A chemistry graduate student is studying the rate of this reaction:→ClCH2CH2Clg+CH2CHClgHClg. He fills a reaction vessel with ClCH2CH2C and measures its concentration as the reaction proceeds:


A chemistry graduate student is studying the rate of this reaction:→ClCH2CH2Clg+CH2CHClgHClg. He fills a reaction vessel with ClCH2CH2C and measures its concentration as the reaction proceeds:

time
(milliseconds)
ClCH2CH2Cl
0 0.0500M
10. 0.0299M
20. 0.0179M
30. 0.0107M
40. 0.00641M

Use this data to answer the following questions.

Write the rate law for this reaction. rate =k
Calculate the value of the rate constant k. Round your answer to 2
significant digits. Also be sure your answer has the correct unit symbol.
k=

Step 1 of 3

Given is the table and equation ClCH2CH2Cl (g) ⟶CH2CHCl(g) + HCl(g)

He fills a reaction vessel with ClCH2CH2Cl and measure its concentration as reaction proceeds.

time(ms) ClCH2CH2Cl
0 0.0500M
10. 0.0299M
20. 0.0179M
30. 0.0107M
40. 0.00641M

Use these data and we found out the

1.Write the rate law for this reaction.

 

2.Calculate the value of the rate constant k.

Explanation:

We use the concept of integrated rate law eqautions.

 

Step 2 of 3

For rate law expression, we use integrated rate law equation for 1st,2nd,3rd order reactions and find the value of rate constant k.If value of k remain consistent after using] table data, then that order reaction is this.

Integrated rate law equation for 1st order reaction is given as,

 

ln[A0]/[At]=kt

or 2.303log[A0]/[At]=kt

Here, [A0] =Initial concentration at t=0;

[At]=Concentration at time t;

t=time;

k=rate constant.

1.At t=10ms,

[A0]=0.0500M

[At]=0.0299M

2.303log0.0500/0.0299=k(10)

 

or k=2.303log0.0500/0.0299(10)

 

or k=2.303log (0.167224)

 

or k=2.303(-0.7767)

 

or k=-1.7887 ms-1

 

or k=-1.8 ms-1

 

2.At t=20ms,

[A0] =0.0500M

[At]=0.0179

2.303log0.0500/0.0179=k(20)

 

or k=2.303log0.0500/0.0179(20)

 

or k=2.303log(0.13966)

 

or k=2.303(-0.854927)

 

or k=-1.9688 ms-1

 

or k=-1.97 ms-1

Step 3 of 3

3.At, t=30ms;

[A0]=0.0500M

[At]=0.0107M

2.303log0.0500/0.0107=k(30)

 

or k=2.303log0.0500/0.0107(30)

 

or k=2.303log(0.15576)

 

or k=2.303(-0.80754)

 

or k=-1.8597ms-1

 

or k=-1.9 ms-1

 

4.At, t=40ms;

[A0]=0.0500M

[At]=0.00641M

2.303log0.0500/0.00641=k(40)

 

or k=2.303log0.0500/0.00641(40)

 

or k=2.303log(0.195007)

 

or k=2.303(-0.7099497)

 

or k=-1.635 ms-1

 

or k=-1.6 ms-1

Explanation:

As the value is consistent remain in the range of -1.6, -1.8 ,-1.9 so this is 1st order reaction .

k S.I. unit is s-1

 

So value of k=1.9 ms-1=1.9 /1000 s-1=0.0019 s-1

Final solution

 

k 1.9ms-1
k 0.0019s-1

 

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