We have obtained absolute integral cross sections (sigma's) for the reactions of spin-orbit-state-selected vanadium cations, V+[a(5)D(J)(J = 0, 2), a(5)F(J) (J = 1, 2), and a(3)F(J) (J = 2, 3)], with a water molecule (H2O) in the center-of-mass collision energy range E-cm = 0.1-10.0 eV. On the basis of these stateselected sigma curves (sigma - versus E-cm plots) observed, three reaction product channels, VO+ + H-2 , VH+ + OH, and VOH+ + H, from the V+ + H2O reaction are unambiguously identified. Contrary to the previous guided ion beam study of the V+(a(5)D(J)) + D2O reaction, we have observed the formation of the VO+ + H(2 )channel from the V+(a(5)D(J)) + H2O ground reactant state at low E-cm's (<3.0 eV). No spin-orbit J-state dependences for the sigma curves of individual electronic states are discernible, indicating that spin-orbit interactions are weak with little effect on chemical reactivity of the titled reaction. For the three product channels identified, the triplet sigma(a(3)F(J)) values are overwhelmingly higher than the quintet sigma(a(5)D(J)) and sigma(a(5)F(J)) values, showing that the reaction is governed by a "weak quintet-triplet spin crossing" mechanism, favoring the conservation of total electron spins. The sigma curves for exothermic product channels are found to exhibit a rapid decreasing profile as E-cm is increased, an observation consistent with the prediction of the charge-dipole and induced-dipole orbiting model. This experiment shows that the V+ + H2O reaction can be controlled effectively to produce predominantly the VO+ + H-2 channel via the V+(a(3)F(J)) + H2O reaction at low E-cm's (<= 0.1 eV) and that the ion-molecule reaction dynamics can be altered readily by selecting the electronic state of r cation. On the basis of the measured E cm thresholds for the sigma(a(5)D(J), a(5)F(J), and a(3)F(J): VH+) and sigma(a(5)D(J), a(5)F(J), and a(3)F(J): VOH+) curves, we have deduced upper bound values of 2.6 +/- 0.2 and 4.3 +/- 0.3 eV for the 0 K bond dissociation energies, D-0 (V+- H) and D-0 (V+-OH), respectively. After correcting for the kinetic energy distribution resulting from the Doppler broadening effect of the H2O molecule, we obtain D-0 (V+-H) = 2.2 +/- 0.2 eV and D-0 (V-+OH) = 4.0 +/- 0.3 eV, which are in agreement with D-0 determinations obtained by sigma curve simulations.