Asked by Anon
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
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
Answers
Answered by
DrBob222
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.
......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.
Answered by
Anon
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?
Answered by
DrBob222
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
Answered by
Anon
How do I find x and the moles?
Answered by
Anon
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?
Answered by
DrBob222
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 = ?
Answered by
Anon
I got it wrong. X doesn't equal .189. I really need help on solving this problem.
Answered by
DrBob222
I've done the chemistry for you. In detail. If you will show your work I'll try to find the error.
Answered by
Anon
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.
Answered by
DrBob222
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.
Answered by
DrBob222
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.
Answered by
Anon
2.422 mols is wrong.
Answered by
DrBob222
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.
Answered by
Your Wrong Bob
Solids are not part of the equation bro
Answered by
Anonymous
1.6 mol
Answered by
guru
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]
Answered by
Adam
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
Answered by
Qi Zi
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
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