<p>The equilibrium constant Kc for the following reaction equals 49 at 230°C.
PCl3(g) + Cl2(g) reverse reaction arrow PCl5(g)
If 0.467 mol each of phosphorus trichloride and chlorine are added to a 4.4-L reaction vessel, what is the equilibrium composition of the mixture at 230°C?
M of PCl3(g):
M of Cl2(g):
M of PCl5(g):</p>
<p>Reverse reaction means <---------- ?</p>
<p>Also you can't ask for help on these boards.</p>
<p>Where can I ask for help then?</p>
<p>i can tell you the method to deal with this question,but I can't calculate for you.the data is ghastly complex,I suspect that you have gotten it wrong.
PCl3 Cl2 PCl5
initial: 0.467mol 0.467mol 0
reacting:x mol x mol xmol
later: (0.467-x)mol (0.467-x)mol x mol
we have known kc at this temperature,so
49=x/4.4/(0.467-x)^2/4.4^2=4.4x/(0.467-x)^2
then you calculate out x,it's done</p>
<p>If they are gases in a closed environment, why wouldn't it establish an equilibrium...
Shouldn't it be double arrows b/c it is a unit in eq?</p>