1.

A first order reaction is 50% complete in 30 minutes at 27^(@)C and in 10 minutes at 47^(@)C. The rate constant at 47^(@)C and energy of activation of the reaction in kJ//mole will be

Answer»

`0.0693, 43.848 kJ MOL^(-1)`
`0.0560,45.621 kJ mol^(-1)`
`0.0625, 42.926 kJ mol^(-1)`
`0.0660, 46.189 kJ mol^(-1)`

Solution :We know that `k = (2.303)/(t) log((a)/(a-x))`
Subtituting the VALUES at the two given conditions
`k_(27^(@)) = (0.0693)/(30) = 0.0231 , k_(47^(@)) = (0.693)/(10)= 0.0693`
We ALSO know that
`"log" (k_(2))/(k_(1)) = (2.303)/(t)"log"((a)/(a-x))`
or `E_(a) = (2.303 R xx T_(1)T_(2))/(T_(2) - T_(1)) "log"(k_(2))/(k_(1))`
`= (2.303 xx 8.314 xx 10^(-3) xx 300 xx 320)/(320 - 300) xx "log" (0.0693)/(0.231)`
`= 43.848 kJ mol^(-1)`


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