What is the equilibrium expression for the reaction P4(s) + 5O2(g) P4O10(s)? (a) Kc = [P4O10] /

Q: Why do we completely ignore solids in the equilibrium expression?

We don't exactly 'ignore' them; we assign them an active mass of $1$. This is because the molar concentration of a pure solid is its density divided by its molar mass. Since both density and molar mass are constant at a given temperature, the concentration of a pure solid remains constant regardless of its quantity.

Q: If I add more solid $P_4$ to the container, will the equilibrium shift forward?

No, it will not. According to Le Chatelier's Principle, changing the amount of a pure solid or liquid does not affect the equilibrium position, provided that at least some of the solid/liquid is present. The equilibrium only responds to changes in concentration, pressure, or temperature.

Q: What if water is a product in a reaction? Do we take its active mass as $1$ too?

It depends on its physical state. If water is present as a pure liquid solvent (e.g., in an aqueous reaction), its concentration is massive and practically constant, so we take it as $1$. However, if water is present as a gas (water vapor) or in a non-aqueous solvent, its concentration can change, and it must be included in the equilibrium expression.

Q: Why is the power of $[O_2]$ exactly $5$?

In the equilibrium expression, the concentration of each species is raised to the power of its stoichiometric coefficient from the balanced chemical equation. Since $5$ moles of $O_2$ react with $1$ mole of $P_4$, the coefficient is $5$.

Q: Is $K_p$ the same as $K_c$ for this reaction?

No, they are related by the equation $K_p = K_c(RT)^{\Delta n_g}$. Here, $\Delta n_g$ is the change in gaseous moles (moles of gaseous products - moles of gaseous reactants). For this reaction, $\Delta n_g = 0 - 5 = -5$. Therefore, $K_p = K_c(RT)^{-5}$.

Question

What is the equilibrium expression for the reaction
$P_4(s) + 5O_2(g) \rightleftharpoons P_4O_{10}(s)$?

Options:

(A) $K_c = \frac{[P_4O_{10}]}{[P_4][O_2]^5}$

(B) $K_c = \frac{[P_4O_{10}]}{5[P_4][O_2]^5}$

(C) $K_c = [O_2]^5$

(D) $K_c = \frac{1}{[O_2]^5}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Let's dive into a fundamental yet crucial concept of Chemical Equilibrium. We are given the reaction:

$$P_4(s) + 5O_2(g) \rightleftharpoons P_4O_{10}(s)$$

At first glance, it looks like a standard equilibrium problem where we just need to write the ratio of products to reactants. However, the key to unlocking this problem lies in carefully observing the **physical states** of the species involved. We have solid phosphorus ($s$) reacting with gaseous oxygen ($g$) to form solid phosphorus pentoxide ($s$).

### The Master Equation and The Catch

According to the **Law of Mass Action**, the equilibrium constant $K_c$ is expressed as the ratio of the product of the active masses (molar concentrations) of the products to that of the reactants, each raised to the power of their respective stoichiometric coefficients.

But here is the catch! The active mass of **pure solids** and **pure liquids** is always taken as unity ($1$).

Why does this happen? The molar concentration is defined as moles per unit volume, which is essentially density divided by molar mass. For a pure solid or a pure liquid, the density is an intensive property—it remains constant regardless of how much substance you have. Therefore, their active mass does not change during the course of the reaction and is conventionally taken as $1$ to simplify the equilibrium expression.

### Final Calculation

Let's apply this golden rule to our specific reaction.

For the reactant phosphorus, $P_4$, it is in the solid state, so its active mass is:
$$[P_4(s)] = 1$$

Similarly, for the product phosphorus pentoxide, $P_4O_{10}$, it is also in the solid state, so its active mass is:
$$[P_4O_{10}(s)] = 1$$

Now, let's construct the equilibrium expression for $K_c$:

$$K_c = \frac{[P_4O_{10}]}{[P_4][O_2]^5}$$

Substituting the active masses of the solids as $1$, the expression simplifies beautifully:

$$K_c = \frac{1}{1 \cdot [O_2]^5}$$

$$K_c = \frac{1}{[O_2]^5}$$

And there we have it! The equilibrium constant depends solely on the concentration of the gaseous oxygen. This elegant simplification is a favorite trap in competitive exams, so always keep an eye on those state symbols!

Animated Solution for Chemistry - Chemical Equilibrium: What is the equilibrium expression for the reaction $P_{4} (s) + 5 O_{2} (g) ⇌ P_{4} O_{10} (s)$ ?

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Visualized Solution

$R e a c t i o n E q u a t i o n$

$L a w o f M a ss A c t i o n$

$A ppl y in g t h e R u l e$

$E q u i l ib r i u m E x p r ess i o n$

$F oo d f or T h o ug h t$

The Sigma Insight: Law of Mass Action

Analyzing the Setup

The Master Equation and The Catch

Final Calculation

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Animated Solution for Chemistry - Chemical Equilibrium: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What is the equilibrium expression for the reaction P4​(s)+5O2​(g)⇌P4​O10​(s)?

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reaction Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Law of Mass Action

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Rule

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equilibrium Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Food for Thought