1. /The decomposition of A is first order, and [A] is monitored. The following d
ID: 557242 • Letter: 1
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1. /The decomposition of A is first order, and [A] is monitored. The following data are recorded: t/min 0 A/IM] 0.100 0.09050.0819 0.0670 a. Calculate k. (What is the rate constant?) b. Calculate the half life. (What is the half life?) c. Calculate [A] when t 5 min. d. Calculate t when [A] 0.0100. (Estimate the time required for 90% of A to decompose.) HINT: AFTER SOLVING SECTION C AND D WRITE THE DATA AND VERIFY THE DATA IN THE LIST TO KNOW THE WORTHINESS OF YOUR DATA 2Given below the initial rate data, determine the rate law and rate constant for the following reaction: 2+ Exp. # [MnO] [CIOS] [H+] Initial Rate (M's) 1 0.10 M 0.10 M 0.10 M 32 x 10 -2 .2 4 0.10 M 0.10M 0.20 M 7.4x 10.3Explanation / Answer
a) Order with respect to each reactant:
For determining order with respect to each reactant we need to take firstly any tw trials
i) Taking trial 1 and 2
Dividing Trial 2 by 1
(0.480/0.480)a = (1.080/ 0.360)b = (1.380/0.460)
on cancelling common terms , we are left with
3b = 31
now we need to compare the 3b = 31
As the base is same 3 , therefore powers are equated and hence order with respect to B is 1
ii) now compare trial 1 and 3
Divide trial 1 by trial 3
(0.600/ 0.200)a = (0.360/0.360)b = (0.460/0.051)
on cancelling common terms , we are left with
3a = 9
now compare 3a = 32
As the base are same , therefore powers are equated and order with respect to A is 2
b) The order with respect to A is 2 and B is 1 , therefore overall order of the reaction is 3
c) Rate = K [A]2 [B]
d) the value of rate constant K can be determined from Rate law expresssion
Rate = k [A]2 [B]
putting the value of Rate , concentration of A and B from any one of teh trial , we can find K
taking the values from trial 1
0.460 = k ( 0.600)2 ( 0.360)
0.460 = kx 0.1296
0.460/0.1296 = k
3.54 = k
the oerall reaction is 3rd order so units are mol2 lit-2 sec-1
therefore k = 3.54 mol2 lit-2 sec-1
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