Question

can anyone explain how to do this problem?

At 673 K, the equilibrium constant for the
reaction
2 HI(g) ⇀↽ H2(g) + I2(g)
is Kc = 0.0200. If 0.5 mol of HI is placed in
a 2.00 liter container and heated to 673 K,
what concentration of H2 will be present at
equilibrium?
Answer in units of M.

Answer 1

(HI) = 0.5mol/2L = 0.25 M
................. 2HI(g) ⇀↽ H2(g) + I2(g)
I..................0.25M.........0...........0
C,,,,,,,,,,,,,,,,,,-2x.............x............x
E,,,,,,,,,,,,,9,25-x..............x............x

Kc = (H2)(I2)/(HI)^2
Plug in the E line and solve for x = (H2) = (I2)
Post your work if you get stuck.