Question

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

Answer 1

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.

Answer 2

For the first Kc I got .514
For the second Kc I got .129.
How do I find the mols afterwards?

Answer 3

There is only one Kc; that is 0.514. If you plug in (0.6)(0.6)/(1.4) you get 0.257 = the reaction quotient. That is smaller than Kc which means the reaction will shift to the right to re-establish equilibrium. I have set up the ICE chart for that equilibrium also.
Kc = 0.514 = (0.6+x)(0.6+x)/(1.4-x).
Solve. Note: I made a typo in the E line of the second ICE chart. It should read, from left to right, as
1.4-x.....0.6+x.....0.6+x

Answer 4

How do I find x and the moles?

Answer 5

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?

Answer 6

I didn't work the problem; however, if x = 0.189, then (A) = 1.4-x = ?M and mols = M x L = M x 2.00 = ?

Answer 7

I got it wrong. X doesn't equal .189. I really need help on solving this problem.

Answer 8

I've done the chemistry for you. In detail. If you will show your work I'll try to find the error.

Answer 9

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.

Answer 10

The equation is
0.514 = (0.6+x)(0.6+x)/(1.4-x)
Solve that and stay focused.

Answer 11

After you obtain x, then 1.4-x = (A) in M and M x L = M x 2L = mols A. 0.189 is the value for x.

Answer 12

2.422 mols is wrong.

Answer 13

Did you type in too many significant figures? You're allowed only three (from 4.00 and 1.00 in the original post)
2.42 mols.

Answer 14

Solids are not part of the equation bro

Answer 15

1.6 mol

Answer 16

it depends on whether the reactant or product is solid or liquid. If it's solid you don't suppose to include in finding equilibrium concentration and the eq. will look like Kc=[B][C]

Answer 17

A ==> B + C

A is (s) and not included in Kc

1.2 + 1.2 = 2.4

4.00 - 2.4 = 1.6 mol left

Answer 18

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

Initial 4mol 00 00

Change -1.2 1.2 1.2

Equilibrium 3.8 1.2 1.2

Kc = [B][C] = (1.2mol/1L)^2 = 1.44


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

Initial 3.8 1.2 1.2

Change -x + x +x

Equilibrium (3.8-x) ( 1.2+x) ( 1.2+x)

Kc = (1.2+x/ 2L)^2

√1.44 = √(1.2+x/2)^2
1.2 = 1.2/2
2.4 = 1.2+x
x = 2.4-1.2 = 1.2

A = 3.8-x = 3.8-1.2=2.6mol