In the formation of sulphur trioxide by the contact process,
`2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)`
The rate of reaction is expressed as
`-(d(O_2))/(dt) = 2.5 xx 10^(-4) mol L^(-1) s^(-1)`
The rate of disappearance of `(SO_(2))` will be

Options

A. `5.0 xx 10^(-4) mol L^(-1) s^(-1)`
B. `-2.25 xx 10^(-4) mol L^(-1) s^(-1)`
C. `3.75 xx 10^(-4) mol L^(-1) s^(-1)`
D. `50.0 xx 10^(-4) mol L^(-1) s^(-1)`

Explanation

To solve the problem, we need to determine the rate of disappearance of \( SO_2 \) based on the given rate of disappearance of \( O_2 \) in the reaction:

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

### Step-by-Step Solution:

1. **Identify the Rate of Reaction**:
The rate of reaction can be expressed in terms of the change in concentration of the reactants and products. The rate of disappearance of \( O_2 \) is given as:
\[ -\frac{d[O_2]}{dt} = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

2. **Relate the Rates Using Stoichiometry**:
From the balanced chemical equation, we can see that the stoichiometric coefficients are:
- For \( SO_2 \): 2
- For \( O_2 \): 1

The relationship between the rates of disappearance of \( SO_2 \) and \( O_2 \) can be established using their stoichiometric coefficients:
\[ -\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{1}{1} \frac{d[O_2]}{dt} \]

3. **Express the Rate of Disappearance of \( SO_2 \)**:
Rearranging the equation gives us:
\[ \frac{d[SO_2]}{dt} = 2 \frac{d[O_2]}{dt} \]

4. **Substituting the Given Rate**:
Now, substitute the value of \( \frac{d[O_2]}{dt} \):
\[ \frac{d[SO_2]}{dt} = 2 \times 2.5 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

5. **Calculate the Rate**:
Performing the multiplication:
\[ \frac{d[SO_2]}{dt} = 5.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

Thus, the rate of disappearance of \( SO_2 \) is:
\[ \frac{d[SO_2]}{dt} = 5.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Final Answer:
The rate of disappearance of \( SO_2 \) is \( 5.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \).