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?
I answered this below. Be sure to note the typo (2.8 should be 1.4 in the second ICE chart.)
•NEED HELP ASAP - Anon, Tuesday, February 5, 2013 at 11:14pm
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?
See below.
Kc = 1.2^2/3.8=0.379
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.
Kc is not 0.379. I thought we went over this. You had Kc right at the beginning as 0.514.
A is solid so it should not be included in the Kc expression guys. And doubling the volume of the container does not effect the concentration of solid
Some people should not be allowed to help others with chemistry...solids do not have concentrations...therefore doubling the volume would have no effect
It does have an effect because B and C form only from A. If A were added to a gas to get B and C, then it would have no effect, but this is not the case.
To find the moles of A remaining after doubling the volume, you can use the concept of equilibrium and the values of the equilibrium constant (Kc) that you calculated.
First, let's review the equilibrium expression for the reaction:
A(s) <-------> B(g) + C(g)
Kc = [B] * [C] / [A]
You have already calculated the values of Kc for the original and the new equilibrium:
Original Kc = 0.514
New Kc = 0.129
Now, let's consider the equilibrium concentrations for the new equilibrium:
For B: The concentration of B initially was 1.20 M. However, when the volume of the container was doubled, the concentration became 0.6 M.
For C: Since B and C are produced in a 1:1 ratio, the concentration of C for the new equilibrium will also be 0.6 M.
Now, we need to calculate the moles of A remaining in both scenarios.
Original equilibrium:
Kc = [B] * [C] / [A]
0.514 = (1.20 M) * (1.20 M) / (4.00 - x)
Solving this equation for x, we get x = 1.2
Since x represents the moles of B and C formed, the moles of A remaining will be:
A = 4.00 - x = 4.00 - 1.2 = 2.8 moles
New equilibrium:
Kc = [B] * [C] / [A]
0.129 = (0.6 M) * (0.6 M) / (1.4 - x)
Solving this equation for x, we get x = 0.48
Similarly, the moles of A remaining will be:
A = 1.4 - x = 1.4 - 0.48 = 0.92 moles
Therefore, after doubling the volume, 0.92 moles of A remain.