Consider the formation of ammonia in two experiments. (a) to a 1.00-l container at 727°c, 1.30 mol of n2 and 1.65 mol of h2 are added. at equilibrium, 0.100 mol of nh3 is present. calculate the equilibrium concentrations of n2 and h2 for the reaction: 2 nh3(g) n2(g) + 3 h2(g). n2 1.25 mol/l h2 1.50 mol/l find kc for the reaction. kc = 422 m2 (b) in a different 1.00-l container at the same temperature, equilibrium is established with 8.34 ✕ 10−2 mol of nh3, 1.50 mol of n2, and 1.25 mol of h2 present. calculate kc for the reaction: nh3(g) 1 2 n2(g) + 3 2 h2(g). 20.5 m (c) what is the relationship between the kc values in parts (a) and (b)? why aren't these values the same?

When it comes to equilibrium reactions, it useful to do ICE analysis. ICE stands for Initial-Change-Equilibrium. You subtract the initial and change to determine the equilibrium amounts which is the basis for Kc. Kc is the equilibrium constant of concentration which is just the ratio of products to reactant. 

Let's do the ICE analysis

      2 NH₃ ⇄ N₂ + 3 H₂
I         0        1.3    1.65
C     +2x       -x      -3x
-------------------------------------
E       0.1        ?        ?

The variable x is the amount of moles of the substances that reacted. You apply the stoichiometric coefficients by multiplying it by x. Now, we can solve x by:

Equilibrium NH₃ = 0.1 = 0 + 2x
x = 0.05 mol
Therefore,
Equilibrium H₂ = 1.65 - 3(0.05) = 1.5 mol
Equilibrium N₂ = 1..3 - 0.05 = 1.25 mol

For the second part, I am confused with the given reaction because the stoichiometric coefficients do not balance which violates the law of conservation of mass. But you should remember that the Kc values might differ because of the stoichiometric coefficient. For a reaction: aA + bB ⇄ cC, the Kc for this is



Hence, Kc could vary depending on the stoichiometric coefficients of the reaction.

Answer) by studying proton collisions hope this : )