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December 4, 2016

Homework Help: Chemistry

Posted by Anon on Tuesday, February 5, 2013 at 10:56pm.

4.00 mol of solid A was placed in a sealed 1.00-L container and allowed to decompose into gaseous B and C. The concentration of B steadily increased until it reached 1.20 M, where it remained constant.

A(s) <-------> B(g)+C(g)

Then, the container volume was doubled and equilibrium was re-established. How many moles of A remain?

? mol A

•Chemistry - DrBob222, Tuesday, February 5, 2013 at 9:38pm
4.00 mol/L = 4M
......A ==> B + C
I.....4......0..0
C....-x......x..x
E....4-x....1.2..x
Since x = 1.2, then (C) = 1.2 and A = 4.00-1.2 = 2.8
Substitute into the Kc expression and solve for Kc.
When the system is double in volume that means concns are halved.
.......A ==> B + C
I....1.4....0.6..0.6
Calculate the reaction quotient to see which way this system will move to re-establish equilibrium. I think it will move to the right.
C.....-x......x......x
E....2.8-x..0.6+x..0.6+x

Substitute into Kc expression and solve.


•Chemistry Help - Anon, Tuesday, February 5, 2013 at 10:19pm
For the first Kc I got .514
For the second Kc I got .129.
How do I find the mols afterwards?

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