Final answer:
The sign of ΔG for an exothermic reaction with a decrease in entropy is typically negative, but cannot be predicted without additional information on the magnitudes of enthalpy change and entropy change, as well as the temperature.
Step-by-step explanation:
The sign of ΔG for an exothermic reaction with a decrease in entropy is typically negative. This is because ΔG, the Gibbs free energy change, is defined by the equation ΔG = ΔH - TΔS. In this equation, ΔH represents the change in enthalpy, T represents the temperature in Kelvin, and ΔS represents the change in entropy. For an exothermic reaction, ΔH is negative, and if there is a decrease in entropy, ΔS is also negative. However, since T is a positive value (temperature in Kelvin), the term -TΔS is positive. This tends to increase ΔG. Whether ΔG becomes positive or negative depends on the relative magnitudes of the enthalpy change and the entropy term multiplied by the temperature.
If we assume a standard condition where temperature and pressure don't significantly alter the outcome, and the magnitude of ΔH is greater than TΔS, the result is that ΔG will be negative, indicating a spontaneous reaction. However, in specific cases where the decrease in entropy is significant, and the temperature is high enough, it may overcome the exothermic nature of the reaction, leading to a positive ΔG which indicates non-spontaneity. Hence, typically, without additional specific information, we cannot predict the sign of ΔG with certainty.