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Benzenediazonium chloride decomposes in water yielding N_{2}(g).
C_{6}H_{5}N_{2}Cl(aq) C_{6}H_{5}Cl(aq) + N_{2}(g)
The data tabulated below were obtained for the decomposition of a 0.071 M solution at 50 ^{o}C (t = corresponds to the completed reaction). To obtain [C_{6}H_{5}N_{2}Cl] as a function of time, note that during the first 3 min, the volume of N_{2}(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 C_{6}H_{5}N_{2}Cl was consumed during the same time.
Time, min |
N_{2}(g), mL |
Time, min |
N_{2}(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 C_{6}H_{5}N_{2}Cl and the formation of N_{2}(g) as a function of time.
(b) What is the initial rate of formation of N_{2}(g)?
(c) What is the rate of disappearance of C_{6}H_{5}N_{2}Cl at t = 20 min?
(d) What is the half-life, t_{½}, of the reaction?
(e) Write the rate law for this reaction, including a value for k.
(a) First, find the concentrations of C_{6}H_{5}N_{2}Cl remaining at each time:
Time, min |
N_{2}(g), mL |
fraction N_{2} = mL/58.3 mL |
C_{6}H_{5}N_{2}Cl, 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 t_{½} ~ 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 t_{½} is the same for the two different time intervals, this means that the reaction is first order.
Rate = k[C_{6}H_{5}N_{2}Cl]
Plotting ln[C_{6}H_{5}N_{2}Cl] 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 t_{½} = 0.693/k = 0.693/0.0664 = 10.4 min.)