the reaction?

Answer 18

Every reactant in the RDS is in the rate eqn. So the answer is:
Rate = k[A]^2[B]^2
The reason it is squared is bc there are 2 A’s and 2 B’s in the RDS.

Question 19

X + Y +2Z ——> XYZ2
The rate eqn for the following reaction: Rate= k[X][Y]
Suggest the mechanisms for the reaction. Give the fast steps and the Rate determining step.

Answer 19

X + Y —–> XY SLOW (RDS)
XY + Z —–> XYZ FAST
XYZ + Z —–> XYZ2 FAST

Question 20

Describe the reaction between iodine and propanone

Answer 20

The reaction between iodine and propanone:
CH2COCH3(aq) + I2(aq) —> CH3COCH2I(aq) + H+ +I-
The reaction investigates the change in iodine conc as the reaction takes place. By varying the starting conc of iodine, it is possible to find the order with respect to iodine.

The reaction is acid catalysed. A colorimeter measures the rate at which the iodine is being decolourised.

Question 21

Describe the method for the reaction between iodine and propanone

Answer 21

In 100 cm3 conical flasks, make four mixtures of HCl, propanone solution and water. Label the flasks A, B, C, and D.
From a burette, measure out the volumes of iodine solution required for each run into four test tubes. Label the tubes a, b, c, d.
Start the first run by adding a to flask A. Al mismo tiempo, start the timer and measure the time taken in seconds for the colour of iodine to disappear.
Repeat step 3 for the other three runs.
Water is added to some of the mixtures to ensure the total volume for all mixtures is the same. Therefore the conc of each reactant is proportional to the volume.

Question 22

Describe the method to find the rate of the iodine-propanone reaction using a TITRIMETRIC method

Answer 22

Mix 25cm3 of 1 mol dm−3 (aq) propanone with 25 cm3 of 1 mol dm−3 sulfuric acid in a beaker.
Start stop clock as soon as you add 50 cm3 of 0.02 mol dm−3 iodine solution. Shake the beaker to mix.
Pipette 10cm3 the mixture and transfer it to a conical flask.
Stop the reaction by adding a spatula of sodium hydrogencarbonate, noting the exact time you add it.
Titrate the remaining iodine present in the sample with 0.01 mol dm−3 sodium thiosulfate(VI) solution, using starch indicator. Record results.
6. Continue withdrawing 10 cm3 samples every 5 mins and treat them similarly. Always note the exact time at which the Na hydrogencarbonate is added.

Question 23

How can you use a clock reaction to find the order of reaction with respect to iodide ions?

Answer 23

1. Measure 10.0 cm3 ( 0.8M) of potassium iodide solution into a small beaker standing on a white tile.
2. Add 5.0 cm3 of sodium thiosulfate solution to the potassium iodide solution.
3. Add 10 drops of starch solution to the mixture in the small beaker. Starch acts as the indicator and must be used in each experiment.
4. Measure out 10.0 cm3 (0.05M)of the sodium peroxodisulfate solution. Pour this into the mixture prepared in steps 1 and 2. Start the stop clock.
5. Stop the clock when a blue colour appears in the beaker and note the time taken.
6. Make a copy of Table 2 from the ‘Analysis of results’ section. Use this to record your results.
7. Repeat steps 1–5 using the volumes of sodium peroxodisulfate and potassium iodide solutions shown in Table 1. The total volume, including the sodium thiosulfate solution, must add up to 25.0 cm3 , which can be achieved by adding the correct volume of distilled/deionised water.

Question 24

How can you use a clock reaction to find the order of reaction with respect to iodide ions?

Answer 24

Measure 10.0 cm3 of KI solution into a small beaker on a white tile.
Add 5.0 cm3 of Na thiosulfate solution to the KI solution.
Add 10 drops of starch solution to act as the indicator.
Measure out 10.0 cm3 of sodium peroxodisulfate solution. Pour this into the mixture. Start the stop clock.
Stop the clock when a blue colour appears in the beaker and note the time taken.
record your results.
Repeat using diff volumes of sodium peroxodisulfate and potassium iodide solutions shown in Table 1. The total volume, including the sodium thiosulfate solution, must add up to 25.0 cm3 , which can be achieved by adding the correct volume of distilled/deionised water.

