Question

The rate constant for a first-order decomposition reaction is 4.60 x 10^-4 at 250 ^oC. If it takes 100.0 kJ/mol for the activated complex to form at that temperature, calculate the rate constant when the temperature is 50 degrees higher.

Answer 1

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

Where

K = rate constant

T = temperature

R = gas constant = 8.314*10^-3 kJ/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' ) ]

Given

K = 4.60*10^-4

K' = ?

T = 25oC = 25+273 = 298 K

T' = 50 oC = 50+273 = 323 K

Ea = activation energy = 100.0 kJ/mol

Plug the values we get

log ( K' / K ) = ( 100.0/ (2.303* 8.314*10^-3 ) x [ ( 1/ 298 ) - ( 1 / 323 ) ]

                   = 1.356

    K'/K = 10 1.356 = 22.7

      K' = 22.7*4.60*10^-4

          = 10.45*10^-3