For the reaction: 2NO(g) + H2(g) <=> N2O(g) + H2O(g) + energy

Write the equilibrium constant expression for the reaction.
I wrote: [N2O][H2O] / [NO]2 [H2]

This reaction takes place at 25°C. At this temperature the concentration of NO is found to be 1.75 mol/L, the concentration of H2 is 3.00 mol/L, the concentration of N2O is 4.00 mol/L, and the concentration of H2O is 1.25 mol/L. Using the balanced chemical equation, calculate the equilibrium constant for this system.

Kat, this is nothing more than plugging the numbers into the Keqp expression and evaluating it. What do you not understand about it. The Keq expression you wrote is correct.

I can plut the numbers into the equation but do I need to do anything with the 25*C?

No. The reaction takes place at 25 C so the Keq will Keq at 25C. The problem doesn't make that clear. If the problem were stated correctly it would say, "Using the balanced chemical equation, calculate the equilibrium constant for this system at 25 C."

O.k So would this be correct?

K=(4.00 mol/L)(1.25 mol/L) /(1.75 mol/L)2 (3.00 mol/L)
K=0.544 mol/L

To calculate the equilibrium constant (K) for this reaction, you need to substitute the given concentrations into the equilibrium constant expression you derived correctly:

K = [N2O][H2O] / ([NO]^2[H2])

Now, substitute the values given into the expression:

K = (4.00 mol/L)(1.25 mol/L) / (1.75 mol/L)^2 (3.00 mol/L)

K = 5.00 mol^2 L^-2 / 3.06 mol^2 L^-2

Simplify:

K ≈ 1.63

Therefore, the equilibrium constant for this system at 25°C is approximately 1.63.