Question

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?

Answer 1

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

Step-by-step 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 =
`3* 0.255\ atm` = 0.765 atm

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