For the formation of ammonia, the equilibrium constant is known to be 5.2 × 10^−5 at 25◦C. After analysis, it is determined that [N2] =3M and [H2]=0.8M, both at equilibrium. How many grams of ammonia are in the 10L reaction vessel at equilibium? Use the equilibrium equation

Question

For the formation of ammonia, the equilibrium constant is known to be 5.2 × 10^−5 at 25◦C. After analysis, it is determined that [N2] =3M and [H2]=0.8M, both at equilibrium. How many grams of ammonia are in the 10L reaction vessel at equilibium? Use the equilibrium equation
N2(g) + 3H2(g) ⇀ ↽ 2NH3(g)
Answer in units of g

DrBob222 · Answer

K = (NH3)^2/(N2)(H2)^3
You know N2 and H2, substitute with Kc and find NH3 in M = mols/L.
Then to find mols in 10L, that is
mols = M x L
Then mol = grams/molar mass. You know molar mass and mols, solve for grams.

J · Answer

1.2 g NH3