What is the equilibrium expression for the reaction `P_(4(s)) + 5O_(2(g))hArrP_(4)O_(10(s))` ?

Options

A. a. `k_(c)=([P_(4)][O_(2)]^(5))/([P_(4)O_(10)])`
B. b. `K_(c)=(1)/([O_(2)]^(5))`
C. c. `K_(c)=([P_(4)O_(10)])/([P_(4)][O_(2)]^(5))`
D. d. `K_(c)=[O_(2)]^(5)`

Explanation

To find the equilibrium expression for the reaction:

\[ P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_{4}O_{10(s)} \]

we follow these steps:

### Step 1: Identify the equilibrium expression format
The general form of the equilibrium constant expression \( K_c \) is given by:

\[ K_c = \frac{[\text{products}]}{[\text{reactants}]} \]

where the concentrations of the products are in the numerator and the concentrations of the reactants are in the denominator.

### Step 2: Write the expression for the given reaction
In this reaction, the products and reactants are:

- Products: \( P_{4}O_{10(s)} \)
- Reactants: \( P_{4(s)} \) and \( O_{2(g)} \)

### Step 3: Consider the states of matter
Since \( P_{4}O_{10 \text{(s)}} \) and \( P_{4 \text{(s)}} \) are solids, their concentrations do not appear in the equilibrium expression. The concentration of solids is considered to be unity (1).

### Step 4: Write the equilibrium expression
Thus, the equilibrium expression for the reaction will only include the gaseous reactants:

\[ K_c = \frac{1}{[O_{2}]^5} \]

### Step 5: Finalize the expression
Since the concentration of solids is considered as 1, we can simplify the expression:

\[ K_c = \frac{1}{[O_{2}]^5} \]

### Conclusion
The equilibrium expression for the reaction is:

\[ K_c = \frac{1}{[O_{2}]^5} \]