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. `4xx10^(-4)`

Answer

To find the equilibrium constant for the reaction \( 2C \rightleftharpoons 2A \), we will use the given equilibrium constants for the reactions \( A \rightleftharpoons B \) and \( B \rightleftharpoons C \).

### Step-by-Step Solution:

1. **Identify the given reactions and their equilibrium constants:**
- For the reaction \( A \rightleftharpoons B \), the equilibrium constant \( K_1 = 2.0 \).
- For the reaction \( B \rightleftharpoons C \), the equilibrium constant \( K_2 = 0.01 \).

2. **Write the equilibrium constant expressions:**
- For \( A \rightleftharpoons B \):
\[ K_1 = \frac{[B]}{[A]} = 2.0 \]
- For \( B \rightleftharpoons C \):
\[ K_2 = \frac{[C]}{[B]} = 0.01 \]

3. **Manipulate the reactions to find the desired reaction:**
- We need to find the equilibrium constant for the reaction \( 2C \rightleftharpoons 2A \).
- To do this, we will first reverse the reactions and then multiply them.

4. **Reverse the first reaction:**
- Reversing \( A \rightleftharpoons B \) gives:
\[ B \rightleftharpoons A \quad \text{with } K' = \frac{1}{K_1} = \frac{1}{2.0} = 0.5 \]

5. **Reverse the second reaction:**
- Reversing \( B \rightleftharpoons C \) gives:
\[ C \rightleftharpoons B \quad \text{with } K'' = \frac{1}{K_2} = \frac{1}{0.01} = 100 \]

6. **Multiply the reversed reactions:**
- Now, we need to multiply both reversed reactions to obtain:
\[ 2C \rightleftharpoons 2B \quad \text{and} \quad 2B \rightleftharpoons 2A \]
- The equilibrium constant for the reaction \( 2C \rightleftharpoons 2A \) will be:
\[ K = K' \times K'' = (0.5)^2 \times (100)^2 \]

7. **Calculate the new equilibrium constant:**
- Calculate \( K' \) for \( 2B \rightleftharpoons 2A \):
\[ K' = (0.5)^2 = 0.25 \]
- Calculate \( K'' \) for \( 2C \rightleftharpoons 2B \):
\[ K'' = (100)^2 = 10000 \]
- Now, multiply these two:
\[ K = K' \times K'' = 0.25 \times 10000 = 2500 \]

### Final Answer:
The equilibrium constant for the reaction \( 2C \rightleftharpoons 2A \) is \( K = 2500 \).

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