Q. At $ 500 \,K $ , the half-life period of a gaseous reaction at an initial pressure of $ 80\, kPa $ is $ 350 \,s $ . When the pressure is $ 40 \,kPa $ , the half-life period is $ 175 \,s $ . The order of the reaction is
Solution:
$ {{P}_{1}}=80\,kPa,({{t}_{1/2}})=350s $
$ {{P}_{2}}=40\,kPa,{{({{t}_{1/2}})}_{2}}=175s $
$ \frac{80}{40}=\frac{350}{175}=2 $
$ \because $ $ \frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{({{t}_{1/2}})}_{1}}}{{{({{t}_{1/2}})}_{2}}}=\frac{{{a}_{1}}}{{{a}_{2}}} $
$ \therefore $ $ ={{t}_{1/2}}\propto a $ (zero order reaction)
Note: For $ {{n}^{th}} $ order reaction $ {{t}_{1/2}}\propto \frac{1}{{{(a)}^{n-1}}} $
