For the reaction, `A+2B hArr C`, the expession for equilibrium constant is

Options

A. `([A][B]^(2))/([C])`
B. `([A][B])/([C])`
C. `([C])/([A][B]^(2))`
D. `([C])/([2B][A])`

Explanation

To derive the expression for the equilibrium constant for the reaction \( A + 2B \rightleftharpoons C \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation is given as:
\[ A + 2B \rightleftharpoons C \]

### Step 2: Identify the products and reactants
In this reaction:
- Reactants: \( A \) and \( B \)
- Product: \( C \)

### Step 3: Write the general form of the equilibrium constant expression
The equilibrium constant \( K_c \) for a reaction is expressed as the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their respective coefficients in the balanced equation.

### Step 4: Apply the equilibrium constant expression to the given reaction
For the reaction \( A + 2B \rightleftharpoons C \), the equilibrium constant expression is:
\[ K_c = \frac{[C]}{[A][B]^2} \]
Here:
- \([C]\) is the concentration of the product \( C \),
- \([A]\) is the concentration of the reactant \( A \),
- \([B]\) is the concentration of the reactant \( B \), and since \( B \) has a coefficient of 2, we square its concentration.

### Step 5: Finalize the expression
Thus, the final expression for the equilibrium constant \( K_c \) for the reaction \( A + 2B \rightleftharpoons C \) is:
\[ K_c = \frac{[C]}{[A][B]^2} \]