Chem. 1A Experiment 11 CSUS Department of Chemistry Chemical Kinetics Name: Chu
ID: 528135 • Letter: C
Question
Chem. 1A Experiment 11 CSUS Department of Chemistry Chemical Kinetics Name: Chu 2 -13 Score: no Section: EXPERIMENT 11 PRE LABORATORY ASSIGNMENT (Turn this in with vou lab) 1. Three runs were used to study the reaction: The following data were obtained (initial concentrations of Cand D o: 2.970M 3.000 5.940M 2.880M (a) Fill in the missing concentrations in the table above based on the reaction stoichiometry. (b) calculate the rate of the reaction in Mrs for each of the three Rate -1xAIA) runs. (c) using the method of initial rates, determine the order with respect to A and B. (d) Calculate the rate constant for the reaction, including units. Rev. F14 Page 1 of 8Explanation / Answer
a) Set up the ICE chart for the reaction as
A + B ------> 3 C + D
initial c1 c2 0 0
change -x -x +3x +x
equilibrium (c1 – x)(c2 – x) 3x x
Consider the first example. Given c1 = 3.000 M and (c1 – x) = 2.970 M, we must have
c1 – (c1 – x) = (3.000 M) – (2.970 M)
===> x = 0.03 M
Therefore, [B] = c2 – x = (2.000 – 0.03) M = 1.970 M.
Fill in the table as below:
[A] initial, M
[A] after 150 s, M
x M
[B], after 150 s = ([B] initial – x) M
[C] = 3x M
3.000
2.970
0.03
1.970
0.09
6.000
5.940
0.06
1.940
0.18
3.000
2.880
0.12
3.880
0.36
b) The rate of the reaction = -(1/a)*[A]/t
Note that a = 1 (since A doesn’t have a stoichiometric co-efficient) and hence, the rate of the reaction can be written as
Rate = -[A]/t
Again, [A] = ([A] after 150 s) – ([A] initial)
Fill in the table as below:
Trial
[A] initial, M
[A] after 150 s, M
Rate of the reaction, M/s
1
3.000
2.970
0.0002
2
6.000
5.940
0.0004
3
3.000
2.880
0.0008
c) Write down the rate law for the reaction as
Rate = k*[A]m*[B]n
k = rate constant for the reaction; m = order of the reaction with respect to A and n = order of the reaction with respect to B.
Compare trials (1) and (2)
(Rate)1/(Rate)2 = k*[A]m*[B]n/(k*[A]m*[B]n)
===> (0.0002 M/s)/(0.0004 M/s) = {k*(3.000 M)m*(2.000 M)n}/{k*(6.000 M)m*(2.000 M)n}
===> ½ = (3.000/6.000)m
===> ½ = (1/2)m
===> m = 1
Again, compare trials (1) and (3)
(Rate)1/(Rate)3 = k*[A]m*[B]n/(k*[A]m*[B]n)
===> (0.0002 M/s)/(0.0008 M/s) = {k*(3.000 M)m*(2.000 M)n}/{k*(3.000 M)m*(4.000 M)n}
===>1/4 = (2.000/4.000)n
===> 1/4 = (1/2)n
===> (1/2)2 = (1/2)n
===> n = 2
The order of the reaction with respect to A and B are 1 and 2 respectively (ans).
d) Plug in values for [A], [B] and rate and write
(0.0002 M/s) = k*(3.000 M)1*(2.000 M)2
===> (0.0002 M/s) = k*(12.000 M3)
===> k = (0.0002 M/s)/(12.000 M3)
===> k = 1.667*10-5 M-2 s-1 1.67*10-5 M-2 s-1 (ans).
[A] initial, M
[A] after 150 s, M
x M
[B], after 150 s = ([B] initial – x) M
[C] = 3x M
3.000
2.970
0.03
1.970
0.09
6.000
5.940
0.06
1.940
0.18
3.000
2.880
0.12
3.880
0.36
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