Please answer 1-6 and 13 As you determined above, the general form of this rate

Question

Please answer 1-6 and 13

As you determined above, the general form of this rate law is: Because the rate constant, k, and the concentration of hydroxide, [OH ] did NOT change from Experiment 1 to Experiment 2. a proportionality can be written: This also means that the ratios determined above follow the same relationship What is the order (value of x) of CIO2 ? As you determined above, the general form of this rate law is: Because the rate constant, k, and the concentration of hydroxide, [CIO2] did NOT change from Experiment 1 to Experiment 2. a proportionality can be written: This also means that the ratios determined above follow the same relationship What is the order (value of y) of OH ?

Explanation / Answer

by observing the exp 1* exp2
raet1(0.0248) = K[0.06]^x [0.03]^y ........1
rate2(0.00276) = K[0.02]^x [0.03]^y .........2

do 1/2 =
0.0248/0.00276 = (0.06/0.02)^x

(9) = (3)^x
(3)^2 = (3)^x
so the order is 2 w.r.t [ClO2]

by observing the exp 2* exp3

rate2(0.00276) = K[0.02]^x [0.03]^y ......3
raet3(0.00828) = K[0.02]^x [0.09]^y .......4

do 3/4 =

(0.00276/0.00828) =(0.03/0.09)^y
(3) = (3)^y
so the order w.r.t [OH-] is 1

rate law is
rate = K3[CLo2-]^2 [OH-]^1
overal order of reaction is 3

0.00276 = K[0.02]^2 [0.03]^1
K = 230 M^-2 sec^-1

units for this 3rd order reaction is M^-2 sec^-1

13) doubling the conc of [ClO2-] and [OH-]

rate = [2*ClO2-]^2 [2OH-]^1
rate = 8 [ClO2-][OH-]

the new rate will become 8 times of initial rate of reaction

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