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 ==> B + C
A is (s) and not included in Kc
1.2 + 1.2 = 2.4
4.00 - 2.4 = 1.6 mol left
Solids are not part of the equation bro
1.6 mol
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
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 got it wrong. X doesn't equal .189. I really need help on solving this problem.
2.422 mols is wrong.
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]
For the first Kc I got .514
For the second Kc I got .129.
How do I find the mols afterwards?