### 15.67?

Posted:

**Sun Mar 11, 2018 6:42 pm**Could someone explain to me very simply what they are asking, and what the solution is telling me to do. I do not understand at all

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=151&t=29416

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Posted: **Sun Mar 11, 2018 6:42 pm**

Could someone explain to me very simply what they are asking, and what the solution is telling me to do. I do not understand at all

Posted: **Sun Mar 11, 2018 9:26 pm**

The solution from Chegg helped me

FOR PART A)

15.67 The presence of a catalyst provides a reaction pathway in which the activation energy (Ea) of a certain reaction is reduced from 125kjmol^-1 to 75kjmol^-1

a) By what factor does the rate of the reaction increase at 298K, all other factors being equal?

You need to use the Arrhenius equation. ln k =ln A -(Ea/RT)

Suppose k1 and k2 are the rate constants in presence and absence of catalyst respectively. The activation energies in the presence and absence of catalyst are Ea1 and Ea2 respectively.

Therefore,

ln k1 =ln A -(Ea1/RT)

ln k2 =ln A -(Ea2/RT)

Then subtract the equations from each other to find how the rate of the reaction changed,

ln2-lnk1 = ln A -(Ea2/RT) -(ln A -(Ea2/RT))

So lnk2/k1 = (Ea1-Ea2)/RT ..... This equation is at constant temperature

The activation energies (Ea1 and Ea2) are given. Plug in values into the above formula and you'll get

k2/k1=6x10^8

k2=(6x10^8)k1

The rate of reaction increased by 6x10^8 factor.

FOR PART A)

15.67 The presence of a catalyst provides a reaction pathway in which the activation energy (Ea) of a certain reaction is reduced from 125kjmol^-1 to 75kjmol^-1

a) By what factor does the rate of the reaction increase at 298K, all other factors being equal?

You need to use the Arrhenius equation. ln k =ln A -(Ea/RT)

Suppose k1 and k2 are the rate constants in presence and absence of catalyst respectively. The activation energies in the presence and absence of catalyst are Ea1 and Ea2 respectively.

Therefore,

ln k1 =ln A -(Ea1/RT)

ln k2 =ln A -(Ea2/RT)

Then subtract the equations from each other to find how the rate of the reaction changed,

ln2-lnk1 = ln A -(Ea2/RT) -(ln A -(Ea2/RT))

So lnk2/k1 = (Ea1-Ea2)/RT ..... This equation is at constant temperature

The activation energies (Ea1 and Ea2) are given. Plug in values into the above formula and you'll get

k2/k1=6x10^8

k2=(6x10^8)k1

The rate of reaction increased by 6x10^8 factor.

Posted: **Sat Mar 17, 2018 3:02 pm**

at the end, why do you multiply by k1/

Posted: **Sat Mar 17, 2018 3:12 pm**

at the end, why do you multiply by k1/

Posted: **Sat Mar 17, 2018 5:36 pm**

in the solutions manual it tells you that the rate of the uncatalyzed reaction is .60. how did they know this??

Posted: **Sat Mar 17, 2018 6:11 pm**

also, why do we set it up so that it is rate of the cat rxn/ rate of the uncat rxn?