The rate of reaction for `A to " products"` is 10 mole, `lit^(-1)` .`min^(-1)` when `t_(1)=2min`. The rate of reaction when `t_(2)=12min`. in the same units is

Options

A. `gt10`
B. `lt10`
C. 10
D. 12

Explanation

To solve the problem, we need to understand how the rate of reaction changes over time. The rate of reaction decreases as the concentration of the reactant decreases.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The rate of reaction at time \( t_1 = 2 \) minutes is given as \( 10 \, \text{mol} \, \text{lit}^{-1} \, \text{min}^{-1} \).

2. **Understand Reaction Dynamics**:
- The rate of reaction generally decreases over time as the reactant is consumed. Therefore, as time progresses from \( t_1 \) to \( t_2 \), the rate at \( t_2 \) will be less than the rate at \( t_1 \).

3. **Determine the Time Interval**:
- We are asked to find the rate at \( t_2 = 12 \) minutes. This is a significant increase in time from \( t_1 \).

4. **Apply the Concept of Decreasing Rate**:
- Since the rate is decreasing, we can conclude that the rate at \( t_2 \) will be less than \( 10 \, \text{mol} \, \text{lit}^{-1} \, \text{min}^{-1} \).

5. **Conclusion**:
- The rate of reaction at \( t_2 = 12 \) minutes must be less than \( 10 \, \text{mol} \, \text{lit}^{-1} \, \text{min}^{-1} \). Therefore, the answer is that the rate at \( t_2 \) is less than \( 10 \).

### Final Answer:
The rate of reaction at \( t_2 = 12 \) minutes is less than \( 10 \, \text{mol} \, \text{lit}^{-1} \, \text{min}^{-1} \).