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Weekly Quiz 3

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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

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