The rate of the reaction for A Products is 10 mol L^–1 min^–1 at time t₁ = 2 minutes. What will be the rate (in mol L^–1 min^–1) at time t₂ = 12 minutes?

Less than 10

The correct answer is option 2. Less than 10.

To determine the rate of the reaction at a different time, we need to understand how the reaction rate changes over time. This depends on the order of the reaction.

Let us analyze the given options based on general knowledge about reaction rates:

Zero-Order Reaction

For a zero-order reaction, the rate is constant and independent of the concentration of reactants. Therefore, if the rate at \( t_1 \) is 10 mol L\(^{-1}\) min\(^{-1}\), the rate at \( t_2 \) will also be the same.

First-Order Reaction

For a first-order reaction, the rate depends on the concentration of the reactant. The rate law can be written as:

\(\text{Rate} = k[A] \)

The concentration of \( A \) decreases exponentially over time, so the rate also decreases over time.

Second-Order Reaction

For a second-order reaction, the rate depends on the square of the concentration of the reactant. The rate law can be written as:

\(\text{Rate} = k[A]^2 \)

The concentration of \( A \) decreases more rapidly than in a first-order reaction, causing the rate to decrease even more significantly over time.

Given that the rate at \( t_1 \) is 10 mol L\(^{-1}\) min\(^{-1}\), and considering typical reaction kinetics where the rate generally decreases over time as the reactant is consumed, the most reasonable assumption is that the rate will be less than 10 mol L\(^{-1}\) min\(^{-1}\) at \( t_2 = 12 \) minutes.

Therefore, the correct answer is: Less than 10