##### 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.

(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.)