Consider a 2nd-order reaction with rate law vr =kr [A]·[B] Consider a 2nd-order reaction with rate law vr =kr [A]·[B]. What happens if [A]0 = [B]0? Nothing special, the same integrated rate law applies. The reaction does not actually run. The integrated rate law becomes

Consider a 2nd-order reaction with rate law vr =kr [A]·[B] Consider a 2nd-order reaction with rate law vr =kr [A]·[B]. What happens if [A]0 = [B]0? Nothing special, the same integrated rate law applies. You cannot use the same int. rate law, since both sides are identical zero! The reaction does not actually run. BALONEY! The integrated rate law becomes Since [A] = [B], the rate law effectively becomes vr = kr [A]2 The integrated rate law becomes