Question

Trial 1

Time	Trans	A=2-log %T	Ln(A)	1/A
1:57	19.0	0.721	-0.327	1.39
2:52	22.0	0.658	-0.419	1.52
3:51	24.5	0.611	-0.493	1.64
4:54	27.3	0.564	-0.573	1.77
5:53	29.7	0.527	-0.641	1.90
6:54	33.6	0.474	-0.747	2.11
7:55	36.2	0.441	-0.819	2.27
8:54	40.0	0.398	-0.921	2.51
9:52	40.5	0.393	-0.934	2.54
10:53	45.3	0.344	-1.07	2.91
11:58	47.7	0.321	-1.14	3.12
13:02	52.2	0.280	-1.27	3.57
14:02	55.3	0.257	-1.36	3.89
14:59	56.0	0.252	-1.38	3.97
15:59	59.4	0.226	-1.49	4.42
16:58	60.4	0.219	-1.52	4.57
18:00	65.2	0.186	-1.68	5.38
19:06	65.9	0.181	-1.71	5.52
20:03	68.8	0.162	-1.82	6.17

Trial 2

Time	Trans	A=2-log % T	Ln(A)	1/A
1:43	25.7	0.590	-0.528	1.69
2:48	30.2	0.520	-0.654	1.92
3:41	45.4	0.343	-1.07	2.92
4:46	43.8	0.358	-1.03	2.79
5:46	47.6	0.322	-1.13	3.11
6:47	50.5	0.297	-1.21	3.37
7:48	56.9	0.245	-1.41	4.08
8:48	61.4	0.212	-1.55	4.72
9:45	62.3	0.206	-1.58	4.85
10:49	66.8	0.175	-1.74	5.71
11:45	72.5	0.140	-1.97	7.14
12:45	71.6	0.145	-1.93	6.90
13:45	76.5	0.116	-2.15	8.62
14:44	78.5	0.105	-2.25	9.52
15:42	83.1	0.080	-2.53	12.5
16:45	83.2	0.080	-2.53	12.5
17:43	88.4	0.054	-2.92	18.5
18:49	90.7	0.042	-3.17	23.8
19:42	84.5	0.073	-2.62	13.7
20:43	89.6	0.048	-3.04	20.8

Trial 1 added 10 ml 0.020 M of NaOH and 10 ml 1.5x10^-5 crystal violet

Trial 2 added 10 ml 0.040 M of NaOH and 10 ml 1.5x10^-5 crystal violet

tested the absorbance of solution for 20 minutes

1. Based on your experimental determined rate law for the reaction of crystal violet and hydroxide ion, how would doubling the concentration of crystal violet affect the reaction rate?

2. Would you have determened an identical rate law for the reaction of crystal violet and hydroxide ion if you had used hydroxide ion concentrations of 0.030 M and 0.050 M instead of 0.020 M and 0.040 M? briefly explain

3.The rate law for a certain reaction is second order with respect to one of the reactants, R. Suppose you study this reaction, observing the absorbance of light at the analytical wavelength for R, and record the data with respect to elapsed time. Also suppose that the concentrations of all the other reactants are in large excess, and that R is the only colored species invilved. Explain which absorbance function, A, LnA, or 1/A, would yeild a straight-line graph when plotted against elapsed time.

4. Suppose that in the reaction Q + R --> P, only the product is colored

b. what happens to the absorbance at the analytical wavelength for P of the mixture as the reaction progresses?

c. Suppose you do an experie=ment involving this reaction in which Q is present in large excess of R. If the reaction is first order whith respect to R, would the grapph of LnA (for P) versus time be a straight line? Briefly explain

Answer 1

1. The Rate law for crystal violet vs. NaOH reaction is given by

R = k[CV]ⁿ [OH]^m

Since the concentration of NaOH used is too high as compared to the concentration of Crystal Violet , the change in [OH^-] is considered to be negligible during the course of the reaction.Thus the rate of reaction only depends on the conecntration of crystal-violet .therefore the rate law becomes

R = k' [CV]ⁿ

where , n is the ORDER of the reaction and can be determined by the nature of the graph obtained from the observed data.

if the graph of ln[CV] vs. Time comes out to be straight line then it is a First-Order Reaction

if the graph of 1/[CV] vs. Time comes out to be straight line then it will be a second order reaction.

the concentration of CV can be found out by using the Beer's Equation

A = ecl

where , e is the molar absorptivity  e for CV is = 5.0 x 10⁴ M^-1 cm^-1

c is the concentration of CV

l is the path length

From data analysis it turns out that the graph of ln[CV] vs. time is a straight line.thus the reaction is first order. Rather it is called aa PSEUDO-FIRST ORDER REACTION as the concentration of NaOH was kept constant.

Hence the value of n turns out to be 1

rate law becomes

R = k' [CV]¹

So now when you alter the concentration of CV ,it casts the same effect on the rate.Doubling the concentration of CV will definitly double the rate of reaction.

2. Yes, the rate law would have been the same as increasing the concentration of base does not has any effect on the rate of the reaction.

3. when all the other reactants are kept at large excess while only the concentration of B matters the rate law expression (second order with respect to B) will be as follows.

R = k' [B]²

where k' is the rate-constant including all the other concentrations.

as it is clear from the graphs , the graph of 1/[B] vs.Time would yield a Straight-line graph as it is a second-order reaction.

4.b) If for the given reaction , the spectrophotometer is set at the wavelength of product P , the observed absorbance will be seem to be increasing as according to beer's law, absorbance is directly proportional to the concentration of colored species

c) If the reaction is again taken to be pseudo-first order reaction where the concentration of Q is kept in excess , the rate law will only contain concentration of R .

as we interpret the zero, first and second order reactions from the graph plotted in concentration , correspondingly the graphs can be interpreted in terms of Absorbance.

The only difference that lies here is that when the absorbance of the reactants is plotted in first-order reaction,the straight line tends to fall(negative slope) ,but here as we are plotting the absorbance for the product P , the absorbance(as i specified in the previous part) should increase.Thus a straight line plot is obtained but with a positive slope.

^.