Final answer:
The correct answer is (a) ∆H° is positive (+) and ∆S° is negative (-).
Step-by-step explanation:
To determine the signs of ∆H° (standard change in enthalpy) and ∆S° (standard change in entropy) for a reaction that is spontaneous at low temperatures but nonspontaneous at high temperatures, we use the Gibbs free energy equation, ∆G = ∆H - T∆S. If a reaction is spontaneous, ∆G must be negative, and if it's nonspontaneous, ∆G must be positive.
At low temperatures (T is small), if ∆G is negative, then ∆H must be large enough and negative to make the ∆G negative since the T∆S term is going to be relatively small. At high temperatures (T is large), the positive T∆S term outweighs the ∆H term, turning ∆G positive; this implies that ∆S must be negative to switch the sign of ∆G with increasing temperature.
Therefore, the correct answer is (a) ∆H° is positive (+) and ∆S° is negative (-).