The substance A is blue, the product is colorless, therefore the reaction can be
ID: 993442 • Letter: T
Question
The substance A is blue, the product is colorless, therefore the reaction can be followed by spectrophotometry. Two runs were performed with the initial hydroxide concentrations [OH-]1,0 = 2.00M and [OH-]2,0 = 1.00M. For both runs the concentration of A was [A]0 = 1.47x10-2M. The concentration of A vs. time data for both runs are given below (copy and paste it into Excel for processing).
a. Determine the order of A in the reaction (x).
X = ___________________
b. Determine the order of [OH-] in the reaction (y).
y = ___________________
c. Determine the rate constant, k, of the reaction.
k = ___________________
d. Interpret the effect of [OH-]>>[A] in terms of the mechanism of the reaction.
time (s)
[A]1 (M)
[A]2 (M)
0
1.47E-02
1.46E-02
92
6.00E-03
8.20E-03
192
3.90E-03
6.00E-03
292
2.60E-03
4.70E-03
392
2.10E-03
3.80E-03
492
1.80E-03
3.20E-03
592
1.40E-03
2.70E-03
692
1.20E-03
2.30E-03
792
1.10E-03
2.10E-03
892
1.00E-03
1.90E-03
992
9.00E-04
1.80E-03
1092
8.00E-04
1.50E-03
1192
8.00E-04
1.40E-03
time (s)
[A]1 (M)
[A]2 (M)
0
1.47E-02
1.46E-02
92
6.00E-03
8.20E-03
192
3.90E-03
6.00E-03
292
2.60E-03
4.70E-03
392
2.10E-03
3.80E-03
492
1.80E-03
3.20E-03
592
1.40E-03
2.70E-03
692
1.20E-03
2.30E-03
792
1.10E-03
2.10E-03
892
1.00E-03
1.90E-03
992
9.00E-04
1.80E-03
1092
8.00E-04
1.50E-03
1192
8.00E-04
1.40E-03
Explanation / Answer
pseudo order reaction
And the rate expression is given by
Rate, r = K’ [A]x, K’ = K [OH-]y
So the order for the reaction r= K’ [A]x need to be determined by integration method. Order is assumed and the rate expression is arrived at
First order when assumed gives ln A= lnA0-K’t
So a plot of lnA Vs t should give a straight line which did not give straight line.
The plot is not fitting into straight line, Hence it is not first order.
Second order reaction is assumed for which the equation is
K’A0*t= XA/(1-XA)
XA = conversion= 1-A/AO
If the plot of XA/(1-XA) Vs t gives a straight line, the equation is second order. The plot is shown below.
For trial-1 the slope is K’A0= 0.015
Ao= 1.47*10-2. K’ =0.015/(1.47*10-2)=1.02
For trial-2, the slope is K’AO= 0.007, K’ = 0.007/(1.47*10-2)=0.4762
The rate now r = K’ [A]2
K’[A]2= K[OH]y [A]2
for trial-1, 1.02* [A]2= K[ 2]y [A]2 (1)
for trail-2, 0.4762*[A]2= K[1]y[A]2 (2)
eq.2/Eq.1, 2= [2]y
y= 1 therefore the reaction is first order with respect to [OH-] and second order with respect to [A]
the rate becomes= K[OH-] [A]2
K[OH-] =K’, K=1.02/2= 0.5 trial-1, trial-2= 0.5/M2.sec
The rate r = 0.5[OH-] [A]2
when [OH] >>A, the reaction is independent of [OH-] so the reaction is pseudo order reaction
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