My question is, is this reaction a order 1 reaction? and if so, is the rate law
ID: 981168 • Letter: M
Question
My question is, is this reaction a order 1 reaction? and if so, is the rate law rate=k[HCI]^1[na2s203]^1
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
0.12M
21.88
21.47
21.67
.046
2
8
6
6
1M
0.15
0.7M
0.11M
46.13
41.72
43.92
.022
3
8
8
4
1M
0.1
0.8M
0.08M
85.91
74.09
80
.016
My question is, is this reaction a order 1 reaction? and if so, is the rate law rate=k[HCI]^1[na2s203]^1
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
0.12M
21.88
21.47
21.67
.046
2
8
6
6
1M
0.15
0.7M
0.11M
46.13
41.72
43.92
.022
3
8
8
4
1M
0.1
0.8M
0.08M
85.91
74.09
80
.016
Explanation / Answer
Keeping [HCl] constant and halving [Na2S2O3] the reaction rate becomes 0.022/0.046 which is nearly 0.5
Thus the order with respect to [Na2S2O3] is 1.
The order with respect to [HCl] cannot be determined using above data since its not being varied only.
It can be assumed that [HCl] is in excess
Hecne the rate law is
rate=k[na2s203]^1
The overall order is 1
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