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?
8 answers
How do I find x and the moles?
•NEED HELP! - Anon, Tuesday, February 5, 2013 at 11:25pm
Is x=0.189? I used the quadratic equation to solve for x from Kc = 0.514 = (0.6+x)(0.6+x)/(1.4-x). How do you find the moles of A?
The container volume is doubled
A(s)<------>B(g)+C(g)
(3.8/2)-----x----x---
0.379 = (x^2)/(1.9 - x)
x^2 + 0.379x - 0.7201 = 0
x=0.679 M
A = 1.9 - 0.679 = 1.22 M
But I got it wrong.
18 answers