Given: Rate of formation of SO3 = 1.6×10³ kg/Lmin

We know that the stoichiometric coefficient of SO2 in the given reaction is 2. Therefore, the rate of decomposition of SO2 will be half of the rate of formation of SO3.

So, the rate of decomposition of SO2 = 1.6×10³/2 = 0.8×10³ kg/Lmin

Therefore, the correct option is (2) 8.0×10² kg/Lmin.

Explanation:

To understand the solution better, let us break it down into the following headings:

- Rate of reaction
- Stoichiometric coefficients
- Relationship between rates of reactants and products

Rate of reaction:

The rate of reaction is the change in concentration of a reactant or a product with respect to time. It is expressed in units of concentration/time, such as mol/L/min or kg/L/min.

In this case, the rate of reaction is given as 1.6×10³ kg/Lmin. This means that for every minute, 1.6×10³ kg of SO3 is formed.

Stoichiometric coefficients:

The stoichiometric coefficients of a balanced chemical equation represent the number of moles of each reactant and product involved in the reaction. These coefficients are used to determine the relationship between the rates of reactants and products.

In this case, the balanced chemical equation is:

2SO2 + O2 → 2SO3

The stoichiometric coefficient of SO2 is 2, which means that for every 2 moles of SO2 consumed, 1 mole of O2 and 2 moles of SO3 are produced.

Relationship between rates of reactants and products:

The rate of formation of a product is related to the rates of the reactants by their stoichiometric coefficients. For example, in the given reaction, the rate of formation of SO3 is related to the rate of consumption of SO2 and O2 by the stoichiometric coefficients of the balanced equation.

In this case, the rate of formation of SO3 is 1.6×10³ kg/Lmin. Since the stoichiometric coefficient of SO2 is 2, the rate of decomposition of SO2 is half of the rate of formation of SO3, which is 0.8×10³ kg/Lmin.

Therefore, the correct option is (2) 8.0×10² kg/Lmin.