The presence of a catalyst provides a reaction pathway in which the activation energy of a reaction is reduced by 51.00 kJ ⋅ mol − 1 51.00 kJ⋅mol−1 . Uncatalyzed: A ⟶ B A⟶B E a = 136.00 kJ ⋅ mol − 1 Ea=136.00 kJ⋅mol−1 Catalyzed: A ⟶ B A⟶B E a = 85.00 k J ⋅ mol − 1 Ea=85.00 kJ⋅mol−1

The given question is incomplete.The complete question is:

The presence of a catalyst provides a reaction pathway in which the activation energy of a reaction is reduced by 51.00 kJ /mol Uncatalyzed: A ⟶ B A⟶B E a = 136.00 kJ/mol Catalyzed: A ⟶ B A⟶B E a = 85.00 k J/mol. determine the factor by which tha catalysed reaction is faster than the uncatalysed reaction at 289.0 K if all other factors are equal.

The factor by which the catalysed reaction is faster than the uncatalysed reaction at 289.0 K is

Explanation:

According to the Arrhenius equation,

The expression used with catalyst and without catalyst is,

where,

= rate of reaction with catalyst

= rate of reaction without catalyst

= activation energy with catalyst

= activation energy without catalyst

R = gas constant =

T = temperature =

Now put all the given values in this formula, we get

Therefore, the factor by which the catalysed reaction is faster than the uncatalysed reaction at 289.0 K is

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying