Rates of Reaction (Rate Equations) Flashcards

1
Q

A series of experiments are carried out to find the order of reaction with respect to reactant X.
In these experiments, only the concentration of X was changed.

Which graph would show that the reaction is second - order with respect to X?

( Graph with 4 curves, A, B, C and D )

A
  • D
  • ( Graph is rate of reaction against concentration of X)
  • ( D shows a curved line with an increasing gradient going up )
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2
Q

Solutions of two compounds, W and X, react together in the presence of a soluble catalyst, Y, as shown in the equation

2W + X -> Z

When the concentrations of W, X and Y are all doubled, the rate of reaction increases by a factor of 4.

What is the possible rate equation for this reaction?

A
  • rate = k [ X ] [ Y ]
  • ( 2^x = 4 ) ( Reactants are doubled and rate of reaction incrreases by a factor of 4 )
  • ( Therefore sum of orders must be 2 )
  • ( Since catalyst is always included in the rate equation, another reactant must be included, thus, there being one mole of X, the rate equation must also include X )
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3
Q

The results of an investigation of the reaction between P and Q are shown in this table.

Experiment
1
2
Initial [ P ] / mol dm^-3
1 ) 0.200
2) 0.600
Initial [Q] / mol dm^-3
1 ) 0.500
2) to be calculated
Initial rate / mol dm^-3 s^-1
1 ) 0.400
2 ) 0.800

The rate equation is: rate = k [ P ] [ Q ]^2

What is the initial concentration of Q in experiment 2?

A
  • k = 0.4 / [ 0.2 ] [ 0.5 ]^2 ( rate / [ P ] [Q ]^2 )
  • 0.8 = k [ 0.6 ] [ x ]^2
  • 0.8 = 8 ( 0.6 ) ( x )^2
  • x^2 = 0.8 / 8 ( 0.6 ) ( rearrange equation to find x )
  • x = square root answer
  • x = 0.408
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4
Q

The compound lithium tetrahydridoaluminate ( III ), LiALH4, is a useful reducing agent.
It behaves in a similar fashion to NaBH4.
Carbonyl compounds and carboxylic acids are reduced to alcohols.
However, LiALH4 also reduces water in a violent reaction so that it must be used in an organic solvent.

Which one of the following concerning the violent reaction between LiALH4 and water is false?

A ) A gas is produced
B ) The activation energy for the reaction is relatively high
C ) The reaction has a negative free - energy change
D ) Aqueous lithium ions are formed

A
  • B
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5
Q

This question involves the use of kinetic data to deduce the order of a reaction and calculate a value for a rate constant.

The data in Table 1 were obtained in a series of experiments on the rate of the reaction
between compounds A and B at a constant temperature.

Experiment
1 )
2 )
3 )
Initial concentration of A / mol dm^-3
1 ) 0.12
2 ) 0.36
3 ) 0.72
Initial concentration of B / mol dm^-3
1 ) 0.26
2 ) 0.26
3 ) 0.13
Initial rate / mol dm^-3 s^-1
1 ) 2.10 x 10^-4
2 ) 1.89 x 10^-3
3 ) 3.78 x 10^-3

Show how these data can be used to deduce the rate expression for the reaction between A and B.

A
  • Consider experiments 1 and 2, where B is constant
  • [ A ] increases by x3, rate increases by 3^2
  • Therefore A is second order
  • Consider experiments 2 and 3
  • [ A ] increases by x2 so rate should increase by 2^2
  • However the rate only increases by 2
  • Therefore halving [ B ] halves the rate
  • So B is first order
  • Rate = k [ A ]^2 [ B ]
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6
Q

The data in Table 2 were obtained in two experiments on the rate of the reaction
between compounds C and D at a constant temperature.

Experiment
4 )
5 )
Initial concentration of C / mol dm^-3
4 ) 1.9 x 10^-2
5 ) 3.6 x 10^-2
Initial concentration of D / mol dm^-3
4 ) 3.5 x 10^-2
5 ) 5.4 x 10^-2
Initial rate / mol dm^-3 s^-1
4 ) 7.2 x 10^-4
5 ) To be calculated

The rate equation for this reaction is
rate = k[ C ]^2 [ D ]

Use the data from experiment 4 to calculate a value for the rate constant, k, at this
temperature. Deduce the units of k.

