In the equilibrium, `2AhArrB+C,` the equilibrium concentrations of `A.B and C at 300K` are `3xx10^(-4)M, 1xx10^(-4)M and 4.5xx10^(-4)M` respectively. Thte vlue of `K_(c)` for the above equilibrium at 300 K is

Options

A. `0.5`
B. `0.05`
C. `5.0`
D. `1.5`

Explanation

To find the equilibrium constant \( K_c \) for the reaction \( 2A \rightleftharpoons B + C \) at 300 K, we can follow these steps:

### Step 1: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is defined as:
\[ K_c = \frac{[B][C]}{[A]^2} \]
where \([A]\), \([B]\), and \([C]\) are the equilibrium concentrations of the respective species.

### Step 2: Substitute the equilibrium concentrations
From the problem, we have:
- \([A] = 3 \times 10^{-4} \, M\)
- \([B] = 1 \times 10^{-4} \, M\)
- \([C] = 4.5 \times 10^{-4} \, M\)

Now we can substitute these values into the \( K_c \) expression:
\[ K_c = \frac{(1 \times 10^{-4})(4.5 \times 10^{-4})}{(3 \times 10^{-4})^2} \]

### Step 3: Calculate the numerator and denominator
First, calculate the numerator:
\[ \text{Numerator} = (1 \times 10^{-4})(4.5 \times 10^{-4}) = 4.5 \times 10^{-8} \]

Now calculate the denominator:
\[ \text{Denominator} = (3 \times 10^{-4})^2 = 9 \times 10^{-8} \]

### Step 4: Divide the numerator by the denominator
Now, we can find \( K_c \):
\[ K_c = \frac{4.5 \times 10^{-8}}{9 \times 10^{-8}} = \frac{4.5}{9} = 0.5 \]

### Final Answer
Thus, the value of \( K_c \) for the equilibrium at 300 K is:
\[ K_c = 0.5 \]

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