Question

I We are doing Kinetics of the Oxidation of a Water Soluble Dye lab, where we test different concentrations of bleach and dye. For the report, I need to answer the following;

Order of Reaction with Respect to Dye

Pseudo Rate Constant with Respect to Dye

Order of Reaction with Respect to Bleach

Here is a data sheet I made with different concentrations, 2 trials each, andinitial rate concentration. Can I determine the order of reaction and pseudo rate constant with this data? Could really use your help.

Percent Absorption (Abs) for bleach/water solutions
Time (Seconds)	2/5 (v/v)		1/5 (v/v)		1/10 (v/v)		1/20 (v/v)
	Run #1	Run #2	Run #1	Run #2	Run #1	Run #2	Run #1	Run #2
10	0.39	0.40	0.44	0.43	0.42	0.43	0.44	0.46
20	0.31	0.35	0.41	0.39	0.40	0.41	0.41	0.44
30	0.26	0.29	0.38	0.36	0.39	0.39	0.41	0.43
40	0.21	0.24	0.36	0.34	0.37	0.37	0.40	0.42
50	0.16	0.21	0.33	0.31	0.31	0.36	0.39	0.41
60	0.13	0.17	0.31	0.28	0.28	0.34	0.38	0.40
70	0.10	0.14	0.28	0.26	0.26	0.33	0.38	0.40
80	0.08	0.12	0.27	0.24	0.24	0.31	0.37	0.39
90	0.06	0.10	0.24	0.21	0.21	0.30	0.36	0.39

Initial rate between 20s – 30s
	2/5 (v/v)	1/5 (v/v)	1/10 (v/v)	1/20 (v/v)
Run #1	-0.005	-0.003	-0.001	0.000
Run #2	-0.006	-0.003	-0.002	-0.001
Average	-0.0055	-0.003	-0.0015	-0.0005

Answer 1

Let the reaction is like this

NaOCl + dye --> Colorless products

Rate = k [NaOCl] ^x [Dye]^y

Now we will use pseudo order in order to determine the rate law with the NaOCl being in excess,as ff NaOCl is present in large excess, its concentration will remain almost constant, and we define a new constant,

Kobs = K [NaOC]^x

so rate = Kobs [Dye]^y

Now we have absrobance values with us, and we will plot a graph between absorbance ( cocentraion ) and Time and a groph between Ln Absorbance and time to know the pattern of graph and hence the kinetics

Percent Absorption (Abs) for bleach/water solutions
Time (Seconds)	2/5 (v/v)				1/5 (v/v)				1/10 (v/v)				1/20 (v/v)
	Run #1	Run #2	average	Ln Abs	Run #1	Run #2	average		Run #1	Run #2	average		Run #1	Run #2	average
10	0.39	0.4	0.395	-0.929	0.44	0.43	0.435	-0.83	0.42	0.43	0.41	-0.8916	0.44	0.46	0.45	-0.8
20	0.31	0.35	0.33	-1.109	0.41	0.39	0.4	-0.92	0.4	0.41	0.395	-0.9289	0.41	0.44	0.425	-0.86
30	0.26	0.29	0.275	-1.291	0.38	0.36	0.37	-0.99	0.39	0.39	0.39	-0.9416	0.41	0.43	0.42	-0.87
40	0.21	0.24	0.225	-1.492	0.36	0.34	0.35	-1.05	0.37	0.37	0.37	-0.9943	0.4	0.42	0.41	-0.89
50	0.16	0.21	0.185	-1.687	0.33	0.31	0.32	-1.14	0.31	0.36	0.335	-1.0936	0.39	0.41	0.4	-0.92
60	0.13	0.17	0.15	-1.897	0.31	0.28	0.295	-1.22	0.28	0.34	0.31	-1.1712	0.38	0.4	0.39	-0.94
70	0.1	0.14	0.12	-2.12	0.28	0.26	0.27	-1.31	0.26	0.33	0.295	-1.2208	0.38	0.4	0.39	-0.94
80	0.08	0.12	0.1	-2.303	0.27	0.24	0.255	-1.37	0.24	0.31	0.275	-1.291	0.37	0.39	0.38	-0.97
90	0.06	0.1	0.08	-2.526	0.24	0.21	0.225	-1.49	0.21	0.3	0.255	-1.3665	0.36	0.39	0.375	-0.98

As per the given data we will take average and will plot graph.

1. Prepare graphs of each of the quantities [Abs], ln[Abs] versus time.

From graphs it is confirmed that the reaction is first order with respect to dye.

1. For the graphdetermine theslope,using add trendline feature in Microsoft excel, right click on data point in the graph add trendline, linear regression, display equation in chart.

2. As the reaction is first order with respect to dye, the rateconstant = K0 = -slope = 0.0389 ( 2/5 v/v ) and 0.0255 (1/v ).

3. The ratio between the two is 0.655 we will round it the nearest whole number 1. So the order with respect to NaOCl =1.

The standard concentration of bleach is not mentioned so if you know that then calculate the concentration of bleach in each trial and can determine the rate constant as Rate constant = K0 X [Bleach]