Trial Determine [FeCl 3 ] in the reaction: C i V i =C f V total Determine [KI] i
ID: 890162 • Letter: T
Question
Trial
Determine [FeCl3] in the reaction:
CiVi=CfVtotal
Determine [KI] in the reaction:
CiVi=CfVtotal
Vtotal(mL)
Initial rate r
(s-1)
Ci(M)
Vi(mL)
Cf(M)
Ci(M)
Vi(mL)
Cf(M)
1
0.020
20.0
.01
0.020
20.0
.01
40.0
.00068
2
0.020
20.0
.01
0.020
10.0
.01
40.0
.00238
3
0.020
10.0
.005
0.020
20.0
.005
40.0
.00165
4
0.020
15.0
.0075
0.020
10.0
.0075
40.0
.00385
5
0.020
10.0
.005
0.020
15.0
.005
40.0
.00340
r = k [FeCl3]m[KI]n
Where r is the initial rate, k is the rate constant, m is the order of the reaction in FeCl3, and n is the order of the reaction in KI. Based on the data table, please determine m and n.
m=
n=
Trial
Determine [FeCl3] in the reaction:
CiVi=CfVtotal
Determine [KI] in the reaction:
CiVi=CfVtotal
Vtotal(mL)
Initial rate r
(s-1)
Ci(M)
Vi(mL)
Cf(M)
Ci(M)
Vi(mL)
Cf(M)
1
0.020
20.0
.01
0.020
20.0
.01
40.0
.00068
2
0.020
20.0
.01
0.020
10.0
.01
40.0
.00238
3
0.020
10.0
.005
0.020
20.0
.005
40.0
.00165
4
0.020
15.0
.0075
0.020
10.0
.0075
40.0
.00385
5
0.020
10.0
.005
0.020
15.0
.005
40.0
.00340
Explanation / Answer
Let us first recalculate Cf for both FeCl3 and KI
in M
Using Cf = CiVi/Vf
Trial FeCl3 Cf (M) KI Cf (M) Initial rate (s-1)
1 0.01 0.01 0.00068
2 0.01 0.005 0.00238
3 0.005 0.01 0.00165
4 0.0075 0.005 0.00385
5 0.005 0.0075 0.00340
Now we have,
r = k[FeCl3]^m[KI]^n
lets take trial 3 and trail 5
r3/r = 0.00068/0.00238 = k1(0.01)^m(0.01)^n/k2(0.01)^m(0.005)^n
0.286 = 2^n
Taking log on both sides,
log(0.20732) = nlog(2)
n = 2
Now take trial 2 and trial 4, th concentration of KI is same so cancels out,
r2/r4 = 0.00238/0.00385 = (0.01/0.0075)^m
again taking log on both side,
m = 6
So the rate law becomes,
r = [FeCl3]^3[KI]^1
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