For a chemical reaction of the type `AhArrB, K=2.0 and BhArrC, K=0.01`. Equilibrium constant for the reaction `2ChArr 2A` is

Options

A. 25
B. 50
C. 2500
D. `4 xx 10^(-4)`

Answer

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

### Step 1: Write the given reactions and their equilibrium constants.
1. \( A \rightleftharpoons B \) with \( K_1 = 2.0 \)
2. \( B \rightleftharpoons C \) with \( K_2 = 0.01 \)

### Step 2: Combine the reactions to find \( A \rightleftharpoons C \).
To find the equilibrium constant for the reaction \( A \rightleftharpoons C \), we can add the two reactions together:

- From \( A \rightleftharpoons B \) (forward reaction), we have \( K_1 = 2.0 \).
- From \( B \rightleftharpoons C \) (forward reaction), we have \( K_2 = 0.01 \).

When we add these reactions, \( B \) cancels out:

\[ A \rightleftharpoons B \quad (K_1 = 2.0) \]
\[ B \rightleftharpoons C \quad (K_2 = 0.01) \]

Adding these gives:
\[ A \rightleftharpoons C \]

The equilibrium constant for this combined reaction \( K_{AC} \) is given by:
\[ K_{AC} = K_1 \times K_2 = 2.0 \times 0.01 = 0.02 \]

### Step 3: Reverse the reaction to find \( C \rightleftharpoons A \).
Now, we need to reverse the reaction \( A \rightleftharpoons C \) to get \( C \rightleftharpoons A \). The equilibrium constant for the reverse reaction is the inverse of \( K_{AC} \):

\[ K_{CA} = \frac{1}{K_{AC}} = \frac{1}{0.02} = 50 \]

### Step 4: Multiply the reaction by 2 to find \( 2C \rightleftharpoons 2A \).
Next, we need to multiply the reaction \( C \rightleftharpoons A \) by 2 to get \( 2C \rightleftharpoons 2A \). When we multiply a reaction by a coefficient, the equilibrium constant is raised to the power of that coefficient:

\[ K_{2C \rightleftharpoons 2A} = K_{CA}^2 = 50^2 = 2500 \]

### Final Answer
Thus, the equilibrium constant for the reaction \( 2C \rightleftharpoons 2A \) is \( 2500 \).

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