The rate law for the given reaction is: \[ \text{rate} = k[A]^2[B] \] Where: - \( k \) is the rate constant, - \( [A] \) and \( [B] \) are the concentrations of reactants \( A \) and \( B \). Given that the concentration of \( A \) is doubled and the concentration of \( B \) is halved, let the initial concentrations of \( A \) and \( B \) be \( [A]_0 \) and \( [B]_0 \), respectively. The new concentrations are: \[ [A] = 2[A]_0 \quad \text{and} \quad [B] = \frac{1}{2}[B]_0 \] Substituting these into the rate law yields: \[ \text{new rate} = k(2[A]_0)^2\left(\frac{1}{2}[B]_0\right) \] \[ \text{new rate} = k \times 4[A]_0^2 \times \frac{1}{2}[B]_0 \] \[ \text{new rate} = 2k[A]_0^2[B]_0 \] Comparing the new rate to the original rate: \[ \text{rate} = k[A]_0^2[B]_0 \] The new rate is twice the original rate, which means the rate is quadrupled.