Homework Answers
9)
see experiment 1 and 3:
[ClO2] doubles
[OH-] is constant
rate becomes 4 times
so, order of ClO2 is 2
see experiment 2 and 1:
[ClO2] is constant
[OH-] doubles
rate doubles
so, order of OH- is 1
Rate law is:
rate = k*[ClO2]^2*[OH-]
Answer: b
10)
Given:
Half life = 8.02 days
use relation between rate constant and half life of 1st order reaction
k = (ln 2) / k
= 0.693/(half life)
= 0.693/(8.02)
= 8.641*10^-2 days-1
we have:
[I]o = 100 [Let initial concentration be 100]
[I] = 95
k = 8.641*10^-2 days-1
use integrated rate law for 1st order reaction
ln[I] = ln[I]o - k*t
ln(95) = ln(1*10^2) - 8.641*10^-2*t
4.554 = 4.605 - 8.641*10^-2*t
8.641*10^-2*t = 5.129*10^-2
t = 0.5936 days
Answer: a
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9. Given the initial rate data below, what is the rate law for the following reaction?...
The following data were obtained for the reaction given below. 2 C102(aq) + 2 OH "(aq)...
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Please assist with Section C
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Please assist with Section C
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