2A k B is a zero-order reaction, where k=1.0 mol L ^ -1 min ^ -1 ....

For a zero-order reaction $2\text{A} \rightarrow \text{B}$, the rate of reaction is expressed as:

$\text{Rate} = -\dfrac{1}{2}\dfrac{d[\text{A}]}{dt} = k$

$\Rightarrow -\dfrac{d[\text{A}]}{dt} = 2k$

Integrating this equation, we get the integrated rate law:

$[\text{A}]_0 - [\text{A}]_t = 2kt$

Given:
Initial concentration, $[\text{A}]_0 = 2$ M
Rate constant, $k = 1.0$ mol L$^{-1}$ min$^{-1}$

For $75\%$ completion of the reaction, the amount of A reacted is:

$[\text{A}]_0 - [\text{A}]_t = \dfrac{75}{100} \times 2 = 1.5$ M

Substituting these values into the integrated rate equation:

$1.5 = 2 \times 1.0 \times t$

$2t = 1.5$

$t = 0.75$ min