Half life of zero order reaction $$A \rightarrow$$ product is 1 hour, when initial concentration of reaction is 2.0 mol L$$^{-1}$$. The time required to decrease concentration of A from 0.50 to 0.25 mol L$$^{-1}$$ is:

For a zero-order reaction $$A \rightarrow \text{products}$$, the rate is independent of concentration:

Rate law : $$\frac{d[A]}{dt} = -k$$

On integrating from $$t = 0$$ (concentration $$[A]_0$$) to any time $$t$$ (concentration $$[A]_t$$) we obtain the integrated equation

$$[A]_t = [A]_0 - k\,t \quad -(1)$$

Half-life $$t_{1/2}$$ is the time at which $$[A]_t = \frac{[A]_0}{2}$$. Substituting into equation $$(1)$$ gives the zero-order half-life formula:

$$t_{1/2} = \frac{[A]_0}{2k} \quad -(2)$$

The question states that for $$[A]_0 = 2.0\;\text{mol L}^{-1}$$, $$t_{1/2} = 1\;\text{h}$$. Using $$-(2)$$ to find the rate constant $$k$$:

$$k = \frac{[A]_0}{2\,t_{1/2}} = \frac{2.0}{2 \times 1\;\text{h}} = 1.0\;\text{mol L}^{-1}\text{h}^{-1}$$

Now we need the time required for the concentration to drop from $$0.50$$ to $$0.25\;\text{mol L}^{-1}$$. Let this time be $$t$$ and take $$[A]_0 = 0.50\;\text{mol L}^{-1}$$ in equation $$(1)$$:

$$[A]_t = 0.50 - k\,t$$

Set $$[A]_t = 0.25\;\text{mol L}^{-1}$$ and $$k = 1.0\;\text{mol L}^{-1}\text{h}^{-1}$$:

$$0.25 = 0.50 - 1.0 \times t$$

$$t = 0.50 - 0.25 = 0.25\;\text{h}$$

Convert $$0.25$$ hour to minutes:

$$0.25\;\text{h} \times 60\;\frac{\text{min}}{\text{h}} = 15\;\text{min}$$

Hence, the time required is 15 minutes.

The correct choice is Option C (15 min).