At `500 K`, the equilibrium constant for reaction `cis-C_(2)H_(2)Cl_(2) hArr transa-C_(2)H_(2)Cl_(2)` is `0.6`. At the same temperature, the equilibrium constant for the reaction `trans-C_(2)H_(2)Cl_(2) hArr cis-C_(2)H_(2)Cl_(2)` will be

Options

A. `1.67`
B. `1.65`
C. `1.06`
D. `1.60`

Answer

To solve the problem, we need to determine the equilibrium constant for the reaction where trans-C₂H₂Cl₂ is converted to cis-C₂H₂Cl₂, given that the equilibrium constant for the reverse reaction is provided.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The equilibrium constant for the reaction:
\[ \text{cis-C}_2\text{H}_2\text{Cl}_2 \rightleftharpoons \text{trans-C}_2\text{H}_2\text{Cl}_2 \]
is given as \( K_1 = 0.6 \) at \( 500 \, K \).

2. **Understand the Relationship Between the Reactions:**
- The reverse reaction is:
\[ \text{trans-C}_2\text{H}_2\text{Cl}_2 \rightleftharpoons \text{cis-C}_2\text{H}_2\text{Cl}_2 \]
- The equilibrium constant for the reverse reaction, \( K_2 \), is related to \( K_1 \) by the following relationship:
\[ K_2 = \frac{1}{K_1} \]

3. **Calculate the Equilibrium Constant for the Reverse Reaction:**
- Substitute the value of \( K_1 \) into the equation:
\[ K_2 = \frac{1}{0.6} \]
- Perform the calculation:
\[ K_2 = \frac{1}{0.6} = 1.6667 \]

4. **Round the Result:**
- Rounding \( 1.6667 \) gives us \( 1.67 \).

5. **Final Answer:**
- Therefore, the equilibrium constant \( K_2 \) for the reaction trans-C₂H₂Cl₂ to cis-C₂H₂Cl₂ is approximately \( 1.67 \).

### Conclusion:
The equilibrium constant for the reaction \( \text{trans-C}_2\text{H}_2\text{Cl}_2 \rightleftharpoons \text{cis-C}_2\text{H}_2\text{Cl}_2 \) is \( K_2 = 1.67 \).