In the following equilibrium reaction,
2A `hArr` B +C
the equilibrium concentrations of A, B and C are `1xx10^(-3)` M , `2xx10^(-3)` M and `3xx10^(-3)` M respectively at 300 K. The value of `K_c` for this equilibrium at the same temperature is

Options

A. `1/6`
B. 6
C. `1/36`
D. 36

Answer

To find the equilibrium constant \( K_c \) for the reaction

\[ 2A \rightleftharpoons B + C \]

given the equilibrium concentrations of A, B, and C, we can follow these steps:

### Step 1: Write the expression for \( K_c \)

The equilibrium constant \( K_c \) is defined as:

\[ K_c = \frac{[B]^{b} \times [C]^{c}}{[A]^{a}} \]

where \( a, b, c \) are the coefficients of the balanced equation. For our reaction:

- \( a = 2 \) (for \( A \))
- \( b = 1 \) (for \( B \))
- \( c = 1 \) (for \( C \))

Thus, the expression for \( K_c \) becomes:

\[ K_c = \frac{[B]^1 \times [C]^1}{[A]^2} \]

### Step 2: Substitute the equilibrium concentrations

From the problem, we have the following equilibrium concentrations:

- \([A] = 1 \times 10^{-3} \, \text{M}\)
- \([B] = 2 \times 10^{-3} \, \text{M}\)
- \([C] = 3 \times 10^{-3} \, \text{M}\)

Substituting these values into the \( K_c \) expression:

\[ K_c = \frac{(2 \times 10^{-3})^1 \times (3 \times 10^{-3})^1}{(1 \times 10^{-3})^2} \]

### Step 3: Calculate the numerator and denominator

Calculating the numerator:

\[ \text{Numerator} = (2 \times 10^{-3}) \times (3 \times 10^{-3}) = 6 \times 10^{-6} \]

Calculating the denominator:

\[ \text{Denominator} = (1 \times 10^{-3})^2 = 1 \times 10^{-6} \]

### Step 4: Calculate \( K_c \)

Now we can calculate \( K_c \):

\[ K_c = \frac{6 \times 10^{-6}}{1 \times 10^{-6}} = 6 \]

### Conclusion

Thus, the value of \( K_c \) for the equilibrium at 300 K is:

\[ \boxed{6} \]

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