A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibrium is established: 2NO(g)+2H2(g)⥫⥬==N2(g)+2H2O(g) At equilibrium [NO]=0.062M. Calculate the equilibrium concentrations of H2, N2, and H2O.
Calculate the equilibrium concentration of N2

The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.

2NO(g) + 2H2(g) ⇒ N2(g) + 2H2O(g)

i mol 0.10 0.050 0.10

c mol -0.038 -0.038 +0019 +0.038

e mol 0.062 0.012 00.019 0.057

Since the volume of the vessel is 1.0 L, the concentrations in molarity are:

[NO] = 0.062 M

[H2] = 0.012 M

[N2] = 0.019 M

[H2O] = 0.057 M

shrutiagrawal1798 shrutiagrawal1798 · Answer 2 · 2022-03-01T13:53:57+01:00

The equilibrium concentration of hydrogen is 0.012 M, nitrogen is 0.019 M, and water is 0.057 M.

What is equilibrium?

Equilibrium is the condition of reaction in which the rate of formation of product, and the reactant breakdown are the same.

The reaction is processed with the change in the stoichiometry of the concentration in the reaction.

The change in the reaction concentration is given in the ICE table attached.

The volume of the compounds in the reaction is given as 1 L.

The concentration of each compound will be:

Nitrogen oxide= 0.062 M
Hydrogen = 0.012 M
Nitrogen = 0.019 M
Water = 0.057 M

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