The Arrhenius equation shows the relationship between the rate constant k and th

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).

A) The activation energy of a certain reaction is 40.1 kJ/mol . At 26  C , the rate constant is 0.0160s1. At what temperature in degrees Celsius would this reaction go twice as fast?

Express your answer with the appropriate units.

B) Given that the initial rate constant is 0.0160s1 at an initial temperature of 26  C , what would the rate constant be at a temperature of 160  C for the same reaction described in Part A?

Express your answer with the appropriate units.

Explanation / Answer

Given : Activation energy of a certain reaction = 40.1 kJ/mol

T = 26 deg C

k = 0.0160 s-1

Here we have to find the temperature at which reaction goes as fast as at 26 deg C

We use Arrhenius equation,

Ln (k2/k1 ) = Ea / R ( 1/ T1 – 1/ T2 )

R is gas constant = 8.314 J per K per mol

Here k2 and k1 are the rate constants at Temperature T1 and temperature T2

Lets k1 is initial rate constant at initial Temperature T1 = 0.0160 s-1 . According to the problem condition, k2 is twice of the k1

Ea is activation energy = 40.1 kJ/ mol = 40100 J/mol

We convert temperature in K

T1 = 26 deg C + 273.15 K = 299.15 K

We plug all the given values in Arrhenius equation.

Ln( 2*0.0160 s-1 / 0.0160 s-1 ) =( 40100 J / mol ) /( 8.314 J K-1 mol-1 ) [( 1/299.15 K )-( 1/ T2) ]

0.693 = 4823.19 * [ T2 – 299.15 / ( 299.15 T2) ]

1.44 E-4 = ( T2 – 299.15 )/ 299.15 T2

0.04298 T2 = T2 – 299.15

299.15 = 0.9570 T2

T2 = 312.58 K

So final temperature in deg C = 312.58 – 273.15 = 39.44 deg C ‘

B)

k = 0.0160 s-1 T = 26 deg C ,

At T = 160 deg . k for this temperature = 160 deg C

Lets use Arrhenius equation

k 1 = 0.0160 s-1 , T1 = 299.15 K

k2 = ? T2 = 160 deg c + 273.15 = 433.15 K

Ln( k2 / 0.0160 s-1 ) =( 40100 J / mol ) /( 8.314 J K-1 mol-1 ) [( 1/299.15 K )-( 1/ 433.15) ]

Ln( k2 / 0.0160 s-1 ) = 4.99

Lets take exp of both side

( k2 / 0.0160 s-1 )= exp ( 4.99)

k2 / 0.0160 s-1 = 146.61

k2 = 2.34 s-1

k2 = 2.34 s-1

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