Question

Given that the initial rate constant is 0.0190s−1 at an initial temperature of 24  ∘C , what would the rate constant be at a temperature of 100.  ∘C for the same reaction described in Part A?

(Part A The activation energy of a certain reaction is 47.0 kJ/mol . At 24  ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast? T2 = 35C)

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' ) ]

K = 0.0190 s^-1

K' = 2K

T = 24 oC = 24+273 = 297 K

T' = ?

Plug the values we get

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

= 1.226x10^-4

T' = 308 K

= 308-273

= 35 oC