Order of Reaction:

Since $ t_{1/2} \propto [A]_0 $, the reaction follows a zero-order kinetics.

Half-life Equation for Zero-order Reaction:

The half-life ($ t_{1/2} $) is calculated as:

$ t_{1/2} = \frac{[A]_0}{2K} $

Given the slope from the graph is $76.92$, which equals $\frac{1}{2K}$. This implies:

$ K = \frac{1}{2 \times 76.92} $

Concentration of A at 10 Minutes:

Apply the zero-order kinetics equation:

$ [A]_{10} = -Kt + [A]_0 $

Substituting the values:

$ [A]_{10} = -\left(\frac{1}{2 \times 76.92}\right) \times 10 + 2.5 = 2.435 \ \text{mol L}^{-1} $

Final Concentration:

Convert to scientific notation:

$ [A]_{10} = 2435 \times 10^{-3} \ \text{mol L}^{-1} $

Thus, the concentration of A at 10 minutes is approximately $ 2435 \times 10^{-3} \ \text{mol L}^{-1} $.