Question

The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as

k=Ae−Ea/RT

where R is the gas constant (8.314 J/mol⋅K), A is a constant called the frequency factor, and Eais the activation energy for the reaction.

However, a more practical form of this equation is

lnk2k1=EaR(1T1−1T2)

which is mathmatically equivalent to

lnk1k2=EaR(1T2−1T1)

where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2).

Part A

The activation energy of a certain reaction is 46.7 kJ/mol . At 26  ∘C , the rate constant is 0.0120s−1. At what temperature in degrees Celsius would this reaction go twice as fast?

Express your answer with the appropriate units.

Part B

Given that the initial rate constant is 0.0120s−1 at an initial temperature of 26  ∘C , what would the rate constant be at a temperature of 180  ∘C for the same reaction described in Part A?

Express your answer with the appropriate units.

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' ) ]

Given E_a = 46.7 kJ/mol

               = 46.7x10³ J/mol

K = 0.0120 s^-1

K' = 2K = 2x0.0120 = 0.0240 s^-1

T = 26 oC = 26+273 = 299 K

T ' = ?

Plug the values we get

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

log ( 2K / K ) = ( (46.7x10³ )/ (2.303x8.314) ) x [ ( 1/ 299 ) - ( 1 / T' ) ]

[ ( 1/ 299 ) - ( 1 / T' ) ] = 1.234x10^-4

T' = 310.4 K

      = 310.4 -273

     = 37.4 ^oC

Part B :-

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

Given E_a = 46.7 kJ/mol

               = 46.7x10³ J/mol

K = 0.0120 s^-1

K' =?

T = 26 oC = 26+273 = 299 K

T ' = 180 oC = 180+273 = 453 K

Plug the values we get

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

log( K'/0.012) = ((46.7x10³ )/(2.303x8.314) x [ (1/299) - (1/453) ]

                      = 2.77

       K' / 0.012 = 10 ^2.77 = 593.05

              K' = 7.12 s^-1