the following data were obtained for the reaction:
2ClO2(aq) + 2OH – (aq) —-> ClO3- (aq) + ClO2- (aq) + H2O (l)
[ClO2]0 [OH-]0 initial rate
(mol/L) (mol/L) (mol/L*s)
0.0500 0.100 5.75*10^ -2
0.100 0.100 2.30*10^ -1
0.100 0.050 1.15*10^ -1
a) determine the rate law and the value of the rate constant
b) what would be the initial rate for an experiment with
[ClO2]0 = 0.175 mol/L and [OH-]0 = 0.0844 mol/L ?
1 Answer
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a)
rate = k[ClO2]^x [OH-]^y
row 2 / row 1:
2.30e-1/ 5.75e-2 = [0.100/0.0500]^x [0.100/0.100]^y
4 = 2^x
x = 2
row 2/ row 3:
2.30e-1/ 1.15e-1 = [0.100/0.100]^2 [0.100/0.050]^y
2 = 2^y
y = 1
Solve for k with row 2:
2.30e-1M/s = k[0.100 M]^2 [0.100 M]
k = 230(s-M^2)^(-1)
rate = (230 (s-M^2)^(-1))[ClO2]^2 [OH-]
b)
Plug values into rate equation found in a):
rate = (230 (s-M^2)^(-1))[0.175 M]^2 [0.0844 M]
rate = 0.594 M/s
