Consider the following data for the reaction: 2ClO(aq) + 2OH-(aq) ----> ClO-3(aq
ID: 850170 • Letter: C
Question
Consider the following data for the reaction:
2ClO(aq) + 2OH-(aq) ----> ClO-3(aq) + ClO-2(aq) + H2O (l)
[ClO]0 (mol/L) [OH-] (mol/L) Initial Rate (mol/L*s)
0.0500 0.100 0.0575
0.100 0.100 0.230
0.100 0.0500 0.115
a.) What is the rate law for this equation?
b.) What is the value of the rate constant?
c.) What would be the initial rate for an experiment where [ClO2]0 = 0.175 mol/L and [OH-]0 = 0.0844 mol/L
Explanation / Answer
if we observe Exp1 & exp2
rate 1 = K[ClO]^x[OH-]^y
0.0575 = K[0.05]^x [0.1]^y .........1
rate2 = K[ClO]^x[OH-]^y
0.230 = K[o.1]^x [0.1]^y ............2
do 1/2
0.25 =[0.5]^x
(0.5)^2 = (0.5)^x
so x = 2
order w.r.t [Clo] is 2
if we observe Exp2 & exp3
rate 3 = K[ClO]^x[OH-]^y
0.23 = K[0.1]^x [0.1]^y .........3
rate 4 = K[ClO]^x[OH-]^y
0.115 = K[0.1]^x [0.05]^y .........4
do 3/4
2 = 2^y
so Y = 1
order w.r.t [OH] is 2
a)
rate = K3[ClO]^2[OH-]^1
b)
K3 = 0.230/(0.1)^2*(0.1)
K3 = 230 M^-2 sec^1
c) rate = 230*(0.175)^2 (0.0844)
new rate = 0.5945 M/sec