Q.i. The correct thermodynamic conditions for the spontaneous reaction at all temperatures are a. $\Delta H < 0$ and $\Delta S > 0$
b. $\Delta H > 0$ and $\Delta S < 0$
c. $\Delta H < 0$ and $\Delta S < 0$
d. $\Delta H < 0$ and $\Delta S = 0$

A spontaneous (or irreversible) process is one with a natural tendency to occur on its own, needing no continuous external influence to make it happen; it proceeds in one particular direction only (the reverse direction needs external intervention), it can be fast or slow (spontaneity says nothing about the RATE of a process), and it continues until the system reaches equilibrium. Everyday examples include water flowing from a higher level to a lower one, hot coffee cooling in a room, ice melting above 0 degC, and the acid-base reaction between NaOH(aq) and HCl(aq) (ΔrH0 = -57 kJ). Because many spontaneous reactions release heat (are exothermic), it might seem that exothermicity alone determines spontaneity, but this is not sufficient on its own -- ice melting and NaCl dissolving in water are both spontaneous even though they absorb heat (are endothermic, e.g. NaCl(s) to Na+(aq) + Cl-(aq), ΔH0 = +3.9 kJ/mol) -- so another factor (entropy) must also be at play. Combining enthalpy and entropy through the Gibbs energy change (ΔG = ΔH - TΔS at constant T and P) resolves this fully: (1) if ΔH is negative and ΔS is positive, ΔG is negative and the reaction is spontaneous at EVERY temperature; (2) if ΔH is positive and ΔS is negative, ΔG is positive and the reaction is nonspontaneous at every temperature; (3) if ΔH and ΔS are both negative, ΔG is negative (spontaneous) only at sufficiently LOW temperatures, where the ΔH term dominates; (4) if ΔH and ΔS are both positive, ΔG is negative (spontaneous) only at sufficiently HIGH temperatures, where the TΔS term dominates. In the two temperature-dependent cases, the crossover temperature at which the process switches between spontaneous and nonspontaneous behaviour is found by setting ΔG = 0, giving T = ΔH/ΔS.