Achieve question #17 week 9 and 10

[Esmeralda Polanco]

Hi could someone help solve question #17 on the Achieve homework from weeks 9 and 10, please.
The presence of a catalyst provides a reaction pathway in which the activation energy of a reaction is reduced by 70.00 kJ • mol-1
Uncatalyzed: A → B Ea = 131.00 kJ • mol-l
Catalyzed:
A → B Ea = 61.00 kJ • mol-1
Determine the factor by which the catalyzed reaction is faster than the uncatalyzed reaction at 290.0 K if all other factors are equal.

Determine the factor by which the catalyzed reaction is faster than the uncatalyzed reaction at 338.0 K if all other factors are equal.

[Maleeha Amir 2E]

The formula for k(cat) would be k(cat) = Ae^(Ea/RT) and k(uncat) = Ae^(Ea/RT). Divide those by each other and you get k(cat)/k(uncat) = e^(-(Ea2-Ea1)/RT)). Make sure you convert R to kJ.

[Rishika_Kanaparthy_1G]

Hi! We will be using the the Arrhenius equation k = Ae^(-Ea/RT)
For k_uncat, we would have k = Ae^(-131/RT)
For k_cat, we’d have k = Ae^(-61/RT)
Then, we can divide them using k_cat/k_uncat
After the A cancels out we can use exponent rules to get e^(74000/RT) >>>> remember to convert kj to joules!
Then we can plug in the 290 for T and use calculator to solve

[705903057]

Hi, so this question deals with the Arrhenius equation: k=Ae^(-Ea/RT)

You’ll want to rewrite the equation in this format: e^((Ea,uncat - Ea,cat)/RT)

At T=290.0K,
Kcat/kuncat = e^(70.00x10^3 J/mol)/(8.314 J/Kxmol x 290.0 K) = ?? Times faster

And then for T= 338
Kcat/kuncat = e^(70.00x10^3 J/mol)/(8.314 J/Kxmol x 338.0 K) = ?? Times faster

[405696752]

so for this question you have to use this equation k=Ae^(-Ea/RT), where R is 8.3145, T is the temp given to you, and Ea is the catalyzed and uncatalyzed numbers. once you plug them in, you solve and convert them into joules and repeat the steps for the second part as well