1. The activation energy of a certain reaction is 41.5kJ/mol . At 20 ?C , the...

1. The activation energy of a certain reaction is 41.5kJ/mol . At 20 ?C , the rate constant is 0.0110s?1. At what temperature in degrees Celsius would this reaction go twice as fast?

2. Given that the initial rate constant is 0.0110s?1 at an initial temperature of 20?C , what would the rate constant be at a temperature of 180?C for the same reaction described in Part 1?

Expert Solution

1. The activation energy of a certain reaction is 41.5kJ/mol . At 20 ?C , the rate constant is 0.0110s?1. At what temperature in degrees Celsius would this reaction go twice as fast?

Given:

Ea = 41.5 kJ/mol = 41500 J/mol

At 20 deg C , Rate constant lets say k1 = 0.0110 s^-1

We have to find out temperature at which reaction twice as fast.

That means we have to find the temperature at which k is twice as the k1 is and this lets say this ka s k2

k1= 0.0110 s^-1 at T1 = 20 deg C = 20+273.15 = 293.15 K

k2 = 2 * k1 = 2* 0.0110 s^-1 = 0.022 s^-1 , T2 = ?

Lets use Arrhenius equation to solve for T2

ln (k2/k1) = -Ea / R ( 1/T2 -1/T1)

Here k2 and k1 are the rate constant of the reaction at T2 and T1 respectively.

Ea is activation energy and R is gas constant and value or R is 8.314 J /K mol

Lets plug all given values in above equation to get T2

ln (0.022/0.0110) = - [( 41500 J/ mol )/ (8.314 J / K mol )] * (1/T2

queen_honey_blossom answered 1 hour ago

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