Question

A certain reaction has an activation energy of 51.55 kJ/mol. At what Kelvin temperature will the reaction proceed 3.00 times faster than it did at 303 K? Number

Answer 1

For this question, the equation which needs to be used is the Arrhenius equation :

-Ba k = Ae RT

Where, k = rate constant,

A = pre-exponential factor or Arrhenius constant, which is a constant for a given reaction,  

E_a = activation energy/mole (in J/mol) ,

R = Universal Gas constant (If E_a is in J/mol, then the value of R used is 8.314 J/ K mol ) ,

T = Temperature in K

This equation is useful for finding variation in rate of a reaction with temperature, since as rate increases, rate constant will also increase directly.

In the given question, we have to find the temperature when the rate of the reaction becomes 3 times faster.

The activation energy will remain constant for all temperatures. The value given is : E_a = 51.55 kJ/mol = 51550 J/mol   [since 1kJ = 1000J]

R = 8.314 J/ K mol

A will also remain constant for a given reaction.

Thus, initially,

let rate constant = k. The initial temperature = 303K

Finally ,

rate = 3 times the initial rate.

Thus, rate constant = 3 times the initial rate constant = 3k. Temperature = T K(needs to be found)

Thus, we can write Arrhenius equation for initial and final conditions:

Initial :              

-51550J/mol k = Ae (8.314 J/Kmol) (303)

Final :                  

-51550J/mol 3k = Ae (.2 16-kmol)(TK)

Dividing initial by final :

- 515502/mot (8-84 x 6ỌC 3030) 3* = Accessori or se trata + 8.31 - (91550)(563-+). Taking natural logarithm on both sides: en(t) = -3150 ( * - +)

- or, t 1.0986 =+ 6200 303 or 1.** X1079 € w = = gos -(1.47X10") or + = 3.12381003 - or T=320.19 ~ 320 -2 So, required temperature = 320.2 k.