I'm not understanding how to find the chemical equation on 5.35. Can someone help me explain?
Thanks
Help on 5.35
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 108
- Joined: Fri Aug 09, 2019 12:16 am
- Been upvoted: 1 time
Re: Help on 5.35
It says in the problem that compound A decomposes into compounds B and C. So the chemical equation would be A (g) --> B (g) + C (g)
-
- Posts: 99
- Joined: Thu Jul 25, 2019 12:17 am
Re: Help on 5.35
You can then use the partial pressures given by the graph along with the reaction A (g) --> B (g) + C (g) to calculate the equilibrium constant.
-
- Posts: 108
- Joined: Sat Jul 20, 2019 12:17 am
-
- Posts: 117
- Joined: Fri Aug 09, 2019 12:17 am
Re: Help on 5.35
Selena Yu 1H wrote:It says in the problem that compound A decomposes into compounds B and C. So the chemical equation would be A (g) --> B (g) + C (g)
The answer key says the chemical equation is 2A (g) --> B (g) + 2C (g)
-
- Posts: 117
- Joined: Fri Aug 09, 2019 12:17 am
Re: Help on 5.35
Rohit Ghosh 4F wrote:You can then use the partial pressures given by the graph along with the reaction A (g) --> B (g) + C (g) to calculate the equilibrium constant.
The answer key says the chemical equation is 2A (g) --> B (g) + 2C (g)
Return to “Equilibrium Constants & Calculating Concentrations”
Who is online
Users browsing this forum: No registered users and 22 guests