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)

Question

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)

Infinity Learn · Accepted Answer

$$k=Ae$$−$$Ea/RTk$$'$$=Ae$$−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 $$=$$−$$EaR=$$−$$8.38.3$$×$$10$$−$$3$$                                                                                         $$=-1000$$

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)

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