Answer
Rate = 0.02 mol/L per second
Work Step by Step
As the reaction follows the SN2 mechanism, the rate law is:
Rate = k[1-bromobutane][sodium methoxide]
When 0.5 M 1-bromobutane reacts with 1.0 M sodium methoxide the rate is 0.05 mol/L per second. Hence, k = 0.05 (mol/L per second)/ 0.5 (mol/L), so k = 0.1 (L/mol per second).
When 0.1 M 1-bromobutane and 2.0 M sodium methoxide are used, rate = 0.1 (L/mol per second) x 0.1 (mol/L) x 2.0 (mol/L). Hence, rate = 0.02 mol/L per second.
