Question

Graph 1: [A565]t vs time (sec)

y=mx+b

y=-0.0277x +0.5852

(graph is curved)   

2	0.644
3	0.61
4	0.479
5	0.434
6	0.39
7	0.353
8	0.319
9	0.291
10	0.261
11	0.238
12	0.207
13	0.182
14	0.161
15	0.146
16	0.132
17	0.117
18	0.111
19	0.095
20	0.086
21	0.071

Graph 2: ln [A565]t vs time (sec)

y=mx+b

y=-0.112x-0.2368 (graph is almost linnear, sloped downward)

2	-0.44
3	-0.494
4	-0.736
5	-0.835
6	-0.942
7	-1.041
8	-1.143
9	-1.234
10	-1.343
11	-1.435
12	-1.575
13	-1.704
14	-1.826
15	-1.924
16	-2.025
17	-2.146
18	-2.198
19	-2.354
20	-2.453
21	-2.645

Graph 3: 1/[A565]t vs Time (sec)

y=mx+b

y=0.5832x-1.0896 (graph is curved)

2	1.553
3	1.639
4	2.088
5	2.304
6	2.564
7	2.833
8	3.135
9	3.436
10	3.831
11	4.202
12	4.831
13	5.495
14	6.211
15	6.849
16	7.576
17	8.547
18	9.009
19	10.526
20	11.628
21	14.085

From this analysis determine, AND SHOW WORK:

A. the reaction order with respect to crystal violet( I believe it is first order )

B. the rate constant using the correct units

Answer 1

we know that

if [A] vs time is a straigt line , then it is zero order

if ln [A] vs time is a straight line , then it is first order

if 1/[A] vs time is a straight line , then it is second order

given

in this case

that

ln [A] vs time is a linear one and sloped downward

so

it is first order

b)

now

we know that

for 1st oder

ln A = -kt + ln Ao

comparing with y = mx + c

we can see that

slope = m = -k

in this case

the line equation is y = -0.112x - 0.2368

so

slope = m = -k = -0.112

k = 0.112

so

the rate constant is 0.112

since its a 1st order reaction , unit is s-1

so

the rate constant is 0.112 s-1