The Arrhenius equation shows the relationship between the rate constant k and th
ID: 908450 • Letter: T
Question
The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as
k=AeEa/RT
where R is the gas constant (8.314 J/molK), 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(1T11T2)
which is mathmatically equivalent to
lnk1k2=EaR(1T21T1)
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 0.0120s1. 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 0.0120s1 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.
Explanation / Answer
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
Ea = activation energy
A = Frequency factor (constant)
Rate constant, K = A e - Ea / RT
log K = log A - ( Ea / 2.303RT ) ---(1)
If we take rate constants at two different temperatures, then
log K = log A - ( Ea / 2.303RT ) --- (2)
& log K' = log A - (Ea / 2.303RT’) ---- (3)
Eq (3 ) - Eq ( 2 ) gives
log ( K' / K ) = ( Ea / 2.303 R ) x [ ( 1/ T ) - ( 1 / T' ) ]
Given Ea = 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 ) = ( Ea / 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 ) = ( Ea / 2.303 R ) x [ ( 1/ T ) - ( 1 / T' ) ]
Given Ea = 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 ) = ( Ea / 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
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