The rate constant for the first order decomposition of H_(2)O_(2) is given by the following equation: log k=14.34-1.25xx10^(4)K//T [Where unit of k is s^(-1)] Calculate E_(a) for this reaction and at what temperature will its half-period be 256 minutes?

The rate constant for the first order decomposition of H_(2)O_(2) is g

1.	The rate constant for the first order decomposition of H_(2)O_(2) is given by the following equation: log k=14.34-1.25xx10^(4)K//T [Where unit of k is s^(-1)] Calculate E_(a) for this reaction and at what temperature will its half-period be 256 minutes?
Answer» SOLUTION :Calculation of `E_(a)`:The REACTION is first order. The decomposition of `H_(2)O_(2)` occurs in reaction .It is given that, log k=`14.34-1.25xx10^(4)K//T` ….(a) log k=`log A-(E_(a))/(2.303 RT)` (Arrhenius(b)) Comparing equation (a) and (b), log k and 14.34 =log A is accepted and removing it, So,`(E_(a))/(2.303RT)=(1.25xx10^(4))/(T)k` `therefore E_(a)=1.25xx104 kxx2,303xx8.314 JK^(-1) mol^(-1)` Calculation of decomposition constant k on the base of `t_((1)/(2))`: `k=(0.693)/(t)` Where ,`t_((1)/(2))=256` MINUTE `=(0.693)/(256xx60s)` Calculation of `t_((1)/(2))` TEMPERATURE: Given equation :log k=14.34-`(1.25xx10^(4)K)/(T)` `therefore log(4.51xx10^(-5))=14.34-(1.25xx10^(4)K)/(T)` `therefore -4.3458-14.34=-(1.25xx10^(4)K)/(T)` `therefore-18.6858=-(1.25xx10^(4)K)/(T)` `therefore T=(1.25xx10^(4)K)/(18.6858)=668.96 K=669K`