Interpretation
At equilibrium, we express the equilibrium constant in terms of concentrations (\(K_c\)) or partial pressures (\(K_p\)) using the law of mass action.
However, there is a crucial refinement students must internalize:
Pure solids and pure liquids do not appear in equilibrium expressions because their concentrations (or activities) remain constant and are taken as unity.
This is not a shortcut—it reflects physical reality. A solid’s “effective concentration” does not change during the reaction, so including it would not affect the equilibrium ratio.
For gases:
Use \(K_c\) in terms of molar concentration
Use \(K_p\) in terms of partial pressures
They are related by:
\(K_p=K_c (RT)^{\Delta n}\)
where \(\Delta n = \text{moles of gaseous products} – \text{moles of gaseous reactants}\)
Concept Application
For each reaction, we will:
Identify phases carefully
Exclude solids and liquids
Include only:
gases (for both \(K_c\) and \(K_p\))
aqueous species (only in \(K_c\))
Compute \(\Delta n\) only using gaseous species (for \(K_p\))
Solution
(1) 3Fe (s) + 4H₂O (g) ⇌ Fe₃O₄ (s) + 4H₂ (g)
Only gases are included: H₂O(g) and H₂(g)
\(K_c = \frac{[H_2]^4}{[H_2O]^4}\)
For \(K_p\):
\(K_p = \frac{(P_{H_2})^4}{(P_{H_2O})^4}\)
Now check \(\Delta n\):
\(\Delta n = 4 – 4 = 0\)
So,
\(K_p = K_c\)
(2) HF (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + F⁻ (aq)
Here:
H₂O is a pure liquid → excluded
All others are aqueous species
\(K_c = \frac{[H_3O^+][F^-]}{[HF]}\)
No \(K_p\) expression exists because there are no gaseous species.
(3) 4NH₃ (g) + 5O₂ (g) ⇌ 4NO (g) + 6H₂O (g)
All species are gases → all are included
\(K_c = \frac{[NO]^4 [H_2O]^6}{[NH_3]^4 [O_2]^5}\)
\(K_p = \frac{(P_{NO})^4 (P_{H_2O})^6}{(P_{NH_3})^4 (P_{O_2})^5}\)
Now compute \(\Delta n\):
\(\Delta n = (4+6) – (4+5) = 10 – 9 = 1\)
So,
\(K_p = K_c (RT)^1 = K_c RT\)
(4) P₄ (s) + 6Cl₂ (g) ⇌ 4PCl₃ (l)
P₄ is a solid → excluded
PCl₃ is a liquid → excluded
Only Cl₂(g) remains
\(K_c = \frac{1}{[Cl_2]^6}\)
\(K_p = \frac{1}{(P_{Cl_2})^6}\)
Now,
\(\Delta n = 0 – 6 = -6\)
So,
\(K_p = K_c (RT)^{-6}\)
Insight
The fastest way to get these right in exams is to build a mental filter:
“If it’s a solid or liquid → ignore it. If it’s gas or aqueous → include it.”
Also remember:
If only one gaseous species appears, it will sit alone in the denominator or numerator—this often surprises students.
When \(\Delta n = 0\), \(K_p = K_c\), which is a powerful shortcut worth spotting instantly.
If you consistently apply these filters, equilibrium expressions stop being memorization—and become almost automatic.
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