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] = 3.1 M and [H2] = 0.8 M, both at equilibrium.
How many grams of ammonia are in
the 10 L reaction vessel at equilibium? Use
the equilibrium equation
N2(g) + 3 H2(g) ⇀↽ 2 NH3(g)
Answer in units of g

Answer 1

.........N2(g) + 3 H2(g) ⇀↽> 2 NH3(g)
E.......3.1M......0.8M..........x

K = 5.2E-5 = (NH3O^2/(N2)(H2)^3
Plug the equilibrium numbers into Keq and solve for (NH3) in mol/L.
Convert to mols in 10 L and convert that to grams.

Answer 2

45