## Crystal Violet Rate Law Lab

January 27, 2016 AP Chemistry pd.4

Rileigh Robertson

The lab originally said to dilute a 25 micro-molar solution as the "pre-lab" but the solution we got was 25.7 micro molar, so I redid the calculations. (below)

5. Calibration curve for dilutions.

## Analysis:

**Something to note: Absorption and concentration are directly related (nearly the same) so I just used Absorbance for my data!

## Trial 1

I did my graphs on excel instead of my graphing calculator because there were so many points, and I wanted to have them all to make my data more accurate.

1-4. Shown Above

5.

1st order: ln[Abs]

2nd order: 1/[Abs]

Zero order: [A] just concentration

1-4. Shown Above

5.

1st order: ln[Abs]

2nd order: 1/[Abs]

Zero order: [A] just concentration

## Trial 2

I did my graphs on excel instead of my graphing calculator because there were so many points, and I wanted to have them all to make my data more accurate.

1-4. Shown Above

5.

1st order: ln[Abs]

2nd order: 1/[Abs]

Zero order: [A] just concentration

6.

[A] - both slopes are in negative, very small, values. trial 2 has the r^2 value closest to one (.956), but trial 1 was till decent at .805

T= trial slope T1/T2= -8.23x10^-4/-1.422x10^-3 = 57.9% accurate

ln[A] - both slopes are in negative, very small (x10-3), values. Trial 1 has the better r^2 value at .888 while trial 2 really dropped (.62)

T= trial slope T2/T1 (since T2 is bigger)= -4.24x10^-3/-7.98x10^-3 = 53.1% accurate

1/[A] - both slopes are in the positive zeros, very small numbers, and trial 1 keeps is consistent with an R62 value of .814, and trial 2 r^2 value drops even more to .317

T= trial slope T2/T1 (since T2 is bigger)= .014/.108 = 13.0% accurate

7.

Overall Rate Law:

rate=k[CV]^1[OH]^-.785

according to a slope to OH ratio

1-4. Shown Above

5.

1st order: ln[Abs]

2nd order: 1/[Abs]

Zero order: [A] just concentration

6.

[A] - both slopes are in negative, very small, values. trial 2 has the r^2 value closest to one (.956), but trial 1 was till decent at .805

T= trial slope T1/T2= -8.23x10^-4/-1.422x10^-3 = 57.9% accurate

ln[A] - both slopes are in negative, very small (x10-3), values. Trial 1 has the better r^2 value at .888 while trial 2 really dropped (.62)

T= trial slope T2/T1 (since T2 is bigger)= -4.24x10^-3/-7.98x10^-3 = 53.1% accurate

1/[A] - both slopes are in the positive zeros, very small numbers, and trial 1 keeps is consistent with an R62 value of .814, and trial 2 r^2 value drops even more to .317

T= trial slope T2/T1 (since T2 is bigger)= .014/.108 = 13.0% accurate

7.

Overall Rate Law:

rate=k[CV]^1[OH]^-.785

according to a slope to OH ratio