Question

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

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

so I did
At= .467 *e ^(-.016*12) = .068 M ^-1 * S^-1

Does that look right?

Answer 6

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

Answer 7

I thought that's what M^-1 meant

Answer 8

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

Answer 9

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: \(\dfrac{1}{[A]}=\dfrac{1}{[A]_0}+kt\)

Answer 10

it should actually work this time.