Chemical Equilibrium in Test 4 #6

Moderators: Chem_Mod, Chem_Admin

Sammy Thatipelli 1B
Posts: 63
Joined: Thu Jul 27, 2017 3:01 am
Been upvoted: 1 time

Chemical Equilibrium in Test 4 #6

Postby Sammy Thatipelli 1B » Sat Dec 09, 2017 4:02 pm

In this question, My answer for x was -1.5 after doing the ice table and solving for x. However, x is negative and I was wondering what you need to do in this question to get a positive number?

Cassandra Mullen 1E
Posts: 54
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 2 times

Re: Chemical Equilibrium in Test 4 #6

Postby Cassandra Mullen 1E » Sat Dec 09, 2017 4:13 pm

This reaction goes in the reverse reaction, which is why the x value you calculated is negative. When you plug this value into your equilibrium values in the ice table, you should get the right answer.

Sammy Thatipelli 1B
Posts: 63
Joined: Thu Jul 27, 2017 3:01 am
Been upvoted: 1 time

Re: Chemical Equilibrium in Test 4 #6

Postby Sammy Thatipelli 1B » Sat Dec 09, 2017 4:29 pm

How do u know the reaction is the reverse reaction though?

Cassandra Mullen 1E
Posts: 54
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 2 times

Re: Chemical Equilibrium in Test 4 #6

Postby Cassandra Mullen 1E » Sat Dec 09, 2017 5:33 pm

If you compare the initial concentrations to the equilibrium concentrations: you will see that on the reactants side, A went from 1.00 M to 2.54 M and B went from 2.00 M to 2.54 M. This means that the reactants increased. C goes from 7.00 M to 5.46 M and D goes from 4.00 to 2.46, so the products decreased. The only way for this to happen is if the reaction is going in the reverse direction. Hope it helps! :)


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 1 guest