At `700K`, for the reaction `2SO_(2)(g)+O_(2)(g)hArr2SO_(3)(g)` the `K_(p) "is" 3.2xx10^(4)`. At the same temperature the `K_(p)` for the reaction `SO_(3)(g)hArrSO_(2)(g)+0.50O_(2)(g)` is:

Options

A. `3.125xx^(-5)`
B. `5.59xx10^(-3)`
C. `1.79xx10^(4)`
D. `1.79xx10^(-2)`

Answer

To solve the problem, we need to find the \( K_p \) for the reaction:

\[ SO_3(g) \rightleftharpoons SO_2(g) + 0.50 O_2(g) \]

given that at \( 700K \), the \( K_p \) for the reaction:

\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]

is \( 3.2 \times 10^4 \).

### Step-by-Step Solution:

1. **Identify the given reaction and its \( K_p \)**:
The given reaction is:
\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]
with \( K_p = 3.2 \times 10^4 \).

2. **Reverse the reaction**:
To find the \( K_p \) for the reaction we are interested in, we first reverse the given reaction:
\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]
When we reverse a reaction, the new \( K_p \) is the reciprocal of the original \( K_p \):
\[ K_p' = \frac{1}{K_p} = \frac{1}{3.2 \times 10^4} \]

3. **Calculate the reciprocal**:
\[ K_p' = \frac{1}{3.2 \times 10^4} = 0.3125 \times 10^{-4} \]

4. **Relate to the desired reaction**:
The reaction we want is:
\[ SO_3(g) \rightleftharpoons SO_2(g) + 0.50 O_2(g) \]
This reaction is half of the reversed reaction. When the stoichiometry of a reaction is halved, the \( K_p \) is the square root of the \( K_p \) of the reversed reaction:
\[ K_p'' = \sqrt{K_p'} \]

5. **Calculate the square root**:
\[ K_p'' = \sqrt{0.3125 \times 10^{-4}} = \sqrt{0.3125} \times 10^{-2} \]
Calculating \( \sqrt{0.3125} \):
\[ \sqrt{0.3125} \approx 0.559 \]
Therefore:
\[ K_p'' \approx 0.559 \times 10^{-2} = 5.59 \times 10^{-3} \]

6. **Final answer**:
The \( K_p \) for the reaction \( SO_3(g) \rightleftharpoons SO_2(g) + 0.50 O_2(g) \) is:
\[ K_p = 5.59 \times 10^{-3} \]