Question

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: 3.20x10-5

10.0: 2.42x10-5

20.0: 1.95x10-5

30.0: 1.63x10-5

40.0: 1.40x10-5

50.0: 1.23x10-5

60.0: 1.10x10-5

Answer 1

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