Calculate the `log_(10)` of the ratio of the catalysed and uncatalysed rate constant at `25^(@)`C. If at this temperature the energy of activation of a catalysed reaction is 162KJ and for the uncatalysed reaction the value is 350 KJ
(Assume frequency factor is same for both reaction)

Options

A. 28.94
B. 32.94
C. 36.94
D. 40.94

Answer

B

Explanation

` K = Ae^( Ea//RT ) rArr log K = log A - (Ea )/( 2.303 RT ) `
` log K _ ("cat.")= log A - (162 xx 10 ^ 3 ) /( 2.303 RT ) `…(1)
` log K _ ("uncat.") = log A - (350 xx10^(3))/(2.303 RT ) `….(2)
eq. (1) - eq. (2)
`log K _ ( "cat. ") - log K _ ("uncat.") = ((350 - 162 ) xx 10 ^ 3 ) /( 2.303 RT ) `
` log (K _ ("cat."))/(K_("uncat."))= 32.94 `