1.

The activation energyfor a reaction at the temperature T K was found to be 2.303 RT J mol^(-1) . The ratio of the rate constant to Arrhenius factor is :

Answer»

`10^(-1)`
`10^(-2)`
`2XX10^(-3)`
`2x10^(-2)`

SOLUTION :(A) According to Arrhenius equation
`k = Ae^(-E_(a)//RT) `
or log k = log `A-(Ea)/(2.303RT)`
`E_(a)= 2.303 RT`
`:. ` log k = log A `-(2.303RT)/(2.303RT)`
log k = log A-1
or log k - log A=-1
log `(k)/(A) =-1`
`(k)/(A) = 10^(-1)`


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