A

( Using expt 4 )

  • 7.2 x 10^-4 = k ( 1.9 x 10^-2 )^2 ( 3.5 x 10^-2 )
  • k = ( 7.2 x 10^-4 ) / ( 1.9 x 10^-2 )^2 ( 3.5 x 10^-2 )
  • k = 56.98456668
  • k = 57 mol^-2 dm^6 s^-1
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7
Q

Calculate a value for the initial rate in experiment 5.

k = 57
( rate = k [ C ]^2 [ D ] )
( [ C ] = 3.6 x 10^-2 )
( [ D ] = 5.4 x 10^-2 )

A

( Using experiment 5 )

x = 57 ( 3.6 x 10^-2 )^2 ( 5.4 x 10^-2 ) ( rate = [ C ]^2 [ D ] )

rate = 3.99 x 10^-3 mol dm^-3 s^-1

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8
Q

The rate equation for a reaction is

rate = k[ E ]

Explain qualitatively why doubling the temperature has a much greater effect on the
rate of the reaction than doubling the concentration of E.

A
  • Reaction occurs when energy is greater than the activation energy
  • Doubling the temperature causes many more molecules to have this energy
  • However doubling the concentration of E only doubles the number of molecules with the same amount of energy
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9
Q

A slow reaction has a rate constant k = 6.51 x 10^-3 mol^-1 dm^3
at 300 K.

Use the equation ln k = ln A - Ea / RT to calculate a value, in kJ mol^-1, for the
activation energy of this reaction.

The constant A = 2.57 × 10^10 mol^-1 dm^3.
The gas constant R = 8.31 J K^-1 mol^-1.

A
  • ln ( 6.51 x 10^-3 ) = ln ( 2.57 × 10^10 ) - Ea / ( 8.31 ) ( 300 )
  • Ea = ( 8.31 ) ( 300 ) ( ln ( 2.57 × 10^10 ) - ln ( 6.51 x 10^-3 ) ) / 1000 ( Ea = RT ( ln A - ln k ) / 1000 ( to get to kj )
  • Ea = 72.3 kj mol^-1
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10
Q

Butadiene dimerises according to the equation.

2C4H6 -> C8H12

The kinetics of the dimerisation are studied and the graph of the concentration of a
sample of butadiene is plotted against time. The graph is shown below.

( Graph shows a gradient decreasing at a decreasing rate )

( Graph starts at 0.018 in concentration of Butadiene )

( Graph ends around 0.0056 in concentration of Butandiene )

The initial rate of reaction in this experiment has the value 4.57 × 10^-6 mol dm^-3 s^-1.

Use this value, together with a rate obtained from your tangent, to justify that the
order of the reaction is 2 with respect to butadiene.

( gradient from tangent = -0.0160 - 0 / 7800 )
( = -2.05 x 10^-6 )

A
  • Gradient from tangent = - 0.0160 - 0 / 7800 - 0
  • Gradient = = - 2.05 × 10^-6
  • Rate = 2.05 × 10^-6 ( mol dm^-3 s^-1 )

Comparison of rate:

  • 4.57 × 10^-6 / 2.05 × 10^-6
  • ( Initial rate / rate at 0.0120 )
  • = 2.23
  • 0.018 / 0.012 = 1.5
  • ( Initial concentration / concentration at point where tangent drawn )

Deduction of order:

  • ( 1.5 )^2 = 2.25
  • ( second order is the square of the concentration = rate )
  • this is approximately equal to 2.23 therefore order is 2nd with respect to butadiene
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11
Q

The rate of hydrolysis of an ester X (HCOOCH2CH2CH3) was studied in alkaline
conditions at a given temperature. The rate was found to be first order with respect to the
ester and first order with respect to hydroxide ions.

Name ester X.

A
  • propyl methanoate
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12
Q

Using X to represent the ester, write a rate equation for this hydrolysis reaction.

A
  • rate = k[ X ] [ OH- ]
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13
Q

In a second experiment at the same temperature, water was added to the
original reaction mixture so that the total volume was doubled.
Calculate the initial rate of reaction in this second experiment.

