Question

A reaction has a rate constant of 1.23×10⁻⁴ s−1 at 26  ∘C and 0.232 s−1 at 79  ∘C .

Part A

Determine the activation barrier for the reaction.

Express your answer in units of kilojoules per mole.

Part B

What is the value of the rate constant at 18  ∘C ?

Express your answer in units of inverse seconds.

Answer 1

Part A :-

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 K = 1.23×10⁻⁴ s−1

K' = 0.232 s−1

T = 26 oC = 26+273 = 299 K

T' = 79 oC = 79+273 = 352 K

Plug the values we get E_a = [(2.303x8.314 x 299x 352) / (352 - 299)] x log (0.232 / (1.23x10^-4))

                                      = 124.5x10³ J/mol

                                      = 124.5 kJ/mol

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Part B :-

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

Given K = 1.23×10⁻⁴ s−1

K' = ?

T = 26 oC = 26+273 = 299 K

T' = 18 oC = 18+273 = 291 K

E_a = 124.5x10³ J/mol

log ( K' / K ) = ( (124.5x10³ )/ (2.303x8.314)) x [ ( 1/ 299 ) - ( 1 / 291 ) ]

                  = -0.598

       K' / K = 10-0.598 = 0.252

            K' = 0.252 x K = 0.252 x 1.23×10⁻⁴ s−1

                = 3.10×10⁻⁵ s−1