Plot the following data and determine the average rate for the loss of reactant at t = 25 sec. Enter the numerical answer with 2 significant figures, no units and in scientific notation. For example; .00126 M/min would be entered as 1.3E-3 (no spaces) 2 A(g) → B 2(g)
Time(sec) [A]: (mol/L)
0.0: 3.20×10-5
10.0: 2.42×10-5
20.0: 1.95×10-5
30.0: 1.63×10-5
40.0: 1.40×10-5
50.0: 1.23×10-5
60.0: 1.10×10-5
1st the order of the reaction need to be determined. for this Rate is -dCA/dt= KCAn, n is the order and K is the rate constant . if CAO= initial concentration, CA= concentration at any time
n=0 is zero order and hence when integrated, CA= CAO-Kt, for zero order, a plot of CA vs t has to be straighe line
n=1 is 1st oder and hence when integrated, CA= CAO*exp(-Kt), so a plot of CA vs t has to be exponential
n=2 is second order and when integrated 1/CA= 1/CAO+Kt, a plot of 1/CA vs t is straight line.
All the three plots ( 1st and 2nd plot combined) are generated and shown below
the plots suggest the reaction to be second order whose slope is
1/CA= 1/CAO+996.7 t
at 25 sec, 1/CA= 31362+996.7*25, CA= 1.78*10-5 mol/l
hence rate of loss (-dCA/dt)= (3.2-1.78)*10–5/25 =5.7*10-7 Mol/l.sec
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