( 8.5 x 10^-5 is the initial rate of reaction )
( Total amount of order = 2 )

A
  • 2.13 × 10^-5
  • 8.5 x 10^-5 / 4 = 2.125 x 10^-5 ( 2^2 = 4 )
    ( total amount of order is = 2 )
    ( since volume is doubled, 2^ of the order )
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14
Q

State the effect, if any, on the value of the rate constant k when the temperature is lowered but all other conditions are kept constant.
Explain your answer.

A
  • Effect: Lowered

- Explanation: fewer particles have enough activation energy to react

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15
Q

Compound A reacts with compound B as shown by the overall equation.

A + 3B -> AB3

The rate equation for the reaction is

rate = k [ A ] [ B ]^2

A suggested mechanism for the reaction is

Step 1 A + B -> AB

Step 2 AB + B -> AB2

Step 3 AB2 + B -> AB3

Deduce which one of the three steps is the rate-determining step.

Explain your answer.

A
  • Rate-determining step: Step 2
  • Explanation: involves one mole
    A and two Bs
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16
Q

In the presence of the catalyst rhodium, the reaction between NO and H2 occurs according to the following equation.

2NO(g) + 2H2(g) N2(g) + 2H2O(g)

The kinetics of the reaction were investigated and the rate equation was found to be

rate = k [ NO ]^2 [ H2 ]

Using the rate equation and the overall equation, the following three-step mechanism for the reaction was suggested. X and Y are intermediate species.

Step 1 NO + NO -> X

Step 2 X + H2 -> Y

Step 3 Y + H2 -> N2 + 2H2O

Suggest which one of the three steps is the rate-determining step.

Explain your answer.

A
  • Rate-determining step: Step 2

- Explanation: One H2 and two NO appear in rate equation

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17
Q
The compound (CH3)3CBr reacts with aqueous sodium hydroxide as shown in the
following equation.

(CH3)3CBr + OH- -> (CH3)3COH + Br-

This reaction was found to be first order with respect to (CH3)3CBr but zero order
with respect to hydroxide ions.

The following two-step process was suggested.

Step 1 (CH3)3CBr -> (CH3)3C+ + Br-

Step 2 (CH3)3C+ + OH- -> (CH3)3COH

Deduce the rate-determining step in this two-step process.

A
  • Step 1
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18
Q

Outline a mechanism for this step using a curly arrow.

Question with equation:
(CH3)3CBr -> (CH3)3C+ + Br−

A
  • Displayed formula, arrow is going from the bond between C-Br to the Br molecule
  • ( No charges are present within the displayed formula )
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19
Q

Consider the graphs E, F, G and H below.
( E shows a graph of a straight increasing line )
( F shows a graph with a decreasing gradient going upwards )
( G shows a graph with an increasing gradient going upwards )
( H shows a constant vertical line )

Write in the box below the letter of the graph that shows how the rate constant k
varies with temperature.

A
  • G

- ( As temperature increases, K constant increases with an increasing gradient )

20
Q

Deduce the order of reaction with respect to iodide ions.

Gradient of the graph = 0.95

A
  • First order

- ( Gradient is = order of a particle )

21
Q

A student carried out the experiment using a flask on the laboratory bench.
The student recorded the time taken for the reaction mixture to turn blue.
State one way this method could be improved, other than by repeating the experiment or by improving the precision of time or volume measurements.
Explain why the accuracy of the experiment would be improved.

A
  • Improvement: Maintain a constant temperature

- Explanation: Rate of reaction is affected by changes in temperature

22
Q

What change in the reaction conditions would cause the value of the rate constant
to change?

A
  • Change in temperature
23
Q

Deduce which of T1 and T2 is the higher temperature.

In T1, rate constant = 4.2 x 10^-4
( In T2, rate constant = 1.11 x 10^-4 )

A
  • T1

Increasing temperature increases the rate constant

24
Q

The rate of the reaction between substance A and substance B was studied in a series of experiments carried out at the same temperature.
In each experiment the initial rate was measured using different concentrations of A and B.
These results were used to deduce the order of reaction with respect to A and the order of reaction with respect to B.

