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What is the equilibrium expression for t...

What is the equilibrium expression for the reaction`P_(4)(s)+50_(2)(g)hArrP_(4)O_(10)(s)`

A

`K_(c)=[O_(2)]^(5)`

B

`K_(c)=[P_(4)O_(10)]//5[P_(4)][O_(2)]`

C

`K_(c)=[P_(4)O_(10)]//[P_(4)][O_(2)]^(5)`

D

`K_(c)=1//[O_(2)]^(5)`

Text Solution

AI Generated Solution

The correct Answer is:
To derive the equilibrium expression for the reaction \( P_4(s) + 5O_2(g) \rightleftharpoons P_4O_{10}(s) \), we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Reaction Components**: The reaction involves: - Reactants: \( P_4 \) (solid) and \( O_2 \) (gas) - Product: \( P_4O_{10} \) (solid) 2. **Understand the Equilibrium Constant Expression**: The equilibrium constant expression, \( K_c \), is defined as the ratio of the concentrations of products to the concentrations of reactants, each raised to the power of their respective stoichiometric coefficients. 3. **Consider the States of Matter**: In equilibrium expressions, concentrations of pure solids and pure liquids are omitted because their activities are considered to be equal to 1 and do not change during the reaction. 4. **Write the Equilibrium Expression**: For the given reaction: - The product \( P_4O_{10} \) is a solid, so it is omitted. - The reactant \( P_4 \) is also a solid, so it is omitted. - The only species that remains in the expression is \( O_2 \) (gas). Therefore, the equilibrium expression is: \[ K_c = \frac{1}{[O_2]^5} \] 5. **Final Expression**: The final equilibrium expression for the reaction is: \[ K_c = \frac{1}{[O_2]^5} \]
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