Syuzanna Abrahamyan
Chem 102
Chemical Kinetics
11/15/18
Purpose: Chemical kinetics is the study of the speed at which chemical and physical processes
take place. In a chemical reaction it is the amount of product that forms in a given interval of
time or it can be defined as the amount of reactant that disappears in a given interval of time.
Scientists that study rates at which processes occur are called kinetics.
Reactions occur at many different rates. Some reactions in geological processes occur at
imperceptibly slow speeds where product change takes place in intervals of years or even
decades. (Patel)
Q-13
Since the reaction takes place in a total volume of 50.50 mL, this volume must be taken into
account in calculating the initial concentration of the two reactants.
For example,
In Experiment 1, since the 20.00 mL of 0.200 M KI added reacts in a total volume of 50.50 m L ,
the initial concentration of [I- ]0 can be calculated as follows:
I
−
= 0.2M KI
o
20.00mL
= 0.08 M KI
50.5 mL
Similarly in experiment 1, the initial concentration of
S O = 0.1M (K S O ) 20.00mL
= 0.04 M K S O
50.5 mL
S O is calculated as follows:
2−
2
8
0
2−
2
8
o
2
2
8
2
2
Similarly in experiment 1, the initial concentration of
8
S O is calculated as follows:
2−
2
3
S O = 0.005M (Na S O ) 10.00mL
= 0.00099 M Na S O
50.5 mL
0
2−
2
3
o
2
2
3
2
2
3
Similarly for all the experiments it is calculated and tabulated below:
S O
2−
[I- ]0
Exp
1
Exp
2
Exp
3
Exp
4
0.2 M KI
2
10.00mL
50.5 mL
8
0
0.1M (K 2 S 2 O 8 )
= 0.04 M
20.00mL
0.2 M KI
50.5 mL
= 0.08 M KI
20.00mL
0.2 M KI
50.5 mL
= 0.08 M KI
20.00mL
0.2 M KI
50.5 mL
= 0.08 M KI
20.00mL
50.5 mL
= 0.04 M
0.1M (K 2 S 2 O 8 )
20.00mL
50.5 mL
= 0.04 M
0.1M (K 2 S 2 O 8 )
10.00mL
50.5 mL
= 0.02 M
0.1M (K 2 S 2 O 8 )
20.00mL
50.5 mL
= 0.04 M
Q-14
Average Elapsed Time
Run 1
Run 2
Run 3
T avg
T1(secs)
60
62
T2(secs)
30
31
61
30.5
T3(secs)
78
71
67
72
T4(secs)
138.01
148.081
143.04
Q-15
Reaction Rates In order to determine the Reaction Rates, the following quantities must be
known:
Initial concentration of [S2O3 2- ] added and completely used up
Concentration of [I2] produced, at the time the deep blue color appears.
S O added
2−
2
3
0
I 2 produced
S O
=
2−
2
3
2
All runs
0.00099
= 0.00049M
2
10.00mL
50.5 mL
= 0.00099 M Na 2S 2 O 3
0.005M (Na 2S 2 O 3 )
Q-16
Experiment
no.
1
temp
[I- ]0
S O
time
Rate(Ms-1)
20.6
0.04
0.04
61
2
19.7
0.08
0.04
30.5
3
20.0
0.08
0.02
72
4
4.3 + 3.5 + 6
= 4.6
3
0.08
0.04
143.04
0.00049
= 8.03310 −6
61
0.00049
=16.0610 −6
30.5
0.00049
= 6.810 −6
70
0.00049
= 3.4310 −6
143.04
Reaction Order with respect to iodide:
From experiment 1 and 2:
Exp1 k 0.04 M 0.04 M
8.03310 −6
=
=
Exp2 k 0.08M m 0.04 M n 16.06610 −6
m
n
0.5 = 2 − m
log 0.5 = −m log 2
m =1
Reaction order with respect to thio-sulfate:
From experiment 2 and 3:
2−
2
8
0
Exp2 k 0.08M 0.04M 16.0610 −6
=
=
Exp3 k 0.08M m 0.02M n
6.810 −6
m
n
2.36 = 2 n
log 2.36 = n log 2
n =1.23 1
Q-17
Rate Law equation: Rate = k I − S 2 O8
2−
Experiment 1
2−
I − = 0.04, S 2O8 = 0.04
Rate 1=8.033X10-6
8.03310 −6
=5.02X10-3 M-1s-1
k=
0.040.04
Experiment 2
2−
I − = 0.08, S 2O8 = 0.04
Rate 1=16.06X10-6
16.0610 −6
=5.018X10-3 M-1sk=
0.080.04
Experiment 3
2−
I − = 0.08, S 2O8 = 0.02
Rate 1=6.8X10-6
6.810 −6
=4.25X10-3 M-1s-1
k=
0.080.02
K avg=
(5.02 + 5.018 + 4.25)10 −3
= 4.7610 −3 M −1 s −1
3
Q-18
Rate law
Rate = 4.7610 −3 I − S 2 O8
Q-19
Experiment 4(ice bath)
I = 0.08, S O = 0.04
2−
−
2
8
Rate 4=3.43X10-6
2−
k=
4.4310 −6
=1.38X10-3 M-1s0.080.04
Using Arrhenius Law
k E 1 1
ln 1 = a −
k 2 R T2 T 1
4.7610 −3 E a
1
1
=
ln
−
−3
1.3810 8.314 273 + 4.6 273 + 20.1
E a = 4510 4 J / mol = 45kJ / mol
Q 1(a) what is the overall order for the reaction?
