FAHD KING UNIVERSITY OF PETROLEUM & MINERALS
CHEMI ST RY DEPART M ENT EXPERIMENTS IN PHYSICAL CHEMISTRY CHEM-31 1
Pr epar ed For
Dr. M . I . W azeer
EX P # 6 k inet ic st udy using spect r oscopy
Done By:
Al-Ghamdi, Kamal S. ID# : 901 432 SEC.#: 56
D a t e Of P er f or m a nce: 1 0 - 4 - 1 9 9 6 D a t e Of Subm i ssi on: 17-4 -19 96
INTRODUCTION
I-Objective:-To obtain the order of the reaction of crystal violet with hydroxide ion. -To obtain the rate constant of the reaction. II-Theory:The reaction of crystal violet ( (CH3)2 N )3 C+ with hydroxide ion ( OH_ ) to produce a colorless product through the following equation:CV + + OH CVOH
Voilet Colorless
and this reaction is first order. So, if we plot ln(2-log%T) vs. time we should get a straight line. III-Important Equations:Rate = d CV + dt
[
] = k [ CV ]
'
+m
k' = k OH A = log %T =
[
]
n
Io = Cl I
Io 100 I
[ CV ] ln [ CV ]
+ +
100 A = log = 2 log( % T ) % T
2 1
= k' t
ln( 2 log %T ) = k' t + Cons tan t
PROCEDURE
As indicated in the experiments' manual.
DATA
Table-1: Raw Data Run#1 t(cm) 2.5 3.9 6.4 8.45 9.3 10.2 Run#2 t(cm) 0.8 1.45 2.15 2.8 3.4 4 Run#3 t(cm) %T 2.05 45 3.4 40 4.75 55 6.05 60 7.4 65 8.85 70 Table-1: Raw Data (Cont.) Run#4 t(cm) 1.4 2.8 4.15 %T 35 40 45 %T 45 40 55 60 65 70 %T 15 20 30 40 45 50
5.4 6.5 7.6 Run#5 t(cm) 1.2 2.5 3.7 4.7 5.8 6.8 Run#6 t(cm) 1 1.8 2.25 2.85 3.35 3.85
50 55 60 %T 40 45 50 55 60 65 %T 70 75 77 80 82 85
CALCULATION
I- Calculating t(sec):After measuring t in (cm), then the unit must be converted to seconds as follows: 1cm = 0 .4 sec. t ( sec.) = 2 .5 cm sec. II- Calculating ln(2-log%T):After finding t in seconds, ln(2-log%T) will be obtained as follows:ln( 2 log %T ) = ln( 2 log 50 ) = -1.200545 So, the values of the two for each run were in Table-2:-
Table-2:Run#1 ln(2-log%T) -0.193695 -0.358147 -0.648405 -0.921454 -1.059043 -1.200545 Run#2 log(2-%T) -1.059043 -0.921454 -1.348469 -1.505759 -1.676183 -1.864962 Run#3 log(2-%T) -1.059043 -0.921454 -1.505759 -1.348469 -1.676183 -1.864962
t(sec.) 1 1.56 2.56 3.38 3.72 4.08 t(sec.) 0.32 0.58 0.86 1.12 1.36 1.6 t(sec.) 0.82 1.36 1.9 2.42 2.96 3.54
Table-2:(Cont.) Run#4 log(2-%T) -0.785411 -0.921454 -1.059043 -1.200545 -1.348469 -1.505759 Run#5 log(2-%T) -0.921454
t(sec.) 0.56 1.12 1.66 2.16 2.6 3.04 t(sec.) 0.48
1 1.48 1.88 2.32 2.72 t(sec.) 0.4 0.72 0.9 1.14 1.34 1.54 III- Finding The Slopes:-
-1.059043 -1.200545 -1.348469 -1.505759 -1.676183 Run#6 log(2-%T) -1.864962 -2.079931 -2.175870 -2.333972 -2.451245 -2.650993
After plotting ln(2-log%T) vs. t, the following equations were obtained:Run#1 Run#2 Run#3 Run#4 Run#5 Run#6 y = - 0.3233x + 0.1482 y = - 0.7171x - 0.698 y = - 0.3408x - 0.6575 y = - 0.2885x - 0.6011 y = - 0.3368x - 0.7306 y = - 0.6656x - 1.5894
So, the averages of the runs 1-3 and 4-6 were obtained as follows:Slopea = Slope b = IV- Solving For n:Slope1 + Slope 2 + Slope 3 = 0 .4604 3 Slope 4 + Slope5 + Slope6 = 0 .4303 3
After the averages of slopes were found, then n must be found as follows: OH n log OH
[ [
] ]
Slope b = log Slopea a
b
0 .0015 0 .4303 n log = log 0 .0025 0 .4604 n ( -0.22185 ) = ( -0.02936 ) n = 0.132342
V- Solving For k:The value of k was found as following:Slopea = k OH
[
]
n a 0 .132342
0 .4604 = k ( 0 .0025 ) k = 1.0174
DISSCUTION
In this experiment, the relation between ln(2-log%T) vs. t was obtained for each run:Run#1 Run#2 Run#3 Run#4 Run#5 y = - 0.3233x + 0.1482 y = - 0.7171x - 0.698 y = - 0.3408x - 0.6575 y = - 0.2885x - 0.6011 y = - 0.3368x - 0.7306
Run#6
y = - 0.6656x - 1.5894
However, the averages of slopes and the values of n and k were found:Slopea = 0 .4604 Slope b = 0 .4303 n = 0.132342
k = 1.0174
These errors were very big and they came from many sources: Human errors. Poor and unclear readings. Wrong measurements.
REFERENCES
1. Experiments in Physical Chemistry KFUPM. Chemistry Department
2. J. C. Turgeon and V. K. LaMer, J. Am. Chem. Soc., 74, 5988 (1952).