For a reversible reaction
`A+BhArrC`
`((dx)/(dt))=2.0xx10^(3) L mol^(-1) s^(-1) [A][B]-1.0xx10^(2) s^(-1) [C]` where `x` is the amount of 'A' dissociated. The value of equilibrium constant `(K_(eq))` is

Options

A. `10`
B. `0.05`
C. `20`
D. Cannot be calculated

Explanation

To find the equilibrium constant \( K_{eq} \) for the reversible reaction \( A + B \rightleftharpoons C \), we start with the rate of change of concentration given by the equation:

\[ \frac{dx}{dt} = 2.0 \times 10^{3} \, L \, mol^{-1} \, s^{-1} \, [A][B] - 1.0 \times 10^{2} \, s^{-1} \, [C] \]

### Step 1: Set the rate of change to zero at equilibrium
At equilibrium, the rate of change of concentration is zero, which means:

\[ \frac{dx}{dt} = 0 \]

Substituting this into the equation gives:

\[ 0 = 2.0 \times 10^{3} \, [A]_{eq} [B]_{eq} - 1.0 \times 10^{2} \, [C]_{eq} \]

### Step 2: Rearrange the equation
Rearranging the equation to isolate the concentration of \( C \):

\[ 1.0 \times 10^{2} \, [C]_{eq} = 2.0 \times 10^{3} \, [A]_{eq} [B]_{eq} \]

### Step 3: Express \( K_{eq} \)
The equilibrium constant \( K_{eq} \) for the reaction is defined as:

\[ K_{eq} = \frac{[C]_{eq}}{[A]_{eq} [B]_{eq}} \]

### Step 4: Substitute the rearranged equation into the expression for \( K_{eq} \)
From the rearranged equation, we can express \( [C]_{eq} \):

\[ [C]_{eq} = \frac{2.0 \times 10^{3}}{1.0 \times 10^{2}} \, [A]_{eq} [B]_{eq} \]

Substituting this into the expression for \( K_{eq} \):

\[ K_{eq} = \frac{\frac{2.0 \times 10^{3}}{1.0 \times 10^{2}} \, [A]_{eq} [B]_{eq}}{[A]_{eq} [B]_{eq}} = \frac{2.0 \times 10^{3}}{1.0 \times 10^{2}} \]

### Step 5: Calculate \( K_{eq} \)
Now, we can simplify the expression:

\[ K_{eq} = \frac{2.0 \times 10^{3}}{1.0 \times 10^{2}} = 20 \]

### Conclusion
Thus, the value of the equilibrium constant \( K_{eq} \) is:

\[ \boxed{20} \]