Table 1) Data Table 1: Varying the Concentration of 1.0 M HCL ---- C o n c e n t
ID: 871342 • Letter: T
Question
Table 1)
Data Table 1: Varying the Concentration of 1.0 M HCL
---- C o n c e n t r a t i o n s ----
#
drops
#
drops
# drops
Initial
Initial
Final
Final
Reaction Time (sec)
Reaction
Well
#
HCl
Water
Na2S2O3
HCl
Na2S2O3
HCl
Na2S2O3
Trial 1
Trial 2
Average
Rate (sec-1)
1
12
0
8
1M
0.3M
0.6M
0.12M
23.46
30.03
26.7
2
6
6
8
0.5M
0.3M
0.15M
0.12M
41.15
31.22
36.2
3
4
8
8
0.33M
0.3M
0.06M
0.12M
46.12
37.12
41.6
Table 2)
Data Table 2: Varying the Concentration of 0.3 M Na2S2O3
---- C o n c e n t r a t i o n s ---
#
drops
#
drops
# drops
Initial
Initial
Final
Final
Reaction Time (sec)
Reaction
Well
#
HCl
water
Na2S2O3
HCl
Na2S2O3
HCl
Na2S2O3
Trial 1
Trial 2
Average
Rate (sec-1)
1
8
0
12
1M
0.3M
0.4M
0.28M
31.12
25.29
43.8
2
8
6
6
1M
0.15
0.4M
0.045M
53.24
57.47
82.0
3
8
8
4
1M
0.1
0.4M
0.02M
89.86
84.93h
129.9
A) Calculate the reaction rate by taking the inverse of the average reaction time, i.e., 1 divided by the average reaction time.
B) Use table 1 to determine the reaction order for HCl.
C) Use table 2 to determine the reaction order for Na2S2O3.
D) Write the rate law for the reaction.
E) Using the rate law, the rate, and the appropriate concentration(s) from one (or more) of your
experiments calculate k.
Data Table 1: Varying the Concentration of 1.0 M HCL
---- C o n c e n t r a t i o n s ----
#
drops
#
drops
# drops
Initial
Initial
Final
Final
Reaction Time (sec)
Reaction
Well
#
HCl
Water
Na2S2O3
HCl
Na2S2O3
HCl
Na2S2O3
Trial 1
Trial 2
Average
Rate (sec-1)
1
12
0
8
1M
0.3M
0.6M
0.12M
23.46
30.03
26.7
2
6
6
8
0.5M
0.3M
0.15M
0.12M
41.15
31.22
36.2
3
4
8
8
0.33M
0.3M
0.06M
0.12M
46.12
37.12
41.6
Explanation / Answer
A) Data 1
s.no Rate(s-1)
1 0.0375
2 0.0276
3 0.0240
Data 2
s.no Rate(s-1)
1 0.0228
2 0.0122
3 0.0077
B) Rate=k * [HCl]^m [Na2S2O3]^n .......by rate law
As in table1,[Na2S2O3]=constant
so Rate=k * [HCl]^m
Rate1 =k * (1M )^m= 0.0375s-1
Also Rate2 =k * (0.5M)^m= 0.0276s-1
So taking ratio of rate1/rate2=(1M )^m/ (0.5M)^m= 0.0375/ 0.0276
2^m=1.359
m log 2 = log 1.359
m=0.1332/0.3010
m=0.44
or zero order approx
C)
As in table,[HCl]=constant
so Rate=k * [Na2S2O3]^n
Rate1 =k' * (0.3M )^n= 0.0228s-1
Also Rate2 =k' * (0.15M)^n= 0.0122s-1
So taking ratio of rate1/rate2=(0.3M )^n/ (0.15M)^m= 0.0228/ 0.0122
2^n=1.869
n=log 1.869/log2=0.272/0.3010=0.903 =first order approximately
D)Rate=k * [HCl]^0 [Na2S2O3]^1
Rate=k * [Na2S2O3] is the rate law
E) k=Rate/[Na2S2O3]
taking value from table 2,
k=0.0228s-1/0.3M=0.076M-1 s-1
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