Question

If a temperature increase from 19.0 ∘C to 37.0 ∘C triples the rate constant for a reaction, what is the value of the activation barrier for the reaction?

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

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' ) ]

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

Given

T = 19.0 ^oC = 19.0+273 = 292 K

T' = 37.0 ^oC = 37.0+273 = 310 K

K' = 3xK

Plug the values we get ,

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

    E_a = [(2.303x8.314 x 292x310) / (310-292)] x log (3K / K)

          = 45941.5 J

           = 45.94 kJ

Therefore the activation barrier for the reaction is 45.94 kJ