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Chemistry
OpenStudy (anonymous):

Given that the rate constant for the decomposition of hypothetical compound X from part A is 1.60 M^−1⋅min^−1, calculate the concentration of X after 12.0min . Time (min) [X](M) 0 0.467 1 0.267 2 0.187 3 0.144 4 0.117 5 0.099 6 0.085 7 0.075

OpenStudy (anonymous):

I'm using formula [X]=1/ kt+1[X]0 and getting 21.34 but I'm not sure about units or if this is right.

OpenStudy (anonymous):

so x = 1 / (1.60 M^−1⋅min^−1) (12) + 1(.467) = 21.34 ?

OpenStudy (aaronq):

think about it, how can you have more than what you started with if it's decomposing?

OpenStudy (aaronq):

use: \(\large \sf A_{t}=A_0*e^{-kt}\), where \(\sf A_{t}\) is the amount after t time elapsed, \(\sf A_0\) is the initial amount. t is time, and k is the decay constant

OpenStudy (anonymous):

so I did At= .467 *e ^(-.016*12) = .068 M ^-1 * S^-1 Does that look right?

OpenStudy (aaronq):

why did you make k=0.016, when it's 1.6

OpenStudy (anonymous):

I thought that's what M^-1 meant

OpenStudy (anonymous):

I got 2.14* 10^-9 but I don't know if that's right.

OpenStudy (aaronq):

Damn, i just noticed it's a 2nd order reaction (from the units of k), the formula i gave is only for a first order. use the 2nd order integrated rate law: \(\dfrac{1}{[A]}=\dfrac{1}{[A]_0}+kt\)

OpenStudy (aaronq):

it should actually work this time.

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