Question

For the system, H2(g) + X2(g) <--> 2HX(g), Kc = 24.4 at 300 K. A system was charged with 2.00 moles of HX in a 3.00 liter container. The catalyst was introduced using a remote unit, and the system was allowed to come to equilibrium. How many moles of H2(g) will be present when the system reaches equilibrium?

I'm actually completely stumped on this one. I know you have to do 2 moles of HX and divide by 3 Liters and you get .67 moles/liter. But they don't give you the other two so what do you have to do? The answer is .288, but I've no idea how to get that. Anyone got an idea??

Answer 1

Right, since Kc is given to 3 places, I would calculate M to 3 places; i.e., 2 mols/3 L = 0.667.

.......H2 + X2 ==> 2HX
I......0....0.......0.667
C.....+x...+x........-2x
E......x....x.......0.667-2x

Substitute the E line into the Kc expression and solve for x, then convert from M to mols. 0.288 is the correct answer for mols. x is 0.09 approx.