Consider the reaction $$ \mathrm{A}+\mathrm{B} \longrightarrow \text { products } $$ From these data obtained at a certain temperature, determine the order of the reaction and calculate the rate constant: $$ \begin{array}{ccc} {[\mathrm{A}](\mathrm{M})} & {[\mathrm{B}](M)} & \text { Rate }(M / \mathrm{s}) \\ \hline 1.50 & 1.50 & 3.20 \times 10^{-1} \\ 1.50 & 2.50 & 3.20 \times 10^{-1} \\ 3.00 & 1.50 & 6.40 \times 10^{-1} \end{array} $$

First, examine how changes in the concentration of each reactant affect the rate of reaction. From the given data, when the concentration of reactant B changes from 1.5M to 2.5M and the concentration of reactant A is held constant at 1.5M, the rate of reaction remains constant at \(3.20 \times 10^{-1}\ M/s\). This indicates that the order of the reaction with respect to reactant B is zero. Now, observe the change in the rate of reaction when the concentration of reactant A is doubled (from 1.5M to 3.0M) and the concentration of reactant B is held constant. The rate of reaction doubles from \(3.20 \times 10^{-1}\ M/s\) to \(6.40 \times 10^{-1}\ M/s\). This indicates that the order of the reaction with respect to reactant A is one, as a doubling of the concentration results in a doubling of the rate. Therefore, the overall order of the reaction is \(1+0 = 1\).

Having determined that the reaction is first order with respect to reactant A, the rate law for the reaction can be expressed as \[Rate = k[A]\] where k is the rate constant, [A] is the concentration of A. To find the value of the rate constant, input any set of rate and concentration data from the table into this equation. For instance, using the first set of rate and concentration data: \(3.20 \times 10^{-1} M/s = k(1.5M)\). By solving for k, we get \(k = \frac{3.20 \times 10^{-1}M/s}{1.5M} = 2.13 s^{-1}\]