Complete the following table. Trial [FeCl 3 ]( M ) [KI] ( M ) Initial rate (unit
ID: 997789 • Letter: C
Question
Complete the following table.
Trial
[FeCl3](M)
[KI] (M)
Initial rate (units of absorbance/s)
Value must have >1 sig figure
1
0.01
0.01
0.00235
2
0.01
0.01
0.00365
3
0.01
0.005
0.00287
4
0.01
0.005
0.00219
5
0.005
0.01
0.00216
6
0.005
0.01
0.00242
1) Compare the average rate from trials 1 and 2 to the average rate of trials 3 and 4 to determine the order in I-. Report the order in I-. Show your work.
2)Compare the average rate from trials 1 and 2 to the average rate of trials 5 and 6 to determine the order in Fe3+. Report the order in Fe3+. Show your work.
3) Write out the rate law.
4) Using the units of rate as “units of absorbance”/s, determine a value for the rate constant. Show your work.
Trial
[FeCl3](M)
[KI] (M)
Initial rate (units of absorbance/s)
Value must have >1 sig figure
1
0.01
0.01
0.00235
2
0.01
0.01
0.00365
3
0.01
0.005
0.00287
4
0.01
0.005
0.00219
5
0.005
0.01
0.00216
6
0.005
0.01
0.00242
Explanation / Answer
Average
average rate
Remarks
FeCl3
KI
Absorbance/s
0.01
0.01
0.003
Average of trial-1 and 2
0.01
0.005
0.00253
Average of trial-3 and 4
0.005
0.01
0.00229
Average of 5 and 6
Let the rate –r= K[FeCl3]a [KI]b
Where a and b are orders with respect to FeCl3 and KI. K is rate constant
From 1 K[0.01]a[0.01]b= 0.003 (1)
From 2 K[0.01]a [0.005]b= 0.00253 (2)
Eq.2/Eq.1 gives (0.005/0.01)b= 0.00253/0.003
0.5b= 0.843
Taking ln
bln(0.5)= ln(0.843), b=0.25
from 3rd data, K [FeCl3]a [KI]b= 0.00229 (3)
Eq.3/Eq.1 gives
(0.5)a= 0.00229/0.003=0.7633
Taking ln
aln(0.5)=ln(0.7633), a=0.4
so the rate becomes –r= K[ Fecl3]0.4 [KI]0.25
from Eq.1
From 1 K[0.010]0.4[0.01]0.25= 0.003 (1)
K= 0.059/sec.(M)0.65
Average
average rate
Remarks
FeCl3
KI
Absorbance/s
0.01
0.01
0.003
Average of trial-1 and 2
0.01
0.005
0.00253
Average of trial-3 and 4
0.005
0.01
0.00229
Average of 5 and 6