Using the comparison (or inspection) method, calculate the order in each reactan

ID: 500524 • Letter: U

Question

Using the comparison (or inspection) method, calculate the order in each reactant from the following data. Show work.

Where x is order (Rate1/Rate 2) = (Concentration 1/Concentration 2)x Use this: Order = Log (rate 1/ rate 2) / Log (Conc1/conc2)

A + 2B+ C --> D+ E

Run Initial [A] Initial [B] Initial [C] Initial rate of formation of E

1 0.20M 0.20M 0.20 M 2.42 x10-6 M/min

2 0.20 M 0.20 M 0.30 M 2.44 x 10-6 M/min

3 0.40 M 0.20 M 0.30 M 9.67 x 10 -6 M/min

4 0.20 M 0.60 M 0.30 M 7.23 x 10-6 M/min

The idea is to pick two runs to compare where only the concentration of one reactant changes:

Comparisons used: Setup of calculation of order:

A runs __ &___

B: runs ___ &___
C: runs ___ &______

Write the overall rate equation using these calculated orders: rate =K

What is the overall order of the reaction?

Solve for rate constant K

rate -r = K[A]a [B]b[C]c, where a, b and c are orders of reactions with respect to A, B and C respectively, from 1st experimental data

2.42*10-6 = K[ 0.2]a[ 0.2]b[ 0.2]c (1)

From 2nd experimental data

2.42*10-6 = K[ 0.2]a[ 0.2]b[ 0.3]c (2)

Eq.2/Eq.1 gives (0.3/0.2)c= 1, c =0

From 3rd experimental data

9.67*10-6 = K[ 0.4]a[ 0.2]b[ 0.3]c (3)

Eq.3/Eq.2 gives 9.67/2.42= 2a, a=2

From experiment 4

7.23*10-6 = K[ 0.2]a[ 0.6]b[ 0.3]c (4)

Eq.2/Eq.4 (2.42/7.23)= K (0.2/0.6)b b= 1

Hence the rate expression becomes –r = K[A]2 [B]1

from the 1st equation

2.42*10-6 = K[ 0.2]2[ 0.2] K= 0.0003025/M2.min

The rate equation becomes –r= 0.0003025[A]2[B]1

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