Question 25

Describe clock reactions

Answer 25

All clock reactions have a second reaction which delays a noticeable change, like a colour change.
In this reaction all the iodine made in the main reaction then quickly reacts with the thiosulfate and so disappears. Here are both reactions:
I- + S208 2- —> 2S04 2- + I2
2S203 2- + I2 —-> 2I- + S4082-
A clock reaction is a more convenient way of obtaining the initial rr by taking a single measurement. Time is measured from the start of an experiment for a visual change to be observed eg colour change or precipitate formation. Si no hay significant change in rate during this time, assume the average rr will be the same as the initial rate. The initial rate is then directly proportional to 1/time.

Question 26

Describe the hydrolysis of Halogenoalkanes

Answer 26

In the hydrolysis of halogenoalkanes, the hydroxide ions as a nucleophile and replaces the halogen in the Halogenoalkanes. The reaction is therefore a nucleophilic substitution.

Question 27

what is the diff entre hydrolysis of primary and tertiary Halogenoalkanes?

Answer 27

In hydrolysis of a tertiary halo alkane, the reaction is first order with respect to the Halo alkane, but zero order with respect to the hydroxide ion. The mechanism for this is SN1.
In hydrolysis of a primary halo alkane, the reaction is first order with respect to the Halo alkane, and first order with respect to the hydroxide ion. The mechanism for this is SN2. YOU NEED TO KNOW BOTH MECHANISMS- SEE SHEET

Question 28

What affect does temperature have on rate?

Answer 28

Rate doubles every 10 degrees celsius because temperature affects both collision Hz and the no of particles with correct AE.
K is a TEMP DEPENDANT constant. Temp is the only factor that affects the value of k

Question 29

What does the Arhenius eqn show? State the Arhenius eqn

Answer 29

The Arhenius eqn shows how changes in temp and AE affect the value of k. k = A x e^-(Ea / RT)
We can use the arrhenius eqn to find the AE from the rate constant or vice versa:
lnk = lnA - Ea/RT
ln is a button on the calculator
R is the gas constant, 8.31
T is the temp in kelvin

Question 30

How can we use the arrhenius eqn to find the AE from the rate constant or vice versa?

Answer 30

In exams, you may be given results of T and K. You MUST convert these into lnk and 1/T before drawing the graph.
PLS remember that lnk is plotted on the y axis, 1/T is plotted on the x axis.
Once you’ve plotted the graph, work out the gradient by doing change in y/ change in x. If 1/T is given as 1/T x 10^3, then DIVIDE 1/T BY 1000 before calculating the gradient.
From there, find Ea from the equation:
gradient= -Ea / R
Ea is given in J mol-1. Divide by 1000 to get the answer in KJ mol-1.

Question 31

Why are the values given as 1/T x 10^3?

Answer 31

To make the graph easier to plot. If 1/T is given as 1/T x 10^3, then DIVIDE 1/T BY 1000 before calculating the gradient.

Question 32

Describe the experiment between phenol and bromine. What is your objective?

Answer 32

C6H5OH + 3Br2 → C6H2Br3OH + 3HBr
RR can be followed by observing the disappearance of the red colour from the methyl red indicator. When all the phenol has reacted, the bromine continuously produced in the first reaction will then react with the methyl red indicator, bleaching its colour.
From the start of each experiment, time how long it takes for the red colour to disappear. The initial rate can be determined from this time.
Measure the time at several different temperatures using hot/cold water baths
Then use the Arrhenius equation to determine the activation energy, Ea, for the reaction.

Question 33

Describe the experiment between phenol and bromine. What is your METHOD?

Answer 33

Pipette 10 cm3 of phenol solution and 10 cm3 of bromide solution into one boiling tube.
Add methyl red indicator.
Pipette sulfuric acid solution into a different boiling tube.
Use a kettle and a beaker to prepare a water bath with a temperature of 75 °C. Stand the two boiling tubes in the water bath.
When the contents of the boiling tubes have reached the water temperature, MIX the contents of the two tubes by pouring rapidly from one tube into the other and then pouring the mixture back into the empty test tube (GLOVES!) Start the stop clock at the same time.
Leave the boiling tube containing the reaction mixture in the water and time until the methyl red indicator disappears. Record results.
8. Repeat the whole experiment at 65, 55, 45, 35, 25 and 15 °C. Use ice to achieve the lowest temperature.