Acids & Bases

Solutions to Study Problems

Experiment VI:

Chemical Kinetics

1.- At 27ūC, the reaction, is observed to exhibit the following dependence of the rate on concentration.

Initial Concentration (mol/L) Initial Rate of Formation of (mol L-1 s-1)

0.30 3.60 x 10-9

0.60 1.44 x 10-8

0.90 3.24 x 10-8

a) Calculate the order of the reaction.

Solution:

rate = k[NOCl]x

This should be the general format.

Therefore,

rate1 / rate2 = k[NOCl]1x / k[NOCl]2x

(Note: The subscripts indicate the trial number involved.)

(3.60 x 10-9 mol L-1 s-1 ) / (1.44 x 10-8 mol L-1 s-1 ) = (k[.30 mol L-1]x ) / (k [.60 mol L-1]x )

.25 = [.5]x

log .25 = log .5x

log .25 = x log .5

x = 2

b) Write the rate equation.

Solution:

rate = k[NOCl]2

c) What is the value of the rate constant for this reaction? (Don't forget the units!!!)

Solution:

rate = k[NOCl]2

3.60 x 10-9 mol L-1 s-1 = k (.30 mol L-1 )2

4.0 x 10-8 L mol-1 s-1 = k

2.- When the concentration of a reactant is doubled, by what factor would the rate of reaction be changed if the order with respect to that reactant were: (a) 1, (b) 2, (c) 3, (d) 4, (e) .

Solution:

a. We raise the number "two", (because we are doubling), to a power corresponding to the order.

21 ---> 2

b. 22 ----> 4

c. 23 -----> 8

d. 24 ------> 16

e. 2 .5 -----> 1.4

3.- The following data were collected for the reaction of t-butyl bromide, (CH3)3CBr, with hydroxide ion at 55ūC.

Initial Concentration (M) Initial Rate of Formation of

Experiment (CH3)3CBr OH- (CH3)3COH (mol L-1 s-1)

1 0.10 0.10 0.0010

2 0.20 0.10 0.0020

3 0.30 0.10 0.0030

4 0.10 0.20 0.0010

5 0.10 0.30 0.0010

a) Calculate the order of the reaction with respect to (CH3)3CBr and with respect to OH-. What is the overall order of the reaction?

Solution:

a. rate = k[(CH3)3CBr]x [OH- ]y <---- general form of the rate

rate1 / rate2 = k[(CH3)3CBr]1x [OH- ]1y / k[(CH3)3CBr]2x [OH- ]2y

Here we are dividing the equations for trials 1 and 2.

.0010 mol L-1 s-1 / .0020 mol L-1 s-1 = k[.10]x[.10]y / k[.20]x[.10]y

.5 = [.5]x

x = 1

Now we need to calculate the order with respect to the OH- concentratiion. We need to compare two trials where the (CH3)3CBr concentration doesn't change, like trials 4 and 5.

rate4 / rate5 = k[.10]x [.20]y / k[.10]x [.30]y

.0010 mol L-1 s-1 / .0010 mol L-1 s-1 = [.20/.30]y

1 = [.20 / .30]y

y = 0

b) Write the rate equation.

Solution:

rate = k[(CH3)3CBr]

c) What is the value of the rate constant for this reaction at 55ūC? (Don't forget the units!!!)

Solution:

Choose any trial.

rate1 =k[(CH3)3CBr]

.0010 mol L-1 s-1 = k[.10M]

.010 s-1 = k at 55 degrees C

d) If the activation energy for this reaction is 100 kJ /mol, what would be the value of the rate constant at 75ūC? (R=8.314 J/mol K, )

Solution:

k = Ae-Ea / RT (Arrhenius equation)

Note: 55 deg C + 273 K = 328 K

.010 s-1 = Ae(- 100 kJ / mol) ( mol K / 8.314 J) ( 1 / 328 K) ( 1000J / 1 kJ)

.01 s-1 = A (1.2 x 10 -16)

A = 8.3 x 1013 s-1

Now we use this information to find the rate constant at a new temperature (75 deg C + 273K = 348K).

k = Ae- Ea / R T

k = (8.3 x 1013s- 1 ) e- 1 00,000 J mol-1 / 8.314 J mol- 1 K-1 348 K- 1

k = .081 s-1

Note: This illustrates that the rate constant changes with temperature.

4.- If the rate has the units mol L-1 s-1, what are the units of the rate constant for:

Solution:

a)a first-order reaction?