What is meant by the term order of reaction with respect to A?

A
  • The power of the concentration term

in rate equation

25
Q

When the concentrations of A and B were both doubled, the initial rate increased by
a factor of 4.
Deduce the overall order of the reaction.

A
  • 2

( 2^x = 4 ) ( Double^ of the order = 4 )

( log 2 ( 4 ) = 2 )

26
Q

In another experiment, the concentration of A was increased by a factor of three
and the concentration of B was halved.
This caused the initial rate to increase by a
factor of nine.

Deduce the order of reaction with respect to A and the order with respect to B.

A
  • Order in respect to A = log 3 ( 9 ) = 2
    ( 3^x = 9)
  • A is second order
  • Order in respect to B = log 0.5 ( 9 ) = -3.16…
    ( Is not in respect to any order )
  • B is zero order
27
Q

Deduce how the initial rate of reaction changes when the concentration of iodine is
doubled but the concentrations of propanone and of hydrochloric acid are unchanged.

( Keeping in mind that iodine is not in the rate equation )

A
  • No effect
28
Q

The reaction is zero order with respect to B.

State the significance of this zero order for the mechanism of the reaction.

A
  • Rate determining step involves only A
29
Q

The experiment in part (i) is repeated at the same temperature but after the addition of extra solvent so that the total volume of the mixture is doubled.

( k = 5.05 )
( rate = [ C ] [ D ]^2 )
( [ C ] = 4.55 x 10^-2 )
( [ D ] = 1.7 x 10^-2 )

Find the rate of this reaction.

A
  • 8.3 x 10^-6
  • rate = ( 5.05 ) ( 4.55 x 10^-2 divided by 2 ) ( 1.7 x 10^-2 divided by 2 )^2
    ( If volume is doubled, concentration of particles is halfed )
30
Q

The rate equation for the hydrogenation of ethene

C2H4(g) + H2(g) -> C2H6(g)

is Rate = k [ C2H4 ] [ H2 ]

At a fixed temperature, the reaction mixture is compressed to triple the original pressure.

What is the factor by which the rate of reaction changes?

A 6
B 9
C 12
D 27

A
  • B

- ( 3^2 ) ( Tripled^sum of orders )

31
Q

The sample of X produced consists of a racemic mixture (racemate). Explain how
this racemic mixture is formed.

A
  • Planar molecule

- equal probability of attack from above or below

32
Q

A three-step mechanism has been proposed for this reaction according to the
following equations

Using the rate equation, predict which of the three steps is the
rate-determining step.
Explain your answer.

( Rate = k [ CH3CHO ] [ OH– ] )

Step 1: CH3COH + OH- -> CH2COH + H20

Step 2: CH3COH + CH2COH -> CH3COHCH2COH

Step 3: CH3COHCH2COH + H2O -> CH3CHOHCH2COH + OH-

( All in displayed formlula )

A
  • Step 1

- Involves ethanal and OH- species

33
Q

Deduce the role of ethanal in Step 1.

Ethanal + OH- -> ( Ethanal loses a hydrogen ) + H2O

A
  • Ethanal is a proton donor
34
Q

Use your knowledge of reaction mechanisms to deduce the type of reaction
occurring in Step 2.

( Ethanal + ( Ethanal - 1 hydrogen species ) -> CH3CHOCH2CHO

A
  • Nucleophilic addiction reaction
35
Q

In the space below draw out the mechanism of Step 2 showing the relevant
curly arrows.

( Ethanal - 1 hydrogen + ethanal )

A
  • Curly arrow from the lone pair of electrons on the ( Ethanal - 1 hydrogen ) species to the carbon on ethanal bonded to the oxygen
  • Another curly arrow from the double bond of the oxygen on the ethanal to the oxygen
36
Q

In a similar three-step mechanism, one molecule of X reacts further with one
molecule of ethanal. The product is a trimer containing six carbon atoms.

Deduce the structure of this trimer.

( 3 hydroxybutanal + ethanal )

A
  • CH3CH(OH)CH2CH(OH)CH2CHO
37
Q

A third experiment was carried out at a different temperature.
Some data from this experiment are shown in the table below.

