anonymous
  • anonymous
24. Suppose that 0.350 mol of A and 0.520 mol of B are placed a 1.50 L container and the following hypothetical equilibrium is established. If the equilibrium amount of C is 0.150 mol, what is the equilibrium constant for this reaction? A(g) +2B(g) --->3C(g)
Chemistry
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

Xishem
  • Xishem
First, you're going to want to set up an ICE chart: \[[A]_0=\frac{0.350mol}{1.5L}=0.23M\]\[[2B]_0=\frac{0.520mol}{1.5L}=0.35M\]\[\Delta [A]=\frac{-1}{3}(0.10M)=-0.033M\]\[\Delta [2B]=\frac{-2}{3}(0.10M)=-0.067M\] Concentration (M) [A] [2B] [3C] Initial 0.23 0.35 0 Change -0.033 -0.067 0.100 Equilibrium 0.200 0.280 0.100 Then, we can go ahead and plug these in to our K_c equation:\[K_c=\frac{[C]^3}{[A][B]^2}=\frac{(0.100)^3}{(0.200)(0.280)^2}=6.38*10^{-2}\]\[K_c=6.38*10^{-2}\]
Xishem
  • Xishem
I forgot to write down a calculation which may cause confusion. I just want to clarify:\[[3C]_E=\frac{0.150mol}{1.5L}=0.100\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.