Creating a "new initial" for the ICE box (Q 11.67)

Moderators: Chem_Mod, Chem_Admin

Anna Nordstrom 1A
Posts: 15
Joined: Fri Sep 25, 2015 3:00 am

Creating a "new initial" for the ICE box (Q 11.67)

Postby Anna Nordstrom 1A » Wed Nov 11, 2015 2:51 pm

Question 11.67 says: The reaction 2HlL(g) = H2(g)+Cl2(g) has K=3.2x10-34 at 298 K. The initial partial pressures are H2, 1.0 bar; HCl 2.0 bar; and Cl2 3.0 bar. At equilibrium there is 1.0 mol H2(g). What is the volume of the container?

In the answer booklet they push the reaction to the left by making the new initials 4, 0 and 2 (for HCl, H2 and Cl2). How did they change the numbers? How do we know when this is allowed? Why is this legal?

Thank you!

Chem_Mod
Posts: 18889
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 716 times

Re: Creating a "new initial" for the ICE box (Q 11.67)

Postby Chem_Mod » Wed Nov 11, 2015 9:45 pm

It is legal because the equilibrium state doesn't "care" where you started from. The only key has only done this to simplify the calculation so that "x" represents H2. x is very small: 2.6*10^-33 bar

Doing the problem "normally" will give 2+2x for HCl, 1-x for H2, 3-x for Cl2. This gives the same answer in the sense that the final value of H2, represented by "1-x", will again be 2.6*10^-33. You may run into problems on the calculator since x is so close to 1, it may just tell you that x=1 which is useless.

Edwin Ng 1G
Posts: 20
Joined: Fri Sep 25, 2015 3:00 am

Re: Creating a "new initial" for the ICE box (Q 11.67)

Postby Edwin Ng 1G » Thu Nov 12, 2015 11:01 pm

Even when I use the new initial values of 4,0 and 2, my calculator spits out x=0 instead of 2.6*10^-33. On a side note, will we be dealing with these very small numbers on quizzes and exams?

Chem_Mod
Posts: 18889
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 716 times

Re: Creating a "new initial" for the ICE box (Q 11.67)

Postby Chem_Mod » Fri Nov 13, 2015 3:51 am

Yes, you may encounter very small numbers on exams and quizzes.


Return to “Non-Equilibrium Conditions & The Reaction Quotient”

Who is online

Users browsing this forum: No registered users and 1 guest