Question

1. In a study of the decomposition of nitrosyl bromide at 10 °C

NOBr ---> NO + ½ Br₂

the following data were obtained:

[NOBr], M	6.71×10-2	3.36×10-2	1.68×10-2	8.40×10-3
seconds	0	20.9	62.8	147

Hint: It is not necessary to graph these data.

(1)
The observed half life for this reaction when the starting concentration is 6.71×10^-2 M is _______ s and when the starting concentration is 3.36×10^-2 M is s.

(2)
The average (1/[NOBr]) / t from t = 0 s to t = 20.9 s is __________ M^-1 s^-1.
The average (1/[NOBr]) / t from t = 20.9 s to t = 62.8 s is ___________ M^-1 s^-1.

(3)
Based on these data, the rate constant for this _______ pick one (zero, first or second) order reaction is _________ M^-1 s^-1.

Answer 1

Given reaction NOBI - NO + Biz [NO B1].M) 6.71X10-2 3-36x10-2 1667X1027.40 x10- Seconds I o I 2019 | 6218 T 147 11 half life is defined as the time period after which the concentration of reactant becomes half of the initial amount now, initial amount = 6.71x10²M half amount = 3.355*10-2M ~ 3.36 x 10-2 this half comcentration of NOBI is at 2019 sec. so, half life : 2019 sec. again, initial concentration = 3.36x102M at 20. 9 sec half concentration 1.68 *10-2M at 6218 sec this half concentration of NOB is in time period of (62.89- 2019 sec : 41.9 sec so, half life = 41.9 sec. avelage rate of reaction - change in concentration of NOB. change in time - (3.36810-2M-6.71x10²m) 20.9 sec 3.35X102M 2019 sec ř 116X003 M/sec. 2)

concentration of NOBR at t=20,9 sec = 3.368102 contentration of No Br at t = 62,8 sec. = 1.68x10²4 average rate of reaction - A NOB] at - (1.68810 M-3.36X10M) 62.8 sec - 2019 sec 1.68x102M 41. gsec. - 4.0810-4 Misec. 3) We can observe from data that for every concentration, their is different half life and different rate of reaction, We know that for zero order reaction, rote of reaction is constant for ist Order reaction , half life is constant. but for and order reaction both changes for every concentration, so, this reaction is and order reaction. now, for and order reaction typ = KTA :k= ti [as] : 20,954. 16.77X10 M) - 0.713 M12