Starting with 0.3500mol Co(g) and 0.05500mol COCL2(g) in a 3.050 L flask at 668K, how many moles of Cl2 are present at equilibrium?
Co(g) = Cl2(g) <-> COCL2(g) Kc = 1.2x10^3
Do you mean Co or CO. Surely you meant CO. You must understand that CO is carbon monoxide and Co is cobalt. Also, I assume the = sign should be a + sign.
Convert to concns in molarity.
(CO) = 0.3500/3.050 = ??
(COCl2) = 0.05500/3.050 = yy
.............CO + Cl2 ==> COCl2
initial..0...??....0.......yy
change.....+x......+x......-x
equil....0.??+x...x........yy-x
Substitute the ICE chart I've prepared and substitute into Kc expression, solve for x. That will give you (Cl2) in moles/L, then multiply by L to convert to moles.
When I converted everything I got
1.2x10^3 = (0.02 + x) / (0.12 - x)(x)
= (0.02 + x) / (0.12x - x^2)
1.2x10^3(0.12x - x^2) = 0.02 + x
144x - 1.2x10^3x^2 = 0.02 + x
Then I brought x to the other side and I got
145x - 1.2x10^3x^2 = 0.02
Then I get confused with the x's and don't know how to continue solving.
Thanks so much for the previous help though
You're rounding far too early. I may make an error by carrying too many places when doing these quadratic equations but I like to carry more than I'm allowed and round at the end.
0.3500/3.050 = 0.11475M = (CO)
0.05500/3.050 = 0.01803M = (COCl2)
First, note that K = 1.29E3 or 1290 and not 1200.
............CO + Cl2 ==> COCl2
initial.0.11475...0.......0.01803
change.....+x......+x......-x
equil..0.11475+x...x.......0.01803-x
Kc = 1290 = (COCl2)/(CO)(Cl2)
1290 = (0.01803-x)/(0.11475+x)(x)
0.01803-x = 1290(0.11475+x)(x)
0.01803-x = 1290x^2 + 148.03x
1290x^2 +149.03x -0.01803 = 0
x = 0.0001208 M = (Cl2)
Then convert to moles and round to the appropriate number of s.f.
OK thanks so much, I appreciate it.
Well, let's do a little chemistry math, shall we? But first, let me put on my lab coat and clown shoes for good measure.
To solve this problem, we need to use the equilibrium constant, Kc. Now, remember, the equilibrium constant is just a fancy way of telling you how many moles of products you get for every mole of reactants.
Given that Kc = 1.2x10^3, we can set up the equation:
Kc = [COCL2(g)] / ([Co(g)] * [Cl2(g)])
Where [COCL2(g)], [Co(g)], and [Cl2(g)] represent the molar concentrations at equilibrium. Since we're dealing with moles here, we need to convert the initial amounts to molar concentrations.
To do that, we divide the number of moles by the volume of the flask (3.050 L):
[Co(g)] = 0.3500 mol / 3.050 L
[Cl2(g)] = ?? (What we're trying to find out!)
[COCL2(g)] = 0.05500 mol / 3.050 L
Now, we can plug these values into our equation:
1.2x10^3 = (0.05500 mol / 3.050 L) / ((0.3500 mol / 3.050 L) * [Cl2(g)])
Now, let's do some algebra magic to isolate [Cl2(g)]:
[Cl2(g)] = (0.05500 mol / 3.050 L) / ((0.3500 mol / 3.050 L) * (1.2x10^3))
Now, let me take off my clown shoes to do the calculation... Hang on a second...
The answer is approximately 0.05548 mol of Cl2.
So, at equilibrium, there are about 0.05548 moles of Cl2 present. Remember, it's always good to double-check my calculations because I have been known to mix up my funny bone with my calculations bone.
Hope this helps!
To determine the number of moles of Cl2 present at equilibrium, we need to calculate the equilibrium concentrations of Co(g), Cl2(g), and COCl2(g) using the given equilibrium constant (Kc) and initial mole amounts.
Step 1: Set up the expression for the equilibrium constant (Kc).
The equilibrium constant expression for the given reaction is as follows:
Kc = [COCl2(g)] / [Co(g)][Cl2(g)]
Step 2: Determine the initial concentrations of Co(g), Cl2(g), and COCl2(g).
The initial concentrations are given as follows:
[Co(g)] = 0.3500 mol / 3.050 L = 0.11475 M
[COCl2(g)] = 0.05500 mol / 3.050 L = 0.01803 M
Step 3: Substitute the initial concentrations into the equilibrium constant expression and solve for [Cl2(g)].
1.2x10^3 = (0.01803 M) / (0.11475 M)([Cl2(g)])
Step 4: Rearrange the equation to solve for [Cl2(g)].
[Cl2(g)] = (0.01803 M) / (1.2x10^3)(0.11475 M)
[Cl2(g)] = 0.13191 M
Step 5: Convert the concentration of Cl2 to moles.
[Cl2(g)] = 0.13191 M x 3.050 L = 0.402 mol
Therefore, there are approximately 0.402 moles of Cl2 present at equilibrium.