Question

The reversible reaction n2(g) + 3h2(g)<-> produces ammonia, which is a fertilizer. At quilibrium, a 1-L flask contains 0.15 mol H2, 0.25 mol N2, and 0.10 mol NH3. Calculate K eq for a reverse reaction?

Please HELPP!

Answer 1

I applaud your writing a question with capital letters to begin sentences and placing comma's, periods, and questions marks appropriately. However, you need to sharpen those skills in writing equations since n2 and h2 have different meanings in lower case versus upper case.
N2(g) + 3H2(g) ==> 2NH3(g)

Write the Keq expression.
Kc = (N2)(H2)^3/(NH3)^2
All you need to do is to calculate the molarity of each component and substitute into the Kc expression.
M = moles/L; therefore,
(H2) = moles/L = 0.15/1 L = 0.15 M
(N2) = 0.25 mol/L etc.