Solid copper (ii) oxide is placed in a sealed container with an excess of ammonia gas. when equilibrium is established the partial pressure of n2 is 0.255atm. the reaction is as follows: 3cuo (s) + 2 nh3 (g) ↔ 3 cu (s) + n2 (g) + 3 h2o (g) a) what is the partial pressure of h2o (g) at equilibrium?

0.765 atm is the partial pressure of H₂O (g) at equilibrium.

Explanation:

Considering the ICE table for the given equilibrium as:-

                        3CuO (s) + 2 NH₃ (g) ↔ 3 Cu (s) + N₂ (g) + 3 H₂O (g)

                       ------------------------------------------------------------------------------

t=teq                 a - 3P           b - 2P              3P          P            3P

Also, given that:-

Partial pressure of N₂ = 0.255 atm

From the above table, x = 0.255 atm

Thus, Partial pressure of H₂O = 3x = = 0.765 atm

0.765 atm is the partial pressure of H₂O (g) at equilibrium.

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