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) Moles of NaOH taken = Molarity X volume = 0.02 x 10 mL = 0.2 millimoles

Moles of crystal violet = Molarity X volume = 1.5x10^-5 X 10mL = 1.5x10^-4 millimoles

so here we have taken NaOH in excess with respect to crystal violet so the rate of reaction will depend only upon the concentration of crystal violet. How the rate depends on crystal violet will be determined by graph.

We will plot graph between

Absrobance (which is equivalent to concentration) v/s time: If straight line then zero order reaction

ln (Absorbance) v/s time: if straight line then first order

1/A v/s time: If straight line then second order

The R2 value shows that the straight line is for Ln A v/s time: so first order with respect to crystal violet

Similarly for trial 2:

Again a straight line with lnA v/s time

So on doubling the concentration of crystal violet the rate of reaction will get doubled

2) Yes we have obtained the same rate law, as the amount of NaOH used will be in excess

3) As explained above, the second order will give a straight line between 1/ A v/s time

4) The absrobance will increase with the formation of product

Yes it will be straight line with a positive slope as the rate of disappearance of reactant will be equal to rate of appearance of product (if it is first order reaction)