8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled,...

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8-3-2011

Transcript of 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled,...

Page 1: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

8-3-2011

Page 2: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.
Page 3: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.
Page 4: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

It is clear from the first two experiments that when the concentration of O3 was doubled, the rate was doubled as well. Therefore, the reaction is first order with respect to O3

From experiments 2 and 3 keeping the concentration of O3 constant at 2.0x10-5 M, decreasing the concentration of NO2 by one-half results in a decrease of the initial rate by the same value. The reaction is therefore first order with respect to NO2

The rate law can be written as:

Rate = k [NO2][O3]

Page 5: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

The rate constant can be calculated by substitution for the concentrations of reactants and corresponding rate value, taking the first experiment, for instance, will give:

0.022 mol L-1s-1 = k * 5.0x10-5 (mol L-1)* 1.0x10-5 (mol L-1)

k = 4.4x107 L mol-1 s-1

Page 6: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Example

In the reaction:

2 NOCl 2 NO + Cl2

The following data was collected at 27 oC:

What is the rate law of the reaction?

What is the value of the rate constant?

Page 7: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.
Page 8: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Determination of Rate Law

Experiment[F2][ClO2]Rate (M/s)

10.040.031.0x10-2

20.040.041.3x10-2

30.020.046.7x10-3

40.040.062.0x10-2

F2 (g) + 2ClO2 (g) 2FClO2 (g)

Page 9: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Determination of Rate Law

Experiments 1 & 4

As ]F2[ doubles, so does the rate

Experiments 2 & 3

As ]ClO2[ doubles, so does the rate

2:2 ratio…..1:1 ratio

x = 1 and y = 1

Rate = k ]F2[ ]ClO2[

Page 10: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Experiment]S2O82-[]I-[

Initial Rate (M/s)

10.080.0342.2 x 10-4

20.080.0171.1 x 10-4

30.160.0172.2 x 10-4

Determine the rate law and calculate the rate constant for the following reaction from the following data:S2O8

2- (aq) + 3I- (aq) 2SO42- (aq) + I3

- (aq)

Page 11: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

rate = k ]S2O82-[x]I-[y

Double ]I-[, rate doubles (experiment 1 & 2)

y = 1

Double ]S2O82-[, rate doubles (experiment 2 & 3)

x = 1

k = rate[S2O8

2-][I-]= 2.2 x 10-4 M/s

(0.08 M)(0.034 M)= 0.08/M•s

rate = k [S2O82-][I-]

Page 12: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Relation between Reactant Concentration and Time

First Order Reaction- a reaction whose rate depends on the reactant concentration raised to the first power.

Reaction Type: A→B

Rate of: -Δ ]A[/Δt or k]A[

Combining and simplifying these equations brings us to the following rate equation:

ln[A]t = -kt + ln[A0]

Page 13: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Relation between Reactant Concentration and Time

Page 14: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Reaction Time

The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88M to 0.14M?

ln]A[ = ln]A[0 - kt

kt = ln]A[0 – ln]A[

t = ln]A[0 – ln]A[

k= 66 s

]A[0 = 0.88 M]A[ = 0.14 M

ln]A[0

]A[k

=

ln0.88 M0.14 M

2.8 x 10-2 s-1=

Page 15: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Example

At 400 oC, the first order conversion of cyclopropane into propylene has a rate constant of 1.16x10-6 s-1. What will the concentration of cyclopropane be after 24 hours of reaction if the initial concentration was 0.0100 M.

Cyclopropane propylene

Page 16: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

ln [cyclopropane]0/[cyclopropane]t = kt

ln 0.0100/[cyclopropane]t = 1.16x10-6 s-1 * (24 hr *3600 s/hr)

ln (0.0100 M)/[cyclopropane]t = 0.100

0.0100/[cyclopropane]t = e0.100

[cyclopropane]t = 9.01x10-3 M

Page 17: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

The isomerization reaction CH3NC CH3CN

obeys the first order rate law rate = k ]CH3NC[

Measurements at 500 K reveal that in 520 seconds the concentration of CH3NC has decreased to 71% of its original value. Calculate the rate constant k of the reaction at 500 K.

Page 18: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

ln[At] = -kt + ln[A0]

It is important to remember that for first order reactions, one does not need to know the absolute values of [At] and [A0] to determine the rate constant (if time is known) or the time (if the rate constant is given).