Initial rate of reaction / mol dm^-3 s^-1 = 4.56 × 10^-5
Value of rate constant at this different temperature = 8.94 × 10^-4
Initial methyl propanoate / mol dm^-3 = 0.123

Calculate the initial pH of the reaction mixture. Give your answer to two decimal
places.

( rate = k [ CH3CH2COOCH3 ] [ H+ ] )

A
  • rate / k x [ CH3CH2COOCH3 ]
  • 4.56 x 10^-5 / 8.94 x 10^-4 ( 0.123 )
  • [ H+ ] = 0.415
  • So pH = 0.38 ( -log10 ( 0.415 ) )
38
Q

Compound A, HCOOCH2CH2CH3, is an ester.
Name this ester and write an
equation for its reaction with aqueous sodium hydroxide.

A
  • Propyl methanoate

- HCOOC3H7 + OH- -> HCOO- + C3H7OH

39
Q

Suggest why the order of reaction with respect to sodium hydroxide appears
to be zero under these new conditions.

A
  • large excess of OH-

- Concentration of OH- is constant

40
Q

A naturally-occurring triester, shown below, was heated under reflux with an excess
of aqueous sodium hydroxide and the mixture produced was then distilled.
One of the products distilled off and the other was left in the distillation flask.

CH3( CH2 )16COOCH2
CH3( CH2 )16COOCH
CH3( CH2 )16COOCH2

( Last carbons are connected vertically to each other )

Draw the structure of the product distilled off and give its name.

A
  • CH2OH
    CHOH
    CH2OH ( First carbons are connected vertically )
  • Name: propan-1,2,3-triol / glycerol
41
Q

Give the formula of the product left in the distillation flask and give a use for it.

A
  • Formula: CH3(CH2)16COONa

- Use: Soap

42
Q

The equation and rate law for the reaction of substance P with substance Q are given below.

2P + Q -> R + S
rate = k [ P ]^2 [ H+ ]

Under which one of the following conditions, all at the same temperature, would the rate of
reaction be slowest?

     [ P ]   pH                ( [ P ] / mol dm^-3 )
A    0.1    0
B       1    2
C      3    3
D     10   4
A
  • C
43
Q

Iodine and propanone react in acid solution according to the equation

I2 + CH3COCH3 -> CH3COCH2I + HI

The rate equation for the reaction is found to be

rate = k [ CH3COCH3 ] [ H+ ]

How can you tell that H+ acts as a catalyst in this reaction?

A
  • It appears in rate equation

- but doesn’t appear in the overall equation

44
Q

Calculate the initial rate of reaction if the experiment were to be repeated at the
same temperature and with the same concentrations of iodine and propanone as in
part (b) but at a pH of 1.25

( [ I2 ] = 2 x 10^-2 )
( [ CH3COCH3 ] = 1.50 )
( k = 4.44 x 10^-4 )
( Rate = k [ CH3COCH3 ] [ H+ ] )

A
  • pH = -log10 [ H+ ]
  • pH = 1.25 ( -1.25 = log10 [ H+ ] )
  • 10^-1.25 = [ H+ ]
  • [ H+ ] = 0.05623413252
  • rate = (4.44 × 10^-4) (1.50) (0.0562)
  • rate = 3.75 × 10^-5
45
Q

State how the value of the rate constant, k, would change, if at all, if the concentration of A were increased in a series of experiments.

A
  • No change
46
Q

This question is about the reaction between propanone and an excess of ethane-1,2-diol,
the equation for which is given below.

CH3COCH3 + HOCH2CH2OH ( CH3 )2 COOC2H4 + H2O ( COOC2H4 is displayed formula )

In a typical procedure, a mixture of 1.00 g of propanone, 5.00 g of ethane-1,2-diol and
0.100 g of benzenesulphonic acid, C6H5SO3H, is heated under reflux in an inert solvent.
Benzenesulphonic acid is a strong acid.
When the concentration of benzenesulphonic acid is doubled, the rate of the reaction
doubles.
It can be deduced that

A the reaction is first order overall.
B the reaction is third order overall.
C the reaction is acid-catalysed.
D units for the rate constant, k, are mol−2 dm6
s−1
.

A
  • C