The equilibrium constant for the reaction
A(s) ⇌ M(s) + 1/2O₂(g)
is Kₚ = 4. At equilibrium, the partial pressure of O₂ is _________ atm. (Round off to the nearest integer)

Solution

Chemistry ma’am · 3 steps

Solids drop out of Kₚ, so only p_(O₂)^(1/2) is left. Set it equal to 4 and square.

1. Leave out solids. A pure solid has a constant activity of 1, so A(s) and M(s) do not appear in Kₚ. Only the gas, O₂, is left, raised to its coefficient 1/2.

   Kₚ = (p_(O₂))^(1/2) = 4

2. Remove the half power. Square both sides: p_(O₂) = 4² = 16 atm.
3. So. The partial pressure of O₂ at equilibrium is 16 atm.

The common slip: 2. Taking the square root of 4 instead of squaring it gives 2. Undoing a power of 1/2 means squaring.

The idea to keep. In a heterogeneous equilibrium, write Kₚ with gases only, each raised to its stoichiometric coefficient. A fractional coefficient becomes a fractional power.

Answer from the NTA final answer key. Solution by PrepWiser.