Which of the following graphs is correct for a second-order reaction?

The half-life equation for an $$n^{\text{th}}$$-order chemical reaction follows the general proportionality:

$$t_{1/2}\propto\frac{1}{a_0^{n-1}}$$

For a second-order reaction, n=2:

$$t_{1/2}=\frac{1}{k a_0}$$

Rearranging in the form y=mx+c:

$$t_{1/2}=\left(\frac{1}{k}\right)\left(\frac{1}{a_0}\right)+0$$

- y-axis: $$t_{1/2}$$

- x-axis: $$\frac{1}{a_0}$$

- Slope: $$\frac{1}{k}$$

- y-intercept: 0

Therefore, the graph of $$t_{1/2}$$ versus $$\frac{1}{a_0}$$ is a straight line passing through the origin with a positive slope.

Correct option: (A)