1.

A firstordergas -phasereaction hasan energyofactivationof 240 kJ mol^(-1). Iffrequencyfactorof thereactionis 1.6 xx 10^(13) s^(-1) , calculateitsrateconstant at 600 K .

Answer»


Solution :Given :ENERGYOF activation` = E_(a) 240 kJ mol^(-1) = 240 XX 10^(3) J mol^(-1)`
Frequencyfactor = A = `1.6 xx 10^(13) s^(-1)`
Temperature`= T= 600 K ` Rateconstant= k = ?
By Arrheniusequation,
`k= A xx e^(-E_(a)//RT) :." In" k = " In" A - (E_(a))/(RT)`
`:. 2.303log _(10) k= 2.303 log_(10)A- (E_(a))/(RT)`
`log_(10)k =log_(10) A - (E_(a))/(2.303 RT) = log_(10)1.6 xx 10^(13)- (240 xx 10^(3))/(2.303 xx 8.314 xx 600)`
`= 13. 204- 20.9= - 7.696`
`:. k= AL-7 .696= ALbar(8).304 = 2.01 xx 10^(-8) s^(-1)`


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