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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry, 4th edition.

Problem 13.78.

Benzenediazonium chloride decomposes in water yielding N2(g).

C6H5N2Cl(aq) C6H5Cl(aq) + N2(g)

The data tabulated below were obtained for the decomposition of a 0.071 M solution at 50 oC (t = corresponds to the completed reaction). To obtain [C6H5N2Cl] as a function of time, note that during the first 3 min, the volume of N2(g) produced was 10.8 mL of a total of 58.3 mL, corresponding to this fraction of the total reaction: 10.8 mL/58.3 mL = 0.185. An equal fraction of the available C6H5N2Cl was consumed during the same time.

Time, min

N2(g), mL

Time, min

N2(g), mL

0

0

18

41.3

3

10.8

21

44.3

6

19.3

24

46.5

9

26.3

27

48.4

12

32.4

30

50.4

15

37.3

58.3

(a) Plot graphs showing the disappearance of C6H5N2Cl and the formation of N2(g) as a function of time.

(b) What is the initial rate of formation of N2(g)?

(c) What is the rate of disappearance of C6H5N2Cl at t = 20 min?

(d) What is the half-life, t½, of the reaction?

(e) Write the rate law for this reaction, including a value for k.






Answer:

(a) First, find the concentrations of C6H5N2Cl remaining at each time:

Time, min

N2(g), mL

fraction N2

= mL/58.3 mL

C6H5N2Cl, M

= 0.071×(1 – fraction)

0

0

0/58.3 = 0.00

0.071(1 – 0) = 0.071

3

10.8

10.8/58.3 = 0.185

0.071(1 – 0.185) = 0.058

6

19.3

19.3/58.3 = 0.331

0.071(1 – 0.331) = 0.047

9

26.3

26.3/58.3 = 0.451

0.071(1 – 0.451) = 0.039

12

32.4

32.4/58.3 = 0.556

0.071(1 – 0.556) = 0.032

15

37.3

37.3/58.3 = 0.640

0.071(1 – 0.640) = 0.026

18

41.3

41.3/58.3 = 0.708

0.071(1 – 0.708) = 0.021

21

44.3

44.3/58.3 = 0.760

0.071(1 – 0.760) = 0.017

24

46.5

46.5/58.3 = 0.798

0.071(1 – 0.798) = 0.014

27

48.4

48.4/58.3 = 0.830

0.071(1 – 0.830) = 0.012

30

50.4

50.4/58.3 = 0.864

0.071(1 – 0.864) = 0.0097

58.3

58.3/58.3 = 1.00

0.071(1 – 1.00) = 0.00


The plots are:

(b) From the definition of rate,


(c) Again, using the definition of rate and approximating t = 20 min using the t = 18 min and t = 21 min points,

(d) The initial concentration of benzenediazonium chloride is 0.071 M, so the first half-life is reached when the concentration has reached 0.0355 M. Using the graph from part (a) gives t½ ~ 10 min.


(e) The order of the reaction can be determined by looking at the time of the second half-life, when the benzenediazonium chloride concentration has been reduced to ½ of its initial value, 0.018 M. This occurs at ~21 min, just twice the time for the first half-life. Since t½ is the same for the two different time intervals, this means that the reaction is first order.

Rate = k[C6H5N2Cl]

Plotting ln[C6H5N2Cl] vs. t confirms this and the slope of the plot gives the rate constant, k.

The slope of the plot = –0.0664, so k = 0.0664 min–1. (This allows a better estimate of the half-life to be t½ = 0.693/k = 0.693/0.0664 = 10.4 min.)