Answer: A, B, D
ln[A] values are −0.500,−1.000,−2.000 at 10, 20, 40 min, i.e. −0.05t throughout, so the reaction is first order with k=0.050 min−1. Then t1/2=0.693/0.050=13.9 min, not 20 min. The instantaneous rate at 20 min is k[A]=0.050×0.3679=0.0184 M min−1.