Help on 5.35

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Tiffany Chao 2H
Posts: 117
Joined: Fri Aug 09, 2019 12:17 am

Help on 5.35

Postby Tiffany Chao 2H » Wed Jan 15, 2020 11:32 pm

I'm not understanding how to find the chemical equation on 5.35. Can someone help me explain?

Thanks

Selena Yu 1H
Posts: 108
Joined: Fri Aug 09, 2019 12:16 am

Re: Help on 5.35

Postby Selena Yu 1H » Thu Jan 16, 2020 12:23 am

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)

Rohit Ghosh 4F
Posts: 99
Joined: Thu Jul 25, 2019 12:17 am

Re: Help on 5.35

Postby Rohit Ghosh 4F » Thu Jan 16, 2020 12:47 pm

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.

KBELTRAMI_1E
Posts: 108
Joined: Sat Jul 20, 2019 12:17 am

Re: Help on 5.35

Postby KBELTRAMI_1E » Thu Jan 16, 2020 2:52 pm

then just simplify and you should have all parts to the question answered

Tiffany Chao 2H
Posts: 117
Joined: Fri Aug 09, 2019 12:17 am

Re: Help on 5.35

Postby Tiffany Chao 2H » Thu Jan 16, 2020 4:50 pm

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)

Tiffany Chao 2H
Posts: 117
Joined: Fri Aug 09, 2019 12:17 am

Re: Help on 5.35

Postby Tiffany Chao 2H » Thu Jan 16, 2020 4:51 pm

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)


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