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.
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
18 answers
For the first Kc I got .514
For the second Kc I got .129.
How do I find the mols afterwards?
For the second Kc I got .129.
How do I find the mols afterwards?
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
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
How do I find x and the moles?
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?
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 = ?
I got it wrong. X doesn't equal .189. I really need help on solving this problem.
I've done the chemistry for you. In detail. If you will show your work I'll try to find the error.
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.
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.
The equation is
0.514 = (0.6+x)(0.6+x)/(1.4-x)
Solve that and stay focused.
0.514 = (0.6+x)(0.6+x)/(1.4-x)
Solve that and stay focused.
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.
2.422 mols is wrong.
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.
2.42 mols.
Solids are not part of the equation bro
1.6 mol
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]
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
A is (s) and not included in Kc
1.2 + 1.2 = 2.4
4.00 - 2.4 = 1.6 mol left
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
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