For the following reaction :
`NO_((g))+O_(3(g))hArrNO_(2(g))+O_(2(g))`
The value of `K_(c)` is `8.2xx10^(4)`. What will be the value of `K_(c)` for the reverse reaction ?

Options

A. (a) `8.2xx10^(4)`
B. (b) `(1)/(8.2xx10^(4))`
C. (c) `(8.2xx10^(4))`
D. (d) `sqrt(8.2xx10^(4))`

Explanation

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

### Step 1: Write the Forward Reaction
The forward reaction is given as:
\[ \text{NO}_{(g)} + \text{O}_3_{(g)} \rightleftharpoons \text{NO}_2_{(g)} + \text{O}_2_{(g)} \]
We are provided with the equilibrium constant \( K_c \) for this reaction, which is:
\[ K_c = 8.2 \times 10^4 \]

### Step 2: Write the Reverse Reaction
The reverse reaction can be written as:
\[ \text{NO}_2_{(g)} + \text{O}_2_{(g)} \rightleftharpoons \text{NO}_{(g)} + \text{O}_3_{(g)} \]

### Step 3: Write the Expression for \( K_c \) for Both Reactions
For the forward reaction, the expression for \( K_c \) is:
\[ K_c = \frac{[\text{NO}_2][\text{O}_2]}{[\text{NO}][\text{O}_3]} \]

For the reverse reaction, the expression for \( K_c' \) is:
\[ K_c' = \frac{[\text{NO}][\text{O}_3]}{[\text{NO}_2][\text{O}_2]} \]

### Step 4: Relate \( K_c' \) to \( K_c \)
From the expressions, we can see that:
\[ K_c' = \frac{1}{K_c} \]

### Step 5: Calculate \( K_c' \)
Now substituting the value of \( K_c \):
\[ K_c' = \frac{1}{8.2 \times 10^4} \]

### Step 6: Perform the Calculation
Calculating the reciprocal:
\[ K_c' = \frac{1}{8.2 \times 10^4} \approx 1.22 \times 10^{-5} \]

### Final Answer
Thus, the value of \( K_c \) for the reverse reaction is:
\[ K_c' \approx 1.22 \times 10^{-5} \]

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