5.35 Part b

Moderators: Chem_Mod, Chem_Admin

Justin Vayakone 1C
Posts: 110
Joined: Sat Sep 07, 2019 12:19 am

5.35 Part b

Postby Justin Vayakone 1C » Tue Jan 07, 2020 5:54 pm

This is the answer to part b: (in attachment). I understand the arrangement of the equilibrium constant and where the values (5, 10, and 18) come from, but I don't understand why these values are over 100. Is it because of units? The graph uses P/pKa, but what does that mean? Pressure per kiloPascal?
Attachments
5.35(PartB).JPG

Ariel Davydov 1C
Posts: 110
Joined: Thu Jul 11, 2019 12:16 am
Been upvoted: 1 time

Re: 5.35 Part b

Postby Ariel Davydov 1C » Tue Jan 07, 2020 6:23 pm

Yes, the division of each factor by 100 is to convert from P/kPa to bar.

Vicki Liu 2L
Posts: 101
Joined: Sat Aug 24, 2019 12:15 am

Re: 5.35 Part b

Postby Vicki Liu 2L » Tue Jan 07, 2020 6:26 pm

I believe the pressure is given in kilopascals and dividing by 100 is to convert the units to bars (since 1 bar= 100 kPa).


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 2 guests