Two reactions, (I)Ararr Product & (II) BrarrProducts follow first order kinetics. The rate reaction-(l) is doubled when the temperature is raised from 300K to 310K. The half-life for this reaction at 310K is 30min. At the same temperature B decompose twice as fast asA. If the energy of activation for reaction- (I)is twice that of reaction-(II) calculate the rate constant of reaction- (II) at 300K.

Two reactions, (I)Ararr Product & (II) BrarrProducts follow first orde

1.	Two reactions, (I)Ararr Product & (II) BrarrProducts follow first order kinetics. The rate reaction-(l) is doubled when the temperature is raised from 300K to 310K. The half-life for this reaction at 310K is 30min. At the same temperature B decompose twice as fast asA. If the energy of activation for reaction- (I)is twice that of reaction-(II) calculate the rate constant of reaction- (II) at 300K.
Answer» Solution :For a FIRST reaction , RATE constant (k) = `(0.693)/("half-life")` and `k_(1_((310)))=(0.693)/(30)=0.0231"min"^(-1)` But `(k_(1_((310))))/(k_(1_((300))))=2 ` (given) We know, `log((k_(2))/(k_(1)))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/T_(2)]""[1]` `THEREFORE`For reaction (I),`log.(k_(1_(310)))/(k_(1_(300)))=(E_(a_(1)))/(2.303R)[(1)/(300)-(1)/(310)]` `therefore`For reaction (I),`log.(k_(1_((310))))/(k_(1_((300))))=(E_(a_(1)))/(2.303R)[(1)/(300)-(1)/(310)]` For reaction [II] ,`log.(k_(2_((310))))/(k_(2_((1)(300))))=(E_(a_(1)))/(2.303R)[(1)/(300)-(1)/(310)]` ` ""[because E_(a_(2))=E_(a_(1))//2]` `thereforelog.(k_(2_((310))))/(k_(2_((300))))=((1)/(2)E_(a_(1)))/(2.303R)[(1)/(300)-(1)/(310)]=log{(k_(1_(310)))/(k_(1_(300)))}^(1//2)` `k_(2(310))/k_(2(300))={k_(1_(310))/k_(1_(300))}^(1//2)=SQRT(2)=1.414` We have , `k_(1_(310))=0.0231"min"^(-1)` `thereforek_(2_(310))=2k_(1_(310))("given")=2xx0.0231=0.0462"min"^(-1)` But, `(k_(2_((310))))/(k_(2_((300))))=1.414 thereforek_(2_(300))=(0.0462)/(1.414)=0.03267"min"^(-1)`