Question

The decomposition of N2O5 occurs with a rate constant of 4.3 x 10–3 sec–1 at 65C and 3.0 x 10–5 sec–1 at 25C. What is the activation energy of the process?

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 that

K' = 4.3x10^-3 s^-1

K = 3.0 x10^-5 s^-1

T' = 65^oC = 65+273 = 338 K

T = 25^oC = 25+273 = 298 K

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 298 x 338) / (338 - 298)] x log ((4.3x10^-3)/(3.0 x10^-5))     

       = 103.9x10³ J/mol

       = 103.9 kJ/mol