Finding half life for first order?

Question

The decomposition of N2O5(g) to NO2(g) and O2(g) obeys the first-order kinetics. Assuming the form of the rate law is
Rate = - D[N2O5] / Dt = k[N2O5]
where k = 3.4 x 10^-5s^-1 at 25°C.
What is the half-life for the reaction described?

Answer 1

Have you learned calculus? If not, just jump over to the next section. This rate equation can be written as:
D[N2O5] / [N2O5] = -kDt
integrate it to get: Ln [N2O5] = -kt + c
or re-write it as: [N2O5] = [N2O5]_0 exp(-kt)

Therefore the half life for the first order reaction can be calculated from:
0.5 = [N2O5] /[N2O5]_0 = exp(-kt)
or: t = (Ln 2)/k =0.693/(3.4 x 10^-5s^-1) = 20387 sec