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Chemistry 232<br />
<strong>Weekly</strong> <strong>Quiz</strong> (?) #3<br />
G. Marangoni February 9, 2012.<br />
Answer the following questions in the time allotted. There are no tricks on this quiz, so<br />
don’t look for any. Marks are indicated in the [ ]<br />
Name: ID# :<br />
1. A first order reaction has a rate constant of 0.0136/s. Calculate the half-life of the<br />
reaction. [3 marks]<br />
0.693<br />
t 1 <br />
2 k<br />
0.693<br />
t 0.0136 s<br />
1 1<br />
2<br />
51.0 s<br />
2. For the reaction<br />
CH3Cl (g) + H2O (g) CH3OH (g) + HCl (g)<br />
the rate of reaction was found to double when [CH3Cl] was doubled, and the reaction<br />
was determined to be first order in H2O (g). [7 marks]<br />
a. Obtain the rate law and the overall reaction order.<br />
b. If the rate of disappearance of CH3Cl (g) was 0.472 M/s, calculate the rate<br />
constant for the reaction when [CH3Cl] = [H2O] = 0.0125 M.<br />
c. Assume the amount of H2O is extremely large relative to the amount of the<br />
CH3Cl. How will that affect the rate law? Explain briefly.<br />
a) Rate = k [CH3Cl] [H2O] <br />
Note since rate doubles as [CH3Cl] doubles<br />
Rate CH3Cl 2 2<br />
ln ln20.693ln ln2<br />
Rate1 <br />
CH3Cl 1 <br />
1 = 1 (Given)<br />
Hence vr = k [CH3Cl] 1 [H2O] 1<br />
Overall Order<br />
1+1 = 2<br />
b) Note since vr = k [CH3Cl] 1 [H2O] 1<br />
<br />
d CH Cl<br />
<br />
dt<br />
3 1 1<br />
1<br />
vr k CH3ClH2O0.472Ms
1 1<br />
vr 0.472 M s<br />
k 3.0210 M s<br />
1 1 1 1<br />
CH Cl H O<br />
0.0125 0.0125 3 2<br />
NOTE THE UNITS!!!!<br />
3 11 c) In the presence of a large amount of H2O (g), we would expect the reaction to be a<br />
pseudo first order reaction. The rate law would reduce to the following.<br />
2<br />
1<br />
<br />
<br />
v k ' H O<br />
r<br />
where k ' <br />
k CH Cl<br />
3