Question

rate law determination of crystal violent reaction

one will be Stream TABLE 1. Concentration of crystal violet versus time in the crystal violet reaction the outerwo (make calculations for the third and fourth column and graph the results vs time) (make calculations for the t o p N use 3 decimal Time (min) [A] (molar conc.) In (A) 1/[A] 0 (initial) 1.000 When doing the 0.893 1-0.113 gruph also ad 0.806 -ba15 1.240 - por 0.735 -0.307 1.3600 0.676 0.3911 1.479 whichever graph has a tomline wa 0.625 -0.470 closest to 1, the 0.581 -0.543 1. is the straight line 0.543 -0,605 1.841 1st oder, and ord 0.510 -0,673 1.960 0.481 -0.731 2.079 0.454 -0.789 wana 11 0.431 -0,841 2.320 0.410 -o gal 2.439 USE MESOF+ EXCEL QUESTIONS: 10 1) is the reaction first-order or second-order? Explain. 2) Calculate the rate constant from the equation for the appropriate graph. 3) Write the correct rate law for the crystal violet reaction. 4) Determine a half-life for the crystal violet reaction by Calculating it: Estimating it from the time difference between two points, where one has an (A) value approximately one-half of the other:

Answer 1

Time (min)	[A]	ln[A]	1/[A]
0	1	0	1
1	0.893	-0.11317	1.119821
2	0.806	-0.21567	1.240695
3	0.735	-0.30788	1.360544
4	0.676	-0.39156	1.47929
5	0.625	-0.47	1.6
6	0.581	-0.543	1.72117
7	0.543	-0.61065	1.841621
8	0.51	-0.67334	1.960784
9	0.481	-0.73189	2.079002
10	0.454	-0.78966	2.202643
11	0.431	-0.84165	2.320186
12	0.41	-0.8916	2.439024

ln[A] vs Time

In[A] vs Time O 8 10 12 14 N In[A] in o n bo ip Time (min)

1/[A] vs time

1/[A] vs Time 1/[A] y = 0.12x + 1.0002 R =1 24 10 12 14 6 8 Time (min)

1) From the above two graphs, drawn in Microsoft excel, the rate of the reaction is second-order.

it is because of the graph between 1/[A] vs time is linear.

2) the equation of the graph is y = 0.12x + 1.0002 (y = mx + c)

where m is the slope, m = rate constant ( integrated rate law for second order is 1/[A] = 1/[Ao] + kt where, y = 1/[A] : x = t)

m = rate constant = 0.12 m-1s-1

3) The correct rate law,

rate = k[A]2, where k = 0.12 m-1s-1

4) Half-life is given by the formula,

t1/2 = 1/k[Ao]

This is derived by,

Using integrated rate law of second order reaction,

1/[A] = 1/[Ao] + kt

By the given hint in the question, [A] value approximately equal to one half of other

So, [A] = [Ao]/2 ; t = t1/2

Substituting in the above equation,

2/[Ao] = 1/[Ao] + kt1/2

By resolving for t1/2

t1/2 = 1/k[Ao]

t1/2 = 1/0.12 m-1s-1 x 1 m

t1/2 = 8.3333 s