The reaction described by the equation ch 3 cl + naoh → ch 3 oh + nacl follows the second-order rate law, rate = k [ ch 3 cl ] [ naoh ] . when this reaction is carried out with starting concentrations [ ch 3 cl ] = 0.2 m and [ naoh ] = 1.0 m , the measured rate is 1 × 10 − 4 mol l − 1 s − 1 . what is the rate after one-half of the ch 3 cl has been consumed? (caution: the initial concentrations of the starting materials are not identical in this experiment. hint: determine how much of the naoh has been consumed at this point and what its new concentration is, compared with its initial concentration.)