<p>Too hard for me, help please? </p>
<p>A(g) + B(g) --> <---- C(g) (it's in equilibruim; i just can't get the arrows right) </p>
<p>K = 130 (equilibrium constant) </p>
<p>Assume that .406 mol of C(g) is placed in a cylinder, w/ temp at 300K and barometric pressure constant at 1 atm. The original volume before decomposition is 10L. What is the volume at equilibrium?</p>
<p>Thanks!</p>
<p>This one is good... bump!</p>
<p>I'll take a shot at this..I haven't taken chem since last year. You said C(g) is placed in a cylinder, so that means that is the starting material? So its C(g)<-->A(g)+B(g)?<br>
Anyways, the concentration of C(g) at equil. would be (.406-n)/(10-V). The equil. concentrations of A and B would be n/v. Therefore, n^2(10-v)/v^2(.406-n)=130. Also, PV=nrT, so V=nR(300). Solve these two equations to get n and v, n being the moles and v being the volume. The answer would be 10-v.</p>
<p>We just finished equilibrium chapter, so I'll give it a shot</p>
<p>the M of C is .406n/10L=.0406M</p>
<p>So
A(g) + B(g) <----> C(g)
Initial 0 0 .0406M
Changes +x +x -x
Equil x x .0406-x</p>
<p>products over reactants
130= x^2/(0.0406-x)
So, x=0.04058</p>
<p>I might have it backwards though, I'm not sure.</p>
<p>juliehoch is correct, its a standard equilibrium problem</p>
<p>0.04058 is the volume? I thought that was molarity, or 0.04058 moles in 1 liter of solution. I'm confused as to how you can get to the volume from this.</p>
<p>try the method i posted...im not sure if it is right but i think it should work; its just an equilibrium set up but im leaving n (moles) and v (volume) so you can solve for v and get the volume.</p>
<p>I not to sure, but if the gases were placed into a cylinder then the volume shouldn't change (right?), but the pressure and molarity would. I am going to go look it up.</p>