Question

In aqueous solution at 29

Answer 1

The units for the rate constant indicates that this reaction follows second order kineics.Apply arrhenius equation separately under the two given conditions , and compare both the conditions using k₂ ( to be found out for the catalysed reaction )

Arrhenius Equation is k = A e^-E_a / ^RT  

The equation is better simplified taking natural logs on each side as ,

lnk = ( - E_a / RT ). + ln A

Changing the base from natural log to (log to the base 10 ), the equation becomes-

log k = log A - E_a  / 2.303RT ----------------( use this equation for calculations as shown below )

First case ( ie. without the use of catalase )-

log  (3.7 x 10^-6 ) = log A - 75.3 / 2.303 RT

Second case ( ie .when catalase is used _

log k₂   = log A - 6.4 / 2.303 RT

Compare the two above equations as ;

log ( k₂ / 3.7 x 10 ^-6  ) = { log A - [ 6.4 / 2.303 RT ] / log A - [ 75.3 / 2.303 RT]

= log A - 6.4 /2.303RT -log A + 75.3 / 2.303 /RT

= [ 75.3 / (2.303 x .008314 x 302) ] - [ 6.4 / (2.303 x 0.008314 x302 )]

= [75.3 / 5.7824 ] - [6.4 / 5.7824]

   = 68.9 / 5.7824

ie. log k₂ - log (3.7 x 10^-6 ) = 11.9154

   Find out K₂   from the above to get the rate constant as desired.