X2 + Y2 <=> 2XY

0.50 mole each of X2 and Y2 are placed in a 1.0 litre vessel and allowed to reach equilibrium at a given temperature. The equilibrium concentrations of XY is found to be 0.025 mol/L. What is the equilibrium cosntant for this reaction?

Do I just get the concentration of X2 and Y2 by dividing 0.5 mol with 1L and then take the answer and the equilibrium concentrations of XY - 0.025 mol/l and divide 0.025 by the concentration of X2 and Y2 in order to get the equilibrium constant?

Your explanation isn't clear to me but it sounds like you have it ok.
initial concentrations:
X2 = 0.5 M
Y2 = 0.5 M
XY = 0

change:
XY = +0.025
X2 = 0.025-(1mol X2/2 mol XY)=1/2*0.025
Y2 = 0.025-(1mol Y2/2 mol XY)=1/2*0.025

eqilibrium:
XY = 0.025
X2 = 0.5 - 1/2(0.025)
Y2 = 0.5 - 1/2(0.025)

Keq = (XY)^2/(X2)(Y2)=
(0.025)^2/(0.4875)^2 = ??

Check my thinking. Check my arithmetic.

Your approach to finding the equilibrium constant is correct. To solve this problem, you need to first determine the initial concentrations of X2 and Y2, which is 0.5 mol/L each since 0.5 moles of each compound are placed in a 1.0 L vessel.

Next, you need to determine the changes in concentrations at equilibrium. Since the balanced chemical equation shows a 1:2 ratio between XY and X2 or Y2, for every 1 mol of XY formed, 0.025 mol of X2 and 0.025 mol of Y2 will be consumed. Therefore, the equilibrium concentrations are:

XY = 0.025 mol/L
X2 = 0.5 mol/L - (0.025 mol XY / 2 mol XY) = 0.4875 mol/L
Y2 = 0.5 mol/L - (0.025 mol XY / 2 mol XY) = 0.4875 mol/L

Finally, you can calculate the equilibrium constant (Keq) using the formula:

Keq = (XY)^2 / (X2)(Y2)
Keq = (0.025 mol/L)^2 / (0.4875 mol/L)(0.4875 mol/L)

Now, you just need to plug in these values and solve for Keq:

Keq = (0.000625 mol^2/L^2) / (0.23750625 mol^2/L^2)
Keq = 0.000625 mol^2/L^2 / 0.23750625 mol^2/L^2
Keq ≈ 0.002628

Therefore, the equilibrium constant for this reaction is approximately 0.002628.