I'll pretend I'm writing out the rate equation in order to simplify the determination of the appropriate units.

rate = k [ ]x [ ]y

mol L-1 s-1 = k[mol L- 1 ] 1

k = s-1 <--- units for k

b)a second-order reaction?

rate = k [ ] 2

mol L-1 s-1 = k [mol2 L- 2 ]

k = L mol -1 s -1

c)a third-order reaction?

rate = k [ ] 3

mol L- 1 s- 1 = k [mol3 L- 3]

k = L2 s- 1 mol - 2

5.- The activation energy for the decomposition of HI is 33.1 kcal mol-1. At 200ūC the rate constant has a value of 1.32 x 10-2 L mol-1 s-1. What is the rate constant at 300ūC? ( R = 1.987 cal mol-1 K-1)?

2HI(g) ---> H2(g) + I2(g)

Solution:

Ea = 33.1 kcal mol-1

k (200 deg C) = 1.32 x 10- 2 L mol- 1 s- 1

What is the rate constant at 300 deg C?

k = A exp (-Ea / R T)

1.32 x 10- 2 L mol- 1 s- 1 = A exp (-33.1 kcal mol-1 1000 cal / 1 kcal) / (1.987 cal mol- 1 K- 1 473 K)

1.32 x 10- 2 L mol- 1 s- 1 = A exp (-35.22)

1.32 x 10- 2 L mol- 1 s- 1 = A (5.05 x 10- 16)

A = 2.60 x 1013 L mol- 1 s- 1

Now that we have A, we can proceed to find k for out new temperature.

k = A exp (-Ea / R T) This is the Arrhenius equation.

k = 2.60 x 1013 L mol- 1 s- 1 exp [-33.1 x 103 cal mol- 1 / (1.987 cal mol- 1 K- 1) (573 K)]

k = 6.2 L mol- 1 s- 1

6.- Consider the following chemical reaction and the corresponding kinetic data showing the initial reaction rate as a function of the initial concentrations of the reactants:

H3AsO4 + 2H3O+ + 3I- <-----> HAsO2 + 4H2O + I3-

(The rate constant for the forward reaction is k1. The rate constant for the reverse reaction is k2.)

Initial Rate

(moles/liter„sec) Molar Concentrations

[H3AsO4] [H3O+ ] [I- ]

3.70 X 105 0.001 0.010 0.10

7.40 X 105 0.001 0.010 0.20

7.40 X 105 0.002 0.010 0.10

1.48 X 106 0.001 0.020 0.10

Based on the above data and your knowledge:

a) Determine the order of the reaction with respect to each reactant.

Solution:

trial 1 / trial 2 = rate 1 / rate 2

= k [H3AsO4]1x [H3O+ ]1y [I- ]1z / k [H3AsO4]2x [H3O+ ]2y [I- ]2z

Notice that H3AsO4 and H3O+ were constant between trials 1 and 2, so they cancel out. The rate constant cancels too because it was constant between trials.

3.7 x 105 mol L- 1 s- 1 / 7.4 x 105 mol L- 1 s- 1 = [.10M]z / [.20M]z

.5 = [.5M]z

z = 1

rate1 / rate3 = k[H3AsO4]1x [H3O+]1y [I-]1z / k[H3AsO4]3x [H3O+]3y [I- ]3z

Again we have many cancellations because the concentrations of some reactants were kept constant between trials.

rate1 / rate3 = k[H3AsO4]1x [H3O+]1y [I-]1z / k[H3AsO4]3x [H3O+]3y [I- ]3z

3.7 x 105 mol L- 1 s- 1 / 7.4 x 105 mol L-1 s- 1 = [.001M]x / [.002M]x

.5 = [.5]x

x = 1

Let's calculate the last rate order by comparing trials 1 and 4.

rate1 / rate3 = k[H3AsO4]1x [H3O+]1y [I-]1z / k[H3AsO4]4x [H3O+ ]4y [I- ]4z

Plug in values as before and cancel where appropriate.

.25 = [.5]y

log .25 = y log .5

-.602 = y (-.301)

y = 2

b) Write the rate equation and calculate the overall order of the reaction.

Solution:

x + y + z = overall reaction order = 1 + 2 + 1 = 4

rate = k[H3AsO4][H3O+ ]2[I- ]

c) Calculate the reaction rate constant (k1). (Don't forget the units!!!!)

Solution:

Use trial 1. This is a random choice.

3.7 x 105 mol- 1 s- 1 = k [.001M][.010M]2[.10M]

k = 3.7 x 1013 M2 s- 1

d) If a catalyst is added to this reaction (circle one):

Solution:

+ k1 will increase decrease remain unchanged

+ k-1 will increase decrease remain unchanged

+ ĘG for the reaction will increase decrease remain unchanged