1.

For a first order reaction the rate constant at 500 K " is " 8xx10^(-4)s^(-1) . Calculate the frequency factor , if the energy of activation for the reaction is 190"kJ mol"^(-1).

Answer»

Solution :`k= 8XX10^(-4) , T = 500 K E_a=190 " kJ MOL "^(-1)A = ? `
According to ARRHENIUS equation ,
`k=Ae^(-E_a//RT)`
ln k - ln a - `E_a/(RT)`
`logk=logA-E_a/(2.303RT)`
`logA=logk+(E_a)/(2.303RT)`
`log(8xx10^(-4))+(190)/(2.303xx8.314xx10^(-3)kJ //K^(-1)xx500)`
`logA=-3.096+190/(9573.57xx10^(-3))`
`logA=16.744`
A = Antilog (16.744)
`A = 5.546xx10^(16)s^(-1)`


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