2NO( g ) + CI 2 ( g ) → 2NOCI ( g )
Solution:
The rate law of this reaction:
Rate = K NO CI 2
2
The reaction is second order in NO and first order in Cl2.The overall order for the reaction is 03.
Q1 (b) How does the reaction rate change when the nitrogen monoxide concentration is doubled
and the chlorine concentration is halved?
Solution:
For experiment 1;
Rate 1 = K NO 1 CI 1
2
For experiment 2;
Rate 2 = K NO
2
2
CI 2
NO
2
NO
2
CL2 1
2CL2 2 1 2 1
Rate1
1
1
=
=
=
=
2
2
CI 2 2 2 1 2
Rate 2 NO CI 2 2 2 2 NO
2
2
1
Thus, rate of experiment-1 ( Rate1 ) is twice the rate of experiment-2 ( Rate 2 ).
Question2: The initial rate of a reaction is found to increase by a factor of eight when the
concentration of one reagent is doubled while all other reagent concentrations are held constant.
What is the order of the reaction with respect to that one reagent? Define terms, set up the rate
Exp2
law ratios and show cancellations for
.
Exp1
Solution 2:
Rate = k A B C …………… (1)
x
y
z
When A is doubled, rate becomes 8 times.
8*Rate = k 2 A B C …………… (2)
x
y
z
From equation (1) and equation (2), we have
A ; 2 x = 8 ; thus x=3
Rate
=
8 Rate 2 Ax
x
Question 3:
Given;
At 593K, rate constant is 5.21 10 −4 and at 673K the rate constant is 7.42 10 −3 .
At 593K, the reactant’s initial concentration was 0.2264 M, the initial reaction rate was identical
to the initial rate when the decomposition was run at 673K with an initial reactant concentration
of 0.05999 M.
Thus,
k 593 0.2264 = k 673 0.05999
x
x
x
k
7.42
0.2264
= 14.2418
= 673 =
k 593 0.521
0.05999
(3.774)x
= 14.2418
x2
Thus, the order of decomposition is 2.
Question 4:
(a)
The rate of reaction for a reactant, X
−
dx
n
= K X …………………..(1)
dt
n is order of reaction is the rate of reaction
dx
n
= K X , when integrated noting
dt
that at t=0, X= Xo, initial concentration and at t=t, X= X, the equation becomes
when n=0 is zero order reaction and when Eq.1 becomes −
X =XO-Kt (2),
For zero order reaction to obey, the plot of X vs. t need to be straight line with a slope of -K
When n=1, the reaction is 1st order and Eq.1 becomes
-dX/dt= K[X]
The integrated equation is:
ln(x)= ln(Xo)-Kt,
so for 1st order reaction to obey, the plot of ln(x) vs. time ,t is straight line.
When n=2, the reaction is 2nd order and eq.1 becomes
-dX/dt= K[X]2,
The integrated equation is
1/X= 1/XO+Kt,
so the plot of 1/X vs t need to be straight line with a slope K.
y = 0.005t + 1.109
R² = 0.997
14
12
y = -0.0004t+ 0.7220
R² = 0.6675
10
X,lnX,1/X
8
6
zero order
4
first order
second order
2
0
-2
0
500
1000
-4
1500
t
2000
2500
y = -0.001t - 0.408
R² = 0.880
(b) Since the second order plot only gives a straight line, the decomposition reaction is second
order with respect to A.
(c)
Second order equation is:
1/X= 1/XO+Kt
Best fit line by graph is :
y = 1.109 + 0.005t
1 / X = 1/X o + Kt
K = 0.005
1/X o = 1.109
X o = 0.902
(d)
Rate law equation
-d X/d t= 0.005[X]2,
Reference: Patel, N.A.; Swope, C. Chemical Kinetics. J. Phys. Chem.
A [Online] 2001. 92, 1038-1050.