On introducing a catalyst at 500 K the rate of a first order reaction increases by 1.718 times. The activation energy in presence of catalyst is 4.15 kJ mol−1. The slope of the plot of ln⁡k(in sec−1) versus 1T (T in Kelvin) in absence of catalyst is (R=8.3 J mol−1 K−1)

k=Ae−Ea/RTk'=Ae−Ea'/RT(where k = rate constant for non catalysed reaction and k' = rate constant for catalysed reaction. Ea=activation energy for non-catalysed reaction and Ea'=activation energy for catalysed reaction) k'k=eEa−E'a/RTAlso given k'=k+1.718k=2.718k∴ 2.718=eEa−Ea'RTloge2.718=Ea−Ea'8.314×10−3×500Ea−Ea'=4.11∴ Ea=8.3 kJ/mol−1k=Ae−Ea/RT∴logek=logeA−EaRTThis is an equation for straight line with slope =−EaR=−8.38.3×10−3                                                                                         =-1000