A study of the system, H2(g) + I2(g) - 2 HI(g), was carried out. Kc = 54.9 at 699.0K for this reaction. A system was charged with 2.50 moles of H2 and 2.50 moles I2 in a 5.00 liter vessel as the only components initially. The system was brought up to 699.0 K and allowed to attain equilibrium. How many moles of H2 should there be in the container at that time?

Question

A study of the system, H2(g) + I2(g)   ->   2 HI(g), was carried out. Kc = 54.9 at 699.0K for this reaction. A system was charged with 2.50 moles of H2 and 2.50 moles I2 in a 5.00 liter vessel as the only components initially. The system was brought up to 699.0 K and allowed to attain equilibrium.   How many moles of H2 should there be in the container at that time?

joannablackwelder · Answer

@abb0t

joannablackwelder · Answer

When I set up the ICE table and plug the numbers into the quadratic formula, I get a negative under the radical. Any ideas anybody?

joannablackwelder · Answer

@aaronq

joannablackwelder · Answer

@chmvijay

jfraser · Answer

from what I did, you don't actually have to use the quadratic approach to solve this, since the RICE table looks like this:$$K = \frac{[HI]^2}{[H_2]*[I_2]} = 54.9$$ so by plugging in the values from the problem you get:$$\sqrt{54.9} = \frac{[2x]}{[0.5 - x]}$$

jfraser · Answer

because the $[H_2]$ and $[I_2]$ are equal, at 0.5M each to start, they both lose the same stoichometric "x" amount, so you have a square on top and on bottom of the fraction, so it simplifies without having to solve it quadratically.

joannablackwelder · Answer

Good call, thanks! I found my mistake! :)

jfraser · Answer

good