1.

A first order reaction is 50% complete in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction (Given: log 2 = 0.3010. log 4 = 0.6021, R = 8.314 JK^–1 mol^–1)

Answer»

Solution :`K_2=0.693//20`,
`k_1=0.693//40`
`LOG k_2/k_1=E_2/2.303 R [1/T_1-1/T_2]`
`k_2//k_1=2`
`log 2 =E_8/(2.303 times 1.314) [(320-300)/(320 times 300)],Ea=27.66 KJ Mol^-1`


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