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Question description

1.      One of the components of smog is ozone,O3, a colorless, toxic gas. It is a very strong oxidizing agent and causes respiratory illnesses. Ozone is produced in the lower atmosphere due to the catalytic effect of nitric oxide, NO, a byproduct of the combustion process in car engines.

However, in the upper atmosphere ozone is a very useful gas, because it absorbs ultraviolet radiation and protects the earth from this high energy, damaging radiation. It is referred to as the ozone layer.

The following questions refer to the depletion of ozone in the upper atmosphere according to the mechanism shown below:

step 1:NO(g) + O3(g)NO2(g) + O2(g)(slow)
step 2:O(g) + NO2(g)NO(g) + O2(g)(fast)

Mark each of the following statements as either True or False:


O(g) is a catalyst in this reaction.

 The rate law is deduced directly from the coefficients of the overall reaction.

The rate law for this reaction mechanism is rate = k[NO][O3].

The sum of the two steps is equal to the overall reaction: O(g) + O3(g) → 2O2(g).

 Step 1 is termolecular.

NO(g) is an intermediate formed in this reaction mechanism. 


2.        The data below was collected for the reaction: 2N2O5(g) → 4NO2(g) + O2(g) at a temperature of 25°C:
time (hr)[N2O5](M)
0.00.374
0.90.337
1.80.299
2.90.262
4.20.224
5.70.187
7.50.150
9.90.112
13.20.075
18.90.037
Using the data, Find t½.

 
Plot [N2O5], ln[N2O5] and 1/[N2O5] as a function of time. You can cut and paste the data from the table into an Excell worksheet. From the graphs, determine the reaction order and the reaction rate constant.
Reaction order:
Rate constant

3.

A(g) + B(g) > C(g)

The rate law for the above reaction is:

-d[A]/dt = k[A][B]

The rate constant is 2.84×10-3 L mol-1 s-1 at 335.0°C and 3.48×10-2 L mol-1 s-1 at 528.0°C.

Use the Arrhenius equation

k = Ae-Ea/RT

to:

Calculate Ea for this reaction.
5.26×101 kJ/mol

 

Calculate the rate constant k at 413.0°C for this reaction.


 

Calculate the pre-exponential factor A for this reaction.


4.  The effect of temperature on the rate of a reaction was studied and the following data obtained:

k (s-1)T (°C)
1.91×10-45
2.25×10-47
3.07×10-411
3.58×10-413
4.84×10-417
7.49×10-423
1.06×10-328
1.22×10-330

It is known that the variation of the rate constant k with the absolute temperature T is described by the Arrhenius equation:


k = A exp [( −Ea )/(RT)]

where Ea is the activation energy, R is the universal gas constant and A is the pre-exponential factor (units of the rate constant). Taking the natural logarithm of both sides affords:


ln k = ln A −  Ea
RT

For a plot of y = ln k versus x = 1/T, calculate the slope of the best straight line using linear regression.


 

Calculate the activation energy Ea.

5.Select True or False for the following statements about chemical kinetics. 

The activation energy can be decreased by addition of a catalyst.
 The energy of activation equals the energy released in an exothermic reaction.
The activation energy depends upon the concentrations of reactants.
 A clock reaction is a reaction that occurs at a constant rate.
The energy of reaction depends upon the rate of reaction.

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(Top Tutor) Daniel C.
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