1.

The rate constant for the decomposition of hydrocarbons is 2.418xx10^(-5)s^(-1) at 546 K.If the energy of activaton is 179.9 KJ/mol, what will be the value of pre-exponetial factor.

Answer»

Solution :According to Arrhenius EQUATION.
log k=log A-`(E_(a))/(2.303 RT)`
log A=log `K+(E_(a))/(2.303RT)`
Where ,T=546 k
k=`2.418xx10^(-5)s^(-1)`
log k=-log `2.418xx10^(-5)=-4.6165`
`E_(a)=179.9 KJ=179900 J MOL^(-1)`
A=(?)
`therefore log A=-4.6165+(179900)/(2.303xx8.314xx546)` `therefore` A=antilog 12.5917
`=3.9057xx10^(12)=3.91xx10^(12)s^(-1)`


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