Rate of formation of SO3 according to the reaction 2SO2+O2→2SO3is1.6×10−3kgmin−1. Hence rate at which SO2 reacts is:

1. Break Down the Problem

We need to find the rate at which $\text{SO}_2$ reacts based on the given rate of formation of $\text{SO}_3$ from the reaction: $2 \text{SO}_2 + \text{O}_2 \rightarrow 2 \text{SO}_3$

2. Relevant Concepts

From stoichiometry, we can determine the relationships between the rates of reaction for the substances involved in the reaction. According to the balanced chemical equation:

• For every 2 moles of $\text{SO}_2$ that react, 2 moles of $\text{SO}_3$ are formed.

Thus, the rate of consumption of $\text{SO}_2$ is equal to the rate of formation of $\text{SO}_3$ times the stoichiometric ratio.

3. Analysis and Detail

Given:

• Rate of formation of $\text{SO}_3 = 1.6 \times 10^{-3} \, \text{kg min}^{-1}$

Using stoichiometry, the rate at which $\text{SO}_2$ reacts can be calculated as follows: $\text{Rate of consumption of } \text{SO}_2 = \text{Rate of formation of } \text{SO}_3 \times \frac{2 \text{ moles of } \text{SO}_2}{2 \text{ moles of } \text{SO}_3}$ Thus: $\text{Rate of consumption of } \text{SO}_2 = 1.6 \times 10^{-3} \, \text{kg min}^{-1}$

The rate of consumption of $\text{SO}_2$ is the same as the rate of formation of $\text{SO}_3$ because the stoichiometric coefficients are equal.

4. Verify and Summarize

Since the stoichiometric coefficients in the balanced equation for $\text{SO}_2$ and $\text{SO}_3$ are both 2, the rate of $\text{SO}_2$ reacts at the same rate as $\text{SO}_3$ is formed.

Final Answer

The rate at which $\text{SO}_2$ reacts is: $1.6 \times 10^{-3} \, \text{kg min}^{-1}$