Lets give it a try.

Here is what I propose:

* At time t=0, reactants are at the point P with an initial CH

_{3}^{.} concentration named

**C**_{0}.

* After 22cm and a speed of 14m/s, time is

**t**_{22}=0.22/14=0.0157 s, and the concentration is

**C**_{22}=C_{0} e^{-kt22} (equation 1)

* After 37cm and a speed of 14m/s, time is

**t**_{37}=0.37/14=0.0264 s, and the concentration is

**C**_{37}=C_{0} e^{-kt37} (equation 2)

* The speed at which antimony is consumed is supposed to be proportional to the concentration of methyl radicals. So, if the second mirror takes 150/45=3.333 more times to disappear than the first one, it's because the concentration is 3.333 times less:

**C**_{37}=C_{22}/3.333 so

**C**_{22}/C_{37}=3.333 (equation 3)

* From (equation 1)/(equation 2), we get:

**C**_{22}/C_{37}=e^{k(t37-t22)} (equation 4)

* From (equation 3) and (equation 4), we get

**e**^{k(t37-t22)}=3.333. Knowing the numerical values for t

_{22} and t

_{37}, we can calculate

**k=1/(t**_{37}-t_{22}) ln(3.333) **k=112.51 s**^{-1}* For half life time t

_{1/2}, we have

**C**_{1/2}=C_{0}/2 and

*C*_{1/2}=C_{0} e^{-kt1/2}. When combined, it gives

*1/2=e*^{-kt1/2} *t*_{1/2}=-ln(1/2)/kThe half life time is

*t*_{1/2}=0.00616 s