isn't half-life a PC game? just kidding.. LOL

the equation

ln([A]1/[A0]) = -kt

Objectivist had given the reaction profile for a first-order reaction, not a 2nd order one.

the rate law is given by:

- dA/dt = k[A]

^{2}-[A]

^{-2} dA/dt = k

integrating both sides with respect to t,

[A]

^{-1} = kt + C (integration constant)

using initial conditions when t = 0, [A] = [Ao] => C = [Ao]

^{-1}hence, the reaction profile is given by:

[A]

^{-1} - [Ao]

^{-1} = kt

When [A]0 is .250 M, the reaction is 30.0% complete in 50.0 minutes.

when t = 50min, [Ao] = .250M, [A] = 70.0% x [Ao] = 0.175

0.175

^{-1} - 0.250

^{-1} = k(50)

k = 0.0343 dm

^{3}mol

^{-1}min

^{-1}to find the half-life of the reaction, ie. when the reactants are 50% are used up, just plug in all the necessary values into the reaction profile equation, given you know k, [Ao], [A] ( = 50% x [Ao])