Question

A certain reaction has an activation energy of 65.39 kJ/mol. At what Kelvin temperature will the reaction proceed 3.50 times faster than it did at 337 K?

Answer 1

According to Arrhenius Equation , K = A e ^{-Ea / RT}

Where

K = rate constant

T = temperature

R = gas constant = 8.314 J/(mol-K)

E_a = activation energy = 65.39 kJ/mol

                                    = 65390 J/mol

A = Frequency factor (constant)

Rate constant, K = A e ^{- Ea / RT}

                  log K = log A - ( E_a / 2.303RT )   ---(1)

If we take rate constants at two different temperatures, then

                log K = log A - ( E_a / 2.303RT )   --- (2)

    &         log K' = log A - (E_a / 2.303RT’)    ---- (3)

Eq (3 ) - Eq ( 2 ) gives

log ( K' / K ) = ( E_a / 2.303 R ) x [ ( 1/ T ) - ( 1 / T' ) ]

Given K' = 3.5 K

         T = 337 K

         T' = ?

       E_a = 65390 J/mol

PLug the values we get [ ( 1/ T ) - ( 1 / T' ) ] = ( 2.303 R / E_a ) x log ( K' / K )

                                                                     = [(2.303 x 8.314)/65390 ] x log ( 3.50K/K)

                                                                     = 1.59x10^-4

                                     (1/337) - ( 1/T' ) = 1.59x10^-4

                                       2.96x10 ^-3 - (1/T' ) = 1.59x10^-4

                                               (1/T' ) = 2.96x10 ^-3 - 1.59x10^-4

                                                                    = 2.80 x10^-3

                                                               T' = 1 / ( 2.80 x10^-3 )

                                                                   = 356 K

Therefore the required temperature is 356 K