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A first order reaction in 50% completed in 30 minutes at 27^@C and in 10 minutes at 47^@C . Calculate the reaction rate constant at 27^@C and the energy of activation of eh reaction in kJ mol^(-1)

Answer»

SOLUTION :For a first order reaction `k=(0.693)/(t_(1//2))`
At `27^@C`
`k_(27^@C)=(0.693)/30=0.0231"min"^(-1)`
At `47^@C`
`k_(47^@C)=(0.693)/10=0.0693"min"^(-1)`
Now applying the following equation:
`log.k_2/k_1=""/(2.303R)[(T_2-T_1)/(T_1T_2)]`
or `log_(10).(0.0231)/(0.0693)=(-E_a)/(2303xx8.314).((320-300)/(320xx300))`
or `-log_(10)0.3333=(E_a)/(19.1471)xx20/96000`
`E_a=(19.1471xx96000)/20xxlog0.3333`
`=-91906xx(-0.4772)`
`=43857"J MOL"^(-1)`
`=43.857"KJ mol"^(-1)`


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