Two isomers (A and B) of a given compound dimerize as follows:

$\begin{array} { l } { 2 \mathrm { A } \longrightarrow \mathrm { A } _ { 2 } } \\ { 2 \mathrm { B } \longrightarrow \mathrm { B } _ { 2 } } \end{array}$

Both processes are known to be second order in the reactant, and $k _ { 1 }$ is known to be $0.250\ \mathrm { L }\ \mathrm { mol } ^ { - 1 }\ \mathrm { s } ^ { - 1 }$ at $25 ^ { \circ } \mathrm { C }.$ In a particular experiment A, and B were placed in separate containers at $25 ^ { \circ } \mathrm { C } ,$ where $[ \mathrm { A } ] _ { 0 } = 1.00 \times 10 ^ { - 2 }\ \mathrm { M }$ and $[ \mathrm { B } ] _ { 0 } = 2.50 \times 10 ^ { - 2 }\ \mathrm { M } .$ After each reaction had progressed for 3.00 min, [A] = 3.00[B]. In this case the rate laws are defined as follows:

$\text {Rate} = - \frac { d [ \mathrm { A } ] } { d t } = k _ { 1 } [ \mathrm { A } ] ^ { 2 } \\ \text {Rate} = - \frac { d [ \mathrm { B } ] } { d t } = k _ { 2 } [ \mathrm { B } ] ^ { 2 }$

Calculate the value of $k _ { 2 }.$