Catalysis 15.69


Moderators: Chem_Mod, Chem_Admin

Michelle Chernyak 1J
Posts: 33
Joined: Fri Sep 29, 2017 7:06 am

Catalysis 15.69

Postby Michelle Chernyak 1J » Mon Mar 12, 2018 8:57 pm

Help! I'm really confused with this problem. I feel like i need to solve for k of the first reaction and then use that to solve for the activation energy of the second but I don't even know if that's right. Also, when they say that all other factors are equal, do they mean to each or to their values from the first reaction?

Thank you

Cali Rauk1D
Posts: 51
Joined: Fri Sep 29, 2017 7:03 am

Re: Catalysis 15.69

Postby Cali Rauk1D » Mon Mar 12, 2018 9:01 pm

In part A use arrehnius equation and then use the temperature in part b and repeat what you did in the previous step

Maeve Gallagher 1J
Posts: 56
Joined: Fri Sep 29, 2017 7:07 am

Re: Catalysis 15.69

Postby Maeve Gallagher 1J » Mon Mar 12, 2018 10:40 pm

You need to solve for the ration if k catalyzed and k uncatalyzed. This ratio is given as 1000, so you set that equal to the Arrhenius equation of the catalyzed reaction divided by the Arrhenius equation of the uncatalyzed reaction. You then use algebra to get the equation Ea(cat) = Ea - RTln1000. When they say all other factors are equal, this means that the other values in the Arrhenius equation are equal to each other: A can be cancelled, R=8.314, T=298. The only value that is different is the activation energy.

Joyce Lee 1C
Posts: 52
Joined: Fri Sep 29, 2017 7:03 am
Been upvoted: 1 time

Re: Catalysis 15.69

Postby Joyce Lee 1C » Tue Mar 13, 2018 12:33 pm

You can set up a proportion of k of the catalyzed reaction and k of the uncatalyzed reaction with the formula k = Ae^-Ea/RT, like how you had to solve for problem #67. In this case, you would make the kcat/kuncat equal to 1000.


Return to “*Enzyme Kinetics”

Who is online

Users browsing this forum: No registered users and 2 guests