question_answer

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

A)  zero 

B)                                         one

C)  two                      

D)         three

E)  half

Correct Answer: A

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                 \[{{t}_{1/2}}\propto \frac{1}{{{(a)}^{n-1}}}\]