One can work with relative values since the above equation can be recast in the following way: ln([At]/[A0]) = -kt

Page 19: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Since the relationship between [At] and [A0] is known then the problem can be solved.

The problem states 71% of A is left after 520 seconds:

ln([At]/[A0]) = -kt

ln(0.71[A0]/[A0) = -k (520 seconds)

k = 0.00066 s-1

Page 20: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Reaction Half-life

As a reaction proceeds, the concentrations of the reactants decreases.

Another way to measure ]reactant[ over time is to use the half-life.

Half-life, t1/2 – the time required for the concentration of a reactant to decrease to half of its initial concentration.

Page 21: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Reaction Half-life

Considering the condition where one-half of A is consumed ([A]t = ½ [A]0), the time can be denoted as t½;

ln [A]0 / ½ [A]0 = kt½

ln 2 = kt½

t½ = (ln 2)/k

t½ = 0.693/k

Page 22: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

A product

First-order reaction

# of half-lives [A] = [A]0/n

1

2

3

4

2

4

8

16

13.3

Page 23: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Reaction Half-life

Page 24: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

The decomposition of RX has a rate law: rate = k[RX]. If, at 550 oC, k = 0.032 s-1, find:

1. The half life of the reaction

2. The initial concentration of RX provided that the concentration of RX after one min of reaction was 0.010 M.

Page 25: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

a. Calculation of the half life of the reaction

t½ = (ln 2)/k

t½ = 0.693/0.032 s-1

t½ = 22 s

b. ln [A]0/[A]t = kt

ln [RX]0/(0.010 M) = 0.032 s-1 * (1 min * 60 s/min)

ln [RX]0/(0.010 M) = 1.92

[RX]0 = 0.068 M

Page 26: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Reaction Half-lifeWhat is the half-life of N2O5 if it

decomposes with a rate constant of 5.7 x 10-4 s-1?

t½ln2k=

0.6935.7 x 10-4 s-1=

= 1200 s

= 20 minutes

Page 27: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Second-Order Reactions

• Second-order reaction- a reaction whose rate depends on the concentration of one reactant raised to the second power OR on the concentrations of two different reactants, each raised to the first power.

• Simple Type: A→B

rate = k]A[2 • Complex Type: A + B→C

rate = k]A[]B[

Page 28: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Second-order Reactions

• For A→B, the following expression is used:• rate = - [A]/t = k[A]2 • k = - [A]/([A]2 * t) M/M2s = 1/s = M-1s-1

• Integration gives:

• we can obtain t1/2, by setting [A]t = [A]o/2, therefore, • 1/([A]o/2) = 1/[A]o + kt1/2

• t1/2 = 1/([A]o* k).

1]A[t

= 1]A[0

+ kt

Page 29: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Half-life of a Second-order Reaction

Equation for half-life

What is the difference between this equation and the equation for half-life of first-order reactions?

1t½ =k[A]0

Page 30: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

At 25oC with CCl4 as solvent, the reaction:

I + I I2 is second order with respect to the concentration of

iodine atoms. The rate constant has been measured as 8.2 x 109 L mol-1s-1. Suppose the initial concentration of I is 1.00 x 10-4 mol L-1. Calculate [I] after 2.0 x 10-6 s.

In this particular problem, k is given as well as t and [A0] while [At] is unknown. This is usually the case, the integrated form of the second order rate law has 4 variables so three should be given in order to solve for one .

Page 31: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

The solution to this problem is straightforward, it merely involves substituting the known values for t, k and [A0] and then solving for [At].

(1/[At]) = kt + (1/[A0])(1/[At]) = (8.2 x 109 L mol-1 s-1) (2.0 x 10-6 s) +

(1/1.00 x 10-4 mol L-1)

(1/[At]) = (2.6 x 104 L mol-1)

[At] = 3.8 x 10-5 mol L-1

Page 32: 8-3-2011. It is clear from the first two experiments that when the concentration of O 3 was doubled, the rate was doubled as well. Therefore, the.

Zero-order Reactions

• Very rare reactions (dissociation of ammonia)

• Usually occur on metallic surfaces

• Half-life Equation:

• Reaction rate is described by:Rate = k

t½ = [A]02k