For the following equilibrium, `K_(c )=6.3xx10^(14) at 1000 K`
`NO(g)+O_(3)(g) hArr NO_(2)(g)+O_(2)(g)`
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is `K_(c )`, for the reverse reaction?

Explanation

To find the equilibrium constant \( K_c \) for the reverse reaction, we can follow these steps:

### Step-by-step Solution:

1. **Identify the Given Reaction and \( K_c \)**:
The forward reaction is:
\[ \text{NO}(g) + \text{O}_3(g) \rightleftharpoons \text{NO}_2(g) + \text{O}_2(g) \]
The equilibrium constant for this reaction at 1000 K is given as:
\[ K_c = 6.3 \times 10^{14} \]

2. **Write the Reverse Reaction**:
The reverse reaction can be written as:
\[ \text{NO}_2(g) + \text{O}_2(g) \rightleftharpoons \text{NO}(g) + \text{O}_3(g) \]

3. **Relate \( K_c \) of Forward and Reverse Reactions**:
The relationship between the equilibrium constants of the forward and reverse reactions is given by:
\[ K_{c, \text{reverse}} = \frac{1}{K_{c, \text{forward}}} \]

4. **Calculate \( K_c \) for the Reverse Reaction**:
Substitute the value of \( K_c \) for the forward reaction into the equation:
\[ K_{c, \text{reverse}} = \frac{1}{6.3 \times 10^{14}} \]

5. **Perform the Calculation**:
\[ K_{c, \text{reverse}} = \frac{1}{6.3 \times 10^{14}} \approx 1.59 \times 10^{-15} \]

6. **Final Answer**:
Therefore, the equilibrium constant \( K_c \) for the reverse reaction is:
\[ K_{c, \text{reverse}} \approx 1.59 \times 10^{-15} \]