In the reaction of formation of sulphur trioxide by contact process $$2SO_2 + O_2 \rightleftharpoons 2SO_3$$ the rate of reaction was measured as $$\frac{d[O_2]}{dt} = -2.5 \times 10^{-4}$$ mol L$$^{-1}$$ s$$^{-1}$$. The rate of reaction in terms of [SO$$_2$$] in mol L$$^{-1}$$ s$$^{-1}$$ will be:

The reaction is given as: $$2SO_2 + O_2 \rightleftharpoons 2SO_3$$

The rate of reaction can be expressed in terms of any reactant or product using the stoichiometric coefficients. For a general reaction: $$aA + bB \rightarrow cC + dD$$, the rate is defined as: $$\text{rate} = -\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$$

For this reaction, the stoichiometric coefficients are:

• For $$SO_2$$, the coefficient is 2.
• For $$O_2$$, the coefficient is 1.
• For $$SO_3$$, the coefficient is 2.

Therefore, the rate of reaction can be written as: $$\text{rate} = -\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{1}{1} \frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt}$$

We are given that $$\frac{d[O_2]}{dt} = -2.5 \times 10^{-4}$$ mol L$$^{-1}$$ s$$^{-1}$$. Using the expression for $$O_2$$: $$\text{rate} = -\frac{d[O_2]}{dt}$$ because the coefficient of $$O_2$$ is 1.

Substituting the given value: $$\text{rate} = - \left( -2.5 \times 10^{-4} \right) = 2.5 \times 10^{-4}$$ mol L$$^{-1}$$ s$$^{-1}$$.

Now, we need to find the rate in terms of $$[SO_2]$$, which is $$\frac{d[SO_2]}{dt}$$. Using the expression for $$SO_2$$: $$\text{rate} = -\frac{1}{2} \frac{d[SO_2]}{dt}$$.

Rearranging for $$\frac{d[SO_2]}{dt}$$: $$\frac{d[SO_2]}{dt} = -2 \times \text{rate}$$.

Substituting the value of rate: $$\frac{d[SO_2]}{dt} = -2 \times \left( 2.5 \times 10^{-4} \right)$$.

Calculating: $$-2 \times 2.5 \times 10^{-4} = -5.0 \times 10^{-4}$$.

So, $$\frac{d[SO_2]}{dt} = -5.00 \times 10^{-4}$$ mol L$$^{-1}$$ s$$^{-1}$$.

Comparing with the options, this matches option D.

Hence, the correct answer is Option D.