I am doing the Crystal Violet reaction with NaOH Kinetic Lab

I am doing the Crystal Violet reaction with NaOH Kinetic Lab

NEED MY WORK CHECKED PLEASE

I am not sure if my answer is correct..

Objective: Determining the initial rates for run 1 and 2

Also, using the initial rates and the concentration, what is the order in hydroxide?

Lastly, calculate the rate constants for this reaction using the data from run 1 and run 2. Report the rate constant as an average value.

I have determined from the data plotted in a graph that the reaction is 1st order in respect to Crystal Violet because the natural log (LN) of the data produces the most linear relationship between time and concentration.

I used the following solutions to complete 2 trials:

Trial

Deionized Water

0.05xxx M NaOH

Crystal Violet Solution

1

15.00 mL

5.00 mL

10.00 mL

2

10.00 mL

10.00 mL

10.00 mL

Here is the data I collected.

Run 1

y = -0.08469x - 11.35596
R² = 0.99968

Run 2

y = -0.1402x - 11.371
R² = 0.9989

Any help in the right direction would be greatly appreciated, thank you!

Here is my work so far..

M₁V₁=M₂V₂ Concentration of OH- = (0.05M NaOH)(5.0mL NaOH)/(30.0mL total solution) = 0.00833

(0.05M NaOH)(10.0mL NaOH)/(30.0mL total solution) = 0.0166

slope run 1 = -0.08469 k^\'=0.08469

slope run 2 = -0.1402 k\'=0.1402

set up: (k\'₁)/(k\'₂)=([NaOH₁]/[NaOH₂] )ⁿ (all raised to the power of n which is what we are solving for)

(0.08469)/(0.1402)=(0.00833)/(0.0166)

0.6040656=0.49998ⁿ

log 0.6040656/ log 0.49998 = n = 0.727 ~ 1 Reaction order for OH^- = 1

CV = 1 OH = 1

Lastly, K=(k\')/([OH^-]) Run 1 : 0.08469/0.00833 =10.1668 Run 2 : 0.1402/0.0166 = 8.4457

average K = 9.3063249

Rate=9.3063[CV⁺][OH^-]

Solution

You performed very well.

All the steps are correct. There\'s no error in your work.

Just make sure you have calibrated the instrument properly.

or else no error. Good luck.