What rate law and integrated rate lets you calculate the time required to let yo

Question

What rate law and integrated rate lets you calculate the time required to let your [CV+] change to half of its value at t=0? Second order rate law and its integrated rate is useful to calculate [CV+] to half life at t =0.

We want to find the rate law that describes how reactant concentrations affect the rate of the reaction shown by: CV+ + OH- --------------- CVOH

The general form of the rate law we will find is: rate = k[CV+]x [OH-]

where x, y are the reaction orders with respect to reactanst, x+y is the overall reaction order and k is the rate constant.

so far, I have found rate = 2.0 * 10-6 and k=1.25.

Now, I need to calculate the time needed for the concentration to be halved.

Explanation / Answer

rate = k[CV+]x [OH-]^y

it is second order reaction

so x = y = 1

rate = k[CV+] [OH-]

2.0 x 10-6 = 1.25 x   [CV+] [OH-]

[CV+] [OH-] = 1.6 x 10^-6

if [CV+] = [OH-]

concentration = 1.3 x 10^-3 M

initial concentration of reactant = 1.3 x 10^-3 M = Ao

half of the concentration = At = 6.5 x 10^-4 M

now for second order reaction

t = 1/k [1/At] - 1/Ao ]

t = 1/ 1.25 [1/ 6.5 x 10^-4 - 1/ 1.3 x 10^-3]

t = 615.38 second

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