the activation energy increases by 5.75 kJ/mol. The ratio of rate constant k2/k1 will be

Sahay Sir > Question Answers > the activation energy increases by 5.75 kJ/mol. The ratio of rate constant k2/k1 will be
For nth order reaction (dx/dt)=Rate = k[A]0n Graph between log (rate) against log[A0] is of the type Lines P, Q, R, S are of the order:
02
Jun
For nth order reaction (dx/dt)=Rate = k[A]0n Graph between log (rate) against log[A0] is of the type Lines P, Q, R, S are of the order: -q For nth order reaction (dx/dt)=Rate = k[A]0n Graph between log (rate) against log[A0] is of the type Lines P R S are of the order: the activation energy [...]
Posted in: AIIMS Last 24 Years Solved 1996 - 2019 Physics and Chemistry Video Solutions , Chapter 17 - Chemical Kinetics , Chemistry ,
Tags: -q , For nth order reaction (dx/dt)=Rate = k[A]0n Graph between log (rate) against log[A0] is of the type Lines P , R , S are of the order: , the activation energy increases by 5.75 kJ/mol. The ratio of rate constant k2/k1 will be ,
On introducing a catalyst at 300 k, the activation energy increases by 5.75 kJ/mol. The ratio of rate constant k2/k1 will be
02
Jun
On introducing a catalyst at 300 k, the activation energy increases by 5.75 kJ/mol. The ratio of rate constant k2/k1 will be On introducing a catalyst at 300 k the activation energy increases by 5.75 kJ/mol. The ratio of rate constant k2/k1 will be June 2, 2021 Category: AIIMS Last 24 Years Solved 1996 - [...]
Posted in: AIIMS Last 24 Years Solved 1996 - 2019 Physics and Chemistry Video Solutions , Chapter 17 - Chemical Kinetics , Chemistry ,
Tags: On introducing a catalyst at 300 k , the activation energy increases by 5.75 kJ/mol. The ratio of rate constant k2/k1 will be ,