The value of `K_(c )` for the reaction `3O_(2)(g) hArr 2O_(3)(g)` is `2.0xx10^(-50)` at `25^(@)C`. If the equilibrium concentration of `O_(2)` in air at `25^(@)C` is `1.6xx10^(-2)`, what is the concentration of `O_(3)`?

Explanation

To solve the problem, we need to find the concentration of \( O_3 \) at equilibrium given the equilibrium constant \( K_c \) and the concentration of \( O_2 \).

### Step-by-Step Solution:

1. **Write the equilibrium expression for the reaction:**
The reaction is given as:
\[ 3O_2(g) \rightleftharpoons 2O_3(g) \]
The equilibrium constant \( K_c \) is expressed as:
\[ K_c = \frac{[O_3]^2}{[O_2]^3} \]

2. **Substitute the known values into the expression:**
We know that:
\[ K_c = 2.0 \times 10^{-50} \]
and the concentration of \( O_2 \) at equilibrium is:
\[ [O_2] = 1.6 \times 10^{-2} \, \text{M} \]
Now, substituting these values into the equilibrium expression:
\[ 2.0 \times 10^{-50} = \frac{[O_3]^2}{(1.6 \times 10^{-2})^3} \]

3. **Calculate \( (1.6 \times 10^{-2})^3 \):**
\[ (1.6 \times 10^{-2})^3 = 1.6^3 \times (10^{-2})^3 = 4.096 \times 10^{-6} \]

4. **Rearrange the equation to solve for \( [O_3]^2 \):**
\[ [O_3]^2 = 2.0 \times 10^{-50} \times 4.096 \times 10^{-6} \]

5. **Calculate \( [O_3]^2 \):**
\[ [O_3]^2 = 8.192 \times 10^{-56} \]

6. **Take the square root to find \( [O_3] \):**
\[ [O_3] = \sqrt{8.192 \times 10^{-56}} = 2.86 \times 10^{-28} \, \text{M} \]

### Final Answer:
The concentration of \( O_3 \) at equilibrium is:
\[ [O_3] = 2.86 \times 10^{-28} \, \text{M} \]