1.

A hydrogenation reaction is carried at 500K. If the same reaction is carried out in presence of the catalyst at the same rate, the temperature required is 400K. If the threshold energy of the reactant is 120 kJ mol^(-1) and the normal energy of the reactant is 20kJ mol^(-1) , calculate the magnitude of activation barried ( in kJ mol^(-1) ) that is lowered by the catalyst.

Answer»

Solution :Let the amount of activation BARRIER LOWERED is x `kJ mol^(-1)` is presenve of catalyst.
`:. K = A .e^(-) ( Ea)/( R T)`
For uncatalysed reaction,
`K_(1) = A. e^(-) ( Ea_(1))/( R T_(1))`
For catalysed reaction
`K_(2) = A. e^(-) ( Ea_(2))/( R T_(2))`
From equation, `K_(1) = K_(2)`
`e^(-) ( Ea_(1))/( RT_(1)) = e^(-)( Ea_(2))/( RT_(2)) RARR( Ea_(1))/( T_(1)) = ( Ea_(2))/( T_(2))`
`(Ea_(1))/( T_(1)) = ( Ea_(1) - x)/( T_(2)) [ :. Ea_(2) = Ea_(1) -x]`
`:. ( 100)/( 500)= ( 100 - x)/( 400)rArr x = 20 kJ mol^(-1)`


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