The following data were obtained for the concentration vs. time for a certain chemical reaction. Values...

The following data were obtained for the concentration vs. time for a certain chemical reaction. Values were measured at 1.0 s intervals, beginning at 0.00 and ending at 20.0 s. Concentrations in mM are:
10.00, 6.91, 4.98, 4.32, 3.55, 3.21, 2.61

2.50, 2.22, 1.91, 1.80, 1.65, 1.52, 1.36
1.42, 1.23, 1.20, 1.13, 1.09, 1.00, 0.92
a) Plot concentration, c, vs. time, t, ln c vs. t, and 1/c vs. t.
b) Decide whether the data best fit zero-order, first-order or second-order kinetics and calculate the rate constant (with units (!!!)).

Expert Solution

a)

Time	Concentration	ln C	1/c
0	10	2.302585	0.1
1	6.91	1.93297	0.144718
2	4.98	1.60543	0.200803
3	4.32	1.463255	0.231481
4	3.55	1.266948	0.28169
5	3.21	1.166271	0.311526
6	2.61	0.95935	0.383142
7	2.5	0.916291	0.4
8	2.22	0.797507	0.45045
9	1.91	0.647103	0.52356
10	1.8	0.587787	0.555556
11	1.65	0.500775	0.606061
12	1.52	0.41871	0.657895
13	1.36	0.307485	0.735294
14	1.42	0.350657	0.704225
15	1.23	0.207014	0.813008
16	1.2	0.182322	0.833333
17	1.13	0.122218	0.884956
18	1.09	0.086178	0.917431
19	1	0	1
20	0.92	-0.08338	1.086957

b) The graph of 1/C vs Time is straight line.therefore the the given data is best fit for second order reaction

The slope of 1/C vs time graph will give us the rate constant.

Therefore slope of 1/C vs time = y/

= (0.144718- 0.1)/(1-0)

= 0.044 (mM^-1 sec)

rate constant K = 0.044 sec/mM

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