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 \( SO_2 \) in the reaction:

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

Given that the rate of disappearance of \( O_2 \) is:

\[ -\frac{d(O_2)}{dt} = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 1: Write the Rate Expressions
From the stoichiometry of the reaction, we can express the rates of disappearance of the reactants and the formation of the product as follows:

\[ -\frac{1}{2} \frac{d(SO_2)}{dt} = -\frac{1}{1} \frac{d(O_2)}{dt} = \frac{1}{2} \frac{d(SO_3)}{dt} \]

### Step 2: Relate the Rates
From the rate expression for \( O_2 \), we can relate the rate of disappearance of \( SO_2 \) to the rate of disappearance of \( O_2 \):

\[ -\frac{d(SO_2)}{dt} = 2 \left(-\frac{d(O_2)}{dt}\right) \]

### Step 3: Substitute the Given Rate
Now, we can substitute the given rate of disappearance of \( O_2 \):

\[ -\frac{d(SO_2)}{dt} = 2 \times (2.5 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1}) \]

### Step 4: Calculate the Rate of Disappearance of \( SO_2 \)
Calculating this gives:

\[ -\frac{d(SO_2)}{dt} = 5.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Step 5: Final Answer
Thus, the rate of disappearance of \( SO_2 \) is:

\[ \frac{d(SO_2)}{dt} = 5.0 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]