In the formation of sulphur trioxide by the contact process,
`2SO_(2)+O_(2) hArr 2SO_(3)`, the rate of reaction was measured as
`(d[O_(2)])/(d t) = 3.0 xx 10^(-4) "mol" L^(-1)s^(-1)`.
The rate of reaction expressed in terms of `SO_(3)` will be

Options

A. `3.0 xx 10^(-4) mol L^(-1) s^(-1)`
B. `6.0 xx 10^(-4) mol L^(-1) s^(-1)`
C. `1.5 xx 10^(-4) mol L^(-1) s^(-1)`
D. `4.5 xx 10^(-4) mol L^(-1) s^(-1)`

Explanation

To express the rate of the reaction in terms of \( SO_3 \), we need to analyze the stoichiometry of the given reaction:

\[ 2SO_2 + O_2 \rightleftharpoons 2SO_3 \]

### Step 1: Understand the Rate of Reaction
The rate of reaction can be expressed in terms of the change in concentration of reactants and products. For the given reaction, the rate can be defined as:

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

### Step 2: Use the Given Rate of \( O_2 \)
We are given the rate of disappearance of \( O_2 \):

\[ \frac{d[O_2]}{dt} = -3.0 \times 10^{-4} \, \text{mol L}^{-1} s^{-1} \]

### Step 3: Relate the Rate of \( O_2 \) to \( SO_3 \)
From the stoichiometry of the reaction, for every 1 mole of \( O_2 \) that disappears, 2 moles of \( SO_3 \) are produced. Therefore, we can express the rate of formation of \( SO_3 \) as:

\[ \frac{d[SO_3]}{dt} = 2 \left(-\frac{d[O_2]}{dt}\right) \]

### Step 4: Substitute the Given Rate
Substituting the value of \( \frac{d[O_2]}{dt} \):

\[ \frac{d[SO_3]}{dt} = 2 \left(3.0 \times 10^{-4} \, \text{mol L}^{-1} s^{-1}\right) \]

### Step 5: Calculate the Rate of \( SO_3 \)
Now, calculating the above expression:

\[ \frac{d[SO_3]}{dt} = 6.0 \times 10^{-4} \, \text{mol L}^{-1} s^{-1} \]

### Final Answer
Thus, the rate of reaction expressed in terms of \( SO_3 \) is:

\[ \frac{d[SO_3]}{dt} = 6.0 \times 10^{-4} \, \text{mol L}^{-1} s^{-